Full article: L9-free groups

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Abstract

In this article we classify all L₉-free finite groups.

KEYWORDS:

2020 MATHEMATICS SUBJECT CLASSIFICATION:

Introduction

There are some algebraic laws that hold in a lattice L if and only if L does not have a sublattice of a specific shape. For example, a lattice is modular if and only if it does not have a sublattice isomorphic to the so-called pentagon L₅.

If L is a lattice, then we call a group L-free if and only if its subgroup lattice does not contain a lattice isomorphic to L. For example, the finite L₅-free groups are exactly the modular groups, and these have been classified by Iwasawa in 1941, see [Citation4]. The subgroup lattice of the dihedral group of order 8, often denoted by L₁₀, and some of its sublattices are of particular interest. One reason is that, if p is a prime number, then a finite p-group is L₅-free if and only if it is L₁₀-free.

There are several sublattices of L₁₀ containing L₅:

In 1999 Baginski and Sakowicz [Citation2] studied finite groups that are L₆-free and L₇-free at the same time, and later Schmidt [Citation8] classified the finite groups that are L₆- or L₇-free. Together with Andreeva and the first author he also characterized, in [Citation1], all finite groups that are L₈-free or M₈-free. Finally, the finite M₉-free groups have been classified by Pölzing and the second author in [Citation6]. Furthermore, there is a general discussion of L₁₀-free groups by Schmidt, which can be found in [Citation9] and [Citation10].

In this paper we investigate finite L₉-free groups. Since L₉ is a sublattice of L₁₀, the groups that we consider are L₁₀-free and therefore we can use Corollary C in [Citation9] as a starting point for our analysis: Every finite L₁₀-free group G has normal Hall subgroups $N_{1} \leq N_{2}$ such that $N_{1} = 〈 P \in Syl (G) | P E G 〉, N_{2} / N_{1}$ is a 2-group and $G / N_{2}$ is meta-cyclic.

Our strategy is to choose $N : = N_{1}$ maximal with respect to the above constraints, and then we show that N has a complement K that is a direct and coprime product of groups of the following structure: cyclic groups, groups isomorphic to Q₈ or semi-direct products $Q ⋊ R$ , where Q has prime order and R is cyclic of prime-power order such that $1 \neq Φ (R) = C_{Q} (R)$ . Furthermore, $[N, K] \cap C_{N} (K)$ is a 2-group and $C_{O_{2} (N)} (K)$ is cyclic or elementary abelian of order 4 and every Sylow subgroup of $[N, K]$ is elementary abelian or isomorphic to Q₈. If the action of K on N satisfies some more conditions, then we say that NK is in class $L$ .

The aim of our article is to prove the following theorem:

Main Theorem. A finite group is in class $L$ if and only if it is L₉-free.

Notation and preliminary results

In this article we mostly follow the notation from Schmidt’s book [Citation7] and from [Citation5]. All groups considered are finite and G always denotes a finite group, moreover p and q always denote prime numbers. We quickly recall some standard concepts:

L(G) denotes the subgroup lattice of G, consisting of the set of subgroups of G with inclusion as the partial ordering. The infimum of two elements $A, B \in L (G)$ is $A \cap B$ (their intersection) and the supremum is $A \cup B = 〈 A, B 〉$ (the subgroup generated by A and B).

If L is any lattice, then G is said to be L-free if and only if L(G) does not have any sub-lattice that is isomorphic to L.

A lattice L is said to be modular if and only if for all $X, Y, Z \in L$ such that $X \leq Z$ , the following (also called the modular law) is true: $(X \cup Y) \cap Z = X \cup (Y \cap Z) .$ We say that a group G is modular if and only if L(G) is modular.

The modular law is similar to Dedekind’s law (see 1.1.11 of [Citation5]). For all $X, Y, Z \leq G$ such that $X \leq Z$ it says that $XY \cap Z = X (Y \cap Z) .$ We will use Dedekind’s law frequently throughout this article without giving an explicit reference each time.

If $N \leq G$ , then we say that an element $g \in G$ induces power automorphisms on N if and only if $U^{g} = U$ for all subgroups U of N. Furthermore ${Pot}_{G} (N) : = {g \in G | U^{g} = U f . a . U \leq N}$ is a subgroup of G because ${Pot}_{G} (N) = \underset{U \leq N}{\cap} N_{G} (U)$ .

Lemma 1.1.

Let K be a finite group that acts coprimely on the p-group P. Then $P = [P, K] C_{P} (K)$ . If P is abelian, then this product is direct. If $[P, K] \leq Φ (P)$ , then $[P, K] = 1$ . Furthermore $[P, K] = [P, K, K]$ and for all K-invariant normal subgroups N of P we have that $C_{P / N} (K) = C_{P} (K) N / N$ .

Proof.

These statements are a collection from 8.2.2, 8.2.7, 8.2.9, and 8.4.2 of [Citation5]. □

Lemma 1.2.

Let $p \in π (G)$ and suppose that G = PK, where P is a normal Sylow p-subgroup of G, $K \leq G$ is a $p'$ -group and $P_{0} : = [P, K] \neq 1$ . Suppose further that $Φ (P_{0}) \leq C_{P} (K)$ and that K acts irreducibly on $P_{0} / Φ (P_{0})$ . If $g \in P ∖ C_{P} (K)$ , then $P_{0} \leq 〈 [g, K], K 〉 \leq 〈 g, K 〉$ .

Proof.

Let $g \in P ∖ C_{P} (K)$ and $R : = 〈 [g, K], K 〉$ . Then we first remark that $R \leq 〈 g, K 〉$ . Lemma 1.1 shows that $P = C_{P} (K) P_{0}$ and hence we have elements $c \in C_{P} (K)$ and $h \in P_{0}$ such that g = ch. We note that $P_{0} E G$ and therefore $Φ (P_{0})$ is a normal subgroup of G, moreover P₀ is a p-group and hence $P_{0} / Φ (P_{0})$ is elementary abelian. Let $- : G \to G / Φ (P_{0})$ denote the natural homomorphism. As $\bar{P_{0}} = [\bar{P}, \bar{K}]$ is abelian and $\bar{K}$ acts coprimely on it, we see that $C_{[\bar{P}, \bar{K}]} (\bar{K}) \cap [\bar{P}, \bar{K}, \bar{K}] = 1$ , again by Lemma 1.1. Therefore $C_{[\bar{P}, \bar{K}]} (\bar{K}) = C_{[\bar{P}, \bar{K}, \bar{K}]} (\bar{K}) = 1$ . We recall that $ch = g \notin C_{P} (K)$ and thus $h \notin C_{P} (K)$ , and then by hypothesis $h \notin Φ (P_{0})$ and in particular $1 \neq \bar{h} \in \bar{P_{0}}$ . It follows that $[\bar{h}, \bar{K}] \neq 1$ because $C_{[\bar{P}, \bar{K}]} (\bar{K}) = 1$ , see above. We conclude that $1 \neq [\bar{h}, \bar{K}] = [\bar{g}, \bar{K}] = \bar{[g, K]} \leq \bar{P_{0}} \cap \bar{R}$ . By hypothesis K acts irreducibly on $\bar{P_{0}}$ , hence $\bar{K}$ does as well and we see that $\bar{P_{0}} = \bar{P_{0} \cap R Φ (P_{0})} = \bar{(P_{0} \cap R) Φ (P_{0})}$ . The main property of the Frattini subgroup (see for example 5.2.3 in [Citation5]) finally gives that $P_{0} = P_{0} \cap R$ . □

Lemma 1.3.

Let Q be a 2-group that is elementary abelian, cyclic or isomorphic to Q₈. Then Q does not admit power automorphisms of odd order.

Proof.

If Q is abelian, then the assertion follows from 2.2.5 of [Citation5]. If $Q ≅ Q_{8}$ , then $Aut (Q) ≅ {Sym}_{4}$ , and any automorphism of order 3 interchanges the maximal subgroups of Q. □

Lemma 1.4.

Suppose that G = NK, where N is a normal Hall subgroup of G and K is a complement. Let $N_{1}, N_{2} \leq N, Q, R \leq K$ and $x \in N$ . Then the following hold:

If $N_{1} Q$ and $N_{2} R^{x}$ are subgroups of G, then $N_{1} Q \cap N_{2} R^{x} = N_{1} (Q \cap R) \cap N_{2} {(Q \cap R)}^{x}$ .
If $N_{2} R^{x} \leq G$ and $〈 x^{Q \cap R} 〉 \cap N_{2} = 1$ , then $Q \cap N_{2} R^{x} = Q \cap R^{x} \leq C_{Q \cap R} (x)$ .
If K is abelian and acts irreducibly on the abelian group $N / Φ (N)$ or if N is abelian and K induces power automorphisms on it, then $C_{K} (N) = C_{K} (x)$ or $x \in Φ (N)$ .
If $Q \leq R$ , then $〈 Q, R^{x} 〉 = 〈 {[x, Q]}^{R^{x}} 〉 R^{x}$ .

Proof.

Suppose that $N_{1} Q$ and $N_{2} R^{x}$ are subgroups of G.

For (a) we do the following calculation: $\begin{matrix} N_{1} Q \cap N_{2} R^{x} & = & (N_{1} Q \cap NQ) \cap N_{2} R^{x} = N_{1} Q \cap (NQ \cap N_{2} R^{x}) = N_{1} Q \cap N_{2} (N Q^{x} \cap R^{x}) \\ = & N_{1} Q \cap N_{2} (N Q^{x} \cap (K^{x} \cap R^{x})) = N_{1} Q \cap N_{2} ((N Q^{x} \cap K^{x}) \cap R^{x}) \\ = & N_{1} Q \cap N_{2} (Q^{x} \cap R^{x}) = N_{1} Q \cap (N {(Q \cap R)}^{x} \cap N_{2} {(Q \cap R)}^{x}) \\ = & (N_{1} Q \cap N (Q \cap R)) \cap N_{2} {(Q \cap R)}^{x} = N_{1} (Q \cap N (Q \cap R)) \cap N_{2} {(Q \cap R)}^{x} \\ = & N_{1} ((Q \cap K) \cap N (Q \cap R)) \cap N_{2} {(Q \cap R)}^{x} \\ = & N_{1} (Q \cap (K \cap N (Q \cap R))) \cap N_{2} {(Q \cap R)}^{x} \\ = & N_{1} (Q \cap (Q \cap R)) \cap N_{2} {(Q \cap R)}^{x} = N_{1} (Q \cap R) \cap N_{2} {(Q \cap R)}^{x} . \end{matrix}$

Then (a) yields that $Q \cap N_{2} R^{x} = (Q \cap R) \cap N_{2} {(Q \cap R)}^{x}$ . Therefore, if $〈 x^{Q \cap R} 〉 \cap N_{2} = 1$ and if $a, b \in Q \cap R$ and $y \in N_{2}$ are such that $a = y b^{x} = y x^{- 1} x^{b^{- 1}} b$ , then the fact that $a b^{- 1} = y x^{- 1} x^{b^{- 1}} \in K \cap N = 1$ implies that a = b and that $y^{- 1} = [x, b^{- 1}] \in N_{2} \cap 〈 x^{Q \cap R} 〉 = 1$ . We deduce that $a = b^{x} = a^{x} \in Q \cap R^{x}$ and that $[x, a] = 1$ . Hence $Q \cap N_{2} R^{x} = (Q \cap R) \cap N_{2} {(Q \cap R)}^{x} = Q \cap R^{x} \leq C_{Q \cap R} (x)$ . This is (b).

If K is abelian, then $x \in C_{N} (C_{K} (x))$ and $C_{N} (C_{K} (x))$ is K-invariant. Thus, if K acts irreducibly on $N / Φ (N)$ and $x \notin Φ (N)$ , then $C_{N} (C_{K} (x)) Φ (N) = N$ , and the fact that $[C_{K} (x), N] = 1$ implies that $C_{K} (x) = C_{K} (N)$ . If K induces power automorphisms on the abelian group N, then 1.5.4 of [Citation7] implies that these are universal. If x = 1, then $x \in Φ (N)$ , and otherwise $x \neq 1$ and every element of K that centralizes x also centralizes N. Altogether (c) holds.

Suppose finally that $Q \leq R$ . Then, for all $g \in Q$ , we have that $g = g^{x} \cdot {(g^{- 1})}^{x} \cdot g \in R^{x} [x, Q] \leq 〈 {[x, Q]}^{R^{x}} 〉 R^{x}$ . It follows that $〈 Q, R^{x} 〉 \leq 〈 {[x, Q]}^{R^{x}} 〉 R^{x}$ .

On the other hand ${(g^{- 1})}^{x} \in Q^{x} \leq R^{x}$ for all $g \in Q$ and therefore $[x, g] = {(g^{- 1})}^{x} g \in 〈 Q, R^{x} 〉$ .

This implies that $[x, Q] \leq 〈 Q, R^{x} 〉$ . Since $R^{x} \leq 〈 Q, R^{x} 〉$ , we deduce that $〈 {[x, Q]}^{R^{x}} 〉 R^{x} \leq 〈 Q, R^{x} 〉$ , which is (d). □

2 Battens and batten groups

Definition 2.1.

We say that G is a batten if and only if G is a cyclic p-group, or isomorphic to Q₈, or G = QR, where Q is a normal subgroup of prime order and R is a cyclic p-group of order coprime to $| Q |$ and such that $C_{R} (Q) = Φ (R) \neq 1$ .
We say that G is a batten group if and only if G is a direct product of battens of pairwise coprime order.
If G is a batten group, then we say that $B \leq G$ is a batten of G if and only if B is a batten that is one of the direct factors of G.

Warning: It is possible for a subgroup of a batten group G to be a batten, abstractly, but not to be a batten of G. This can happen when it is a p-subgroup for some prime p of a batten as in the third case of Definition 2.1.

Example 2.2.

Suppose that $Q : = 〈 x 〉$ is a group of order 19 and that $R = 〈 y 〉$ is a subgroup of $Aut (X)$ of order 27. Further suppose that $x^{y} : = x^{7}$ . Then $B : = QR$ is a non-nilpotent batten. For this we calculate that $x^{y^{3}} = {(x^{7})}^{y^{2}} = {(x^{49})}^{y} = {(x^{11})}^{y} = x^{77} = x$ . Then the fact that $x^{y} = x^{7} \neq x$ implies that $C_{R} (Q) = 〈 y^{3} 〉 = Φ (R)$ . We note that B is a batten group and that Q is a subgroup of B that is a batten, but not a batten of B because $[Q, R] \neq 1$ .
Let B = QR be as in (a), let $T ≅ Q_{8}$ and let S be a cyclic group of order 625. Then $B \times T \times S$ is a batten group.

Remark 2.3.

Let G be a batten group and let B be a batten of G such that $| π (B) | = 2$ .

B is not nilpotent, but B has a unique normal Sylow subgroup.
A Sylow subgroup Q of B is cyclic, and therefore Q is batten. But Q is not a direct factor of G and hence Q is not a batten of G.

Definition 2.4.

Suppose that G is a non-nilpotent batten. Then there is a unique prime $q \in π (G)$ such that G has a normal Sylow q-subgroup Q, and Q is cyclic of order q. In this case we set $B (G) : = Q$ .

From the definition we can immediately see that $B (G)$ is a characteristic subgroup of a non-nilpotent batten G and that it has prime order.

Lemma 2.5.

Suppose that G is a non-nilpotent batten, that $r \in π (G)$ and that $R \in {Syl}_{r} (G)$ has order at least r ². Then $Z (G) = C_{R} (B (G)) = Φ (R) = O_{r} (G)$ .

Proof.

Since $| R | \geq r^{2}$ , we see that $R \neq B (G)$ . Then Definition 2.1 implies that R is cyclic and that there is a prime $q \in π (G) ∖ {r}$ such that $Q : = B (G) \in {Syl}_{q} (G)$ .

Now $G = Q ⋊ R$ and $C_{R} (B (G)) = Φ (R)$ . We recall that R is cyclic, and then this implies that $Φ (R) \leq Z (G)$ . Since G is not nilpotent, we see that $B (G)$ is not contained in Z(G). Thus Z(G) is an r-group, because $B (G)$ has prime order. In addition $R ≰ Z (G)$ , because G is not nilpotent. Since $Φ (R)$ is a maximal subgroup of the cyclic group R, it follows that $Φ (R) = Z (G)$ . Moreover we have that $[Q, O_{r} (G)] \leq Q \cap R = 1$ , whence $O_{r} (G) \leq C_{R} (Q) = Φ (R) = Z (G) \leq C_{R} (Q)$ . This proves all statements. □

Lemma 2.6.

If G is a batten group and $P \leq G$ is a Sylow p-subgroup of G for some prime p, then G has a subgroup of order p. In addition, $Ω_{1} (P) \leq Z (G)$ or there is some non-nilpotent batten B of G such that $Ω_{1} (P) = P = B (B)$ .

Proof.

Since $P \leq G$ , it follows that P is cyclic or isomorphic to Q₈. Therefore $Ω_{1} (P)$ has order p.

If P is a batten, then $Ω_{1} (P) \leq Z (P) \leq Z (G)$ . In particular $Ω_{1} (P)$ is the unique subgroup of G of its order. Otherwise there is a non-nilpotent batten B of G such that $P \leq B$ . If P has order p, then $P = Ω_{1} (P) = B (B)$ is a normal subgroup of B and so of G. Again $Ω_{1} (P)$ is the unique subgroup of K of order q. Otherwise we have that $Ω_{1} (P) \leq Φ (P) = Z (B) \leq Z (G)$ by Lemma 2.5. In particular $Ω_{1} (P)$ is the unique subgroup of K of its order. □

Lemma 2.7.

Suppose that H is a batten and that $U ≨ H$ . Then U is a cyclic batten group. Furthermore, all subgroups of batten groups are batten groups.

Proof.

Assume for a contradiction that U is not a cyclic batten group. Then U is not a cyclic batten, and therefore H is neither cyclic of prime power order nor isomorphic to Q₈. Thus H is not nilpotent, in particular $| π (H) | = 2$ and all Sylow subgroups of H are cyclic. This implies that U is not a p-group. Let $π (H) = {q, r}$ , let $Q : = B (H)$ and $R \in {Syl}_{r} (H)$ be such that H = QR and $C_{R} (Q) = Φ (R) \neq 1$ . Now $π (U) = {q, r}$ as well and therefore $B (H) \leq U$ . Then Dedekind’s law gives that $U = B (H) \cdot (U \cap R)$ is a proper subgroup of $H = B (H) \cdot R$ , and it follows that $U \cap R$ is a proper subgroup of R. In particular, since R is cyclic, we have that $1 \neq U \cap R \leq Φ (R)$ . Then Lemma 2.5 gives that $Φ (R) = Z (H)$ . Altogether $U \leq B (H) Φ (R) = B (H) \times Φ (R)$ . But then U is a direct product of cyclic groups of prime power order, i.e. a cyclic batten group, and this is a contradiction.

Next suppose that G is a batten group and that U is a subgroup of G. Then U is a direct product of subgroups of the battens of G whose orders are pairwise coprime. Consequently U is a batten group as well, by the arguments above. □

We remark that sections of battens, or batten groups, are not necessarily batten groups. For example, $Q_{8} / Z (Q_{8})$ is not a batten group.

Lemma 2.8.

Suppose that K is a batten group and that $Q \leq K$ is a q-group.

If Q is not normal in K, then there is a non-nilpotent batten B of K such that $B = B (B) Q$ and $N_{K} (Q) = C_{K} (Q)$ .

If $Q E K$ , then $| K : C_{K} (Q) | \in {1, 4}$ or this index is a prime number.

Proof.

Since K is a batten group, there is a batten B of K such that $Q \leq B$ . Moreover there is a subgroup L of K such that $K = L \times B$ . Then $L \leq C_{K} (B) \leq C_{K} (Q)$ (*).

We first suppose that Q is not a normal subgroup of K. Since K is a direct product of battens, it follows that Q is not normal in B. Thus B is neither abelian nor hamiltonian (otherwise all subgroups of B would be normal), and it follows that B is not nilpotent. Now Q is a proper subgroup of B because $Q E B$ . We conclude from Lemma 2.5 that neither $Q \leq B (B)$ nor $Q \leq Z (B)$ , whence $B = B (B) Q$ and therefore $N_{B} (Q) = Q = C_{B} (Q)$ . Consequently (*) and Dedekind’s modular law yield that $N_{K} (Q) = L N_{B} (Q) = L C_{B} (Q) = C_{K} (Q)$ .

Suppose now that $Q E K$ . Using (*) we see that $| K : C_{K} (Q) | = | B : C_{B} (Q) |$ . Hence we may suppose that B is not abelian. If B is not nilpotent, then Z(B) and a Sylow q-subgroup of B centralize Q. Thus Lemma 2.5 yields that $| K : C_{K} (Q) | = | B : C_{B} (Q) |$ equals the prime in $π (B) ∖ {q}$ . Let $B ≅ Q_{8}$ and suppose that $Q ≰ Z (B)$ . Then Q has order 4 or 8. In the first case $| K : C_{K} (Q) | = | B : C_{B} (Q) | = | B : Q | = 2$ and in the second case $| K : C_{K} (Q) | = | B : C_{B} (Q) | = | B : Z (B) | = 4$ , which completes the proof. □

3 L₁₀ and its sublattices

Throughout this article we will use the notation from the next definition whenever we refer to L₁₀ and its sublattices:

Definition 3.1.

The lattice L₁₀ is defined to be isomorphic to $L (D_{8})$ , with notation as indicated in the picture.

Now we define

$L_{5} : = {E, S, U, A, F}$ ,
$L_{6} : = L_{5} \cup {T}$ ,
$L_{7} : = L_{5} \cup {D, C}$ ,
$L_{8} : = L_{10} ∖ {B, V}$ ,
$M_{8} : = L_{10} ∖ {T, V}$ ,
$L_{9} : = L_{10} ∖ {V}$ ,
$M_{9} : = L_{10} ∖ {B}$ ,

with the corresponding inclusion relations induced from the lattice L₁₀.

Definition 3.2.

Let L be a lattice. An equivalence relation $\equiv$ on L is called a congruence relation if and only if, for all $A, B, C, D \in L$ such that $A \equiv B$ and $C \equiv D$ , we have that $〈 A, C 〉 \equiv 〈 B, D 〉$ and $A \cap C \equiv B \cap D$ .

Lemma 3.3.

Let $\equiv$ be a congruence relation on $L_{9} = {A, B, C, D, E, F, U, T, S}$ as in Definition 3.1 and suppose that $\equiv$ is not equality. Then $E \equiv D$ .

Proof.

Let $\equiv$ be a congruence relation on L₉ and suppose that $E \equiv D$ .

If $X \in {A, B, C, F}$ , then $E \cap D = E \equiv D = X \cap D$ and therefore $E \equiv X$ .

First we assume that $X_{0} \in L_{9} ∖ {F}$ is such that $F \equiv X_{0}$ . Then there is some $X \in {A, B, C}$ such that $X_{0} \leq X$ , and then $X = 〈 X_{0}, X 〉 \equiv 〈 F, X 〉 = F$ . We choose $Y \in {T, U}$ such that $X \cap Y = E$ . Then $E = X \cap Y \equiv F \cap Y = Y$ and therefore, if $Z \in {T, U} ∖ {Y}$ , then $Z = 〈 Z, E 〉 \equiv 〈 Z, Y 〉 = F$ . Now it follows that $E = Z \cap D \equiv F \cap D = D$ , which gives a contradiction.

We have seen that E is not congruent to any of the elements $A, B, C, D, F$ . Next we assume that $X \in {S, T, U}$ is such that $E \equiv X$ and we choose $Y \in {A, C}$ such that $X ≰ Y$ . Then $Y = 〈 Y, E 〉 \equiv 〈 Y, X 〉 = F$ , which gives another contradiction. We conclude that {E} and {F} are singleton classes with respect to $\equiv$ .

If there are $X, Y \in L ∖ {E, F}$ such that $X \neq Y$ and $X \equiv Y, X \cap Y = E$ or $〈 X, Y 〉 = F$ , then this implies that $X = X \cap X \equiv X \cap Y = E$ or $X = 〈 X, X 〉 \equiv 〈 X, Y 〉 = F$ . As this is impossible, we conclude that such elements $X, Y$ do not exist.

In particular $A, B, C$ are pairwise non-congruent and $D, S, T, U$ are also pairwise non-congruent.

We assume that $D \equiv X \in {A, B, C}$ . Then we choose $Y \in {T, U}$ such that $Y ≰ X$ , and this gives the contradiction $F \neq 〈 Y, D 〉 \equiv 〈 Y, X 〉 = F$ . Hence {D} is a singleton as well.

Finally, we assume that there are $X \in {A, B, C}$ and $Y \in {T, S, U}$ such that $X \equiv Y$ . We have seen that $X \cap Y \neq E$ and then it follows that $Y \leq X$ by the structure of L₉. Now $D = X \cap D \equiv Y \cap D = E$ , which is impossible. In conclusion, for all $X, Y \in L_{9}$ , we have that $X \equiv Y$ if and only if X = Y. This means that $\equiv$ is equality. □

In the following lemma we argue similarly to Lemma 2.2 in [Citation8].

Lemma 3.4.

Suppose that $n \in N$ and that $G_{1}, \dots, G_{n}$ are normal subgroups of G of pair-wise coprime order such that $G = G_{1} \times \dots \times G_{n}$ . Then G is L₉-free if and only if, for every $i \in {1, \dots, n}$ , the group G_i is L₉-free.

Proof.

Since subgroups of L₉-free groups are L₉-free we just need to verify the “if” part.

Suppose that G₁,…,G_n are L₉-free. Then Lemma 1.6.4 of [Citation7] implies that $L (G) ≅ L (G_{1}) \times \dots \times L (G_{n})$ . By induction we may suppose that n = 2. Assume that $L = {E, T, S, U, D, A, B, C, F}$ is a sublattice of $L (G) ≅ L (G_{1}) \times L (G_{2})$ that is isomorphic to L₉ as in Definition 3.1. Then the projections $φ_{1}$ and $φ_{2}$ of L into $L (G_{1})$ and $L (G_{2})$ , respectively, are not injective, because $L (G_{1})$ and $L (G_{2})$ are L₉-free. Let $i \in {1, 2}$ and define, for all $X, Y \in L$ : $X \equiv_{i} Y : \Leftrightarrow φ_{i} (X) = φ_{i} (Y) .$

Then $\equiv_{i}$ is a congruence relation on L, because $φ_{i}$ is a lattice homomorphism, but it is not equality because $ϕ_{i}$ is not injective. Then Lemma 3.3 implies that $φ_{1} (D) = φ_{1} (E)$ and $φ_{2} (D) = φ_{2} (E)$ , and hence D = E. This is a contradiction. □

Lemma 3.5.

$L_{9} = {A, B, C, D, E, F, U, T, D}$ as in Definition 3.1 is completely characterized by the following:

L9(i) $D \neq E$ .

L9(ii) $〈 S, T 〉 = 〈 S, D 〉 = 〈 T, D 〉 = A$ and $S \cap T = S \cap D = T \cap D = E$ .

L9(iii) $〈 D, U 〉 = C$ and $D \cap U = E$ .

L9(iv) $〈 S, U 〉 = 〈 T, U 〉 = F$ .

L9(v) $〈 A, B 〉 = 〈 B, C 〉 = F$ and $A \cap B = A \cap C = B \cap C = D$ .

Proof.

We first remark that L₉ satisfies the relations given in L9 (i) – L9 (v).

Suppose conversely that a lattice $L = {A, B, C, D, E, F, S, T, U}$ satisfies the relations given in L9 (i) – L9 (v). Then we see that $E \leq A, B, C, D, S, T, U \leq F$ and $D \leq A, B, C$ as well as $S, T \leq A$ and $U \leq C$ . If these are the unique inclusions and $| L | = 9$ , then $L ≅ L_{9}$ .

For all $X, Y \in L$ such that $X \leq Y$ we have that $X \cap Y = X$ and $〈 X, Y 〉 = Y$ . Thus L9 (ii) shows that S, T, and D are pair-wise not subgroups of each other.

Using L9 (iii) we obtain that $D ≰ U$ and $U ≰ D$ , and L9 (iv) gives that also S, T and U are pair-wise not subgroups of each other. In addition, by L9 (v), we have that A, B and C are pair-wise not subgroups of each other. Together with the fact that $S, T \leq A$ and $U \leq C$ , this implies that $A ≰ U, C ≰ S, T$ and $B ≰ S, T, U$ . Moreover we have that $D \leq A, B, C$ and $S, T \leq A$ and $U \leq C$ and $A \neq F \neq B$ and $C \neq F$ . Together with L9 (ii) and L9 (iii), this information yields that $A ≱ U$ and $C ≱ S, T$ as well as $B ≱ S, T, U$ .

We conclude that there is a lattice homomorphism $φ$ from L₉ to L. Hence we obtain a congruence relation $\equiv$ on L₉ by defining that $X \equiv Y$ if and only if $φ (X) = φ (Y)$ , for all $X, Y \in L_{9}$ .

If $φ$ is not injective, then Lemma 3.3 implies that E = D. This contradicts L9 (i). Consequently $φ$ is injective and $L ≅ L_{9}$ . □

The next lemma gives an example of a group that is not L₉-free.

Lemma 3.6.

D₁₂ is not L₉-free.

Proof.

Let G be isomorphic to D₁₂ and let $a, b \in G$ be such that o(a) = 6, o(b) = 2 and $G = 〈 a, b 〉$ . Then we find a sublattice in L(G) isomorphic to L₉ by checking the equations from Lemma 3.5.

We let $L : = {1, 〈 b 〉, 〈 a^{2} b 〉, 〈 a^{2} 〉, 〈 ab 〉, 〈 a^{2}, b 〉, 〈 a 〉, 〈 a^{2}, ab 〉, G}$ and we define $A : = 〈 a^{2}, b 〉$ and $C : = 〈 a^{2}, ab 〉$ .

L9(i)We see that $a^{2} \neq 1$ and hence $〈 a 〉 \neq 1$ .

L9(ii)We notice that $A \leq G$ is isomorphic to ${Sym}_{3}$ with cyclic normal subgroup $〈 a^{2} 〉$ of order 3 and distinct subgroups $〈 b 〉, 〈 a^{2} b 〉$ of order 2. Then $〈〈 b 〉, 〈 a^{2} b 〉〉 = 〈〈 b 〉, 〈 a^{2} 〉〉 = 〈〈 a^{2} b 〉, 〈 a^{2} 〉〉 = A$ and $〈 b 〉 \cap 〈 a^{2} b 〉 = 〈 b 〉 \cap 〈 a^{2} 〉 = 〈 a^{2} b 〉 \cap 〈 a^{2} 〉 = 1$ .

L9(iii)The subgroup C is also isomorphic to ${Sym}_{3}$ , the subgroup $〈 ab 〉 \leq C$ has order 2, and moreover $〈〈 a^{2} 〉, 〈 ab 〉〉 = C$ and $〈 a^{2} 〉 \cap 〈 ab 〉 = 1$ .

L9(iv)We first see that $〈〈 b 〉, 〈 ab 〉〉 = 〈 a, b 〉 = G$ and then $〈〈 a^{2} b 〉, 〈 ab 〉〉 = 〈 a^{2} b {(ab)}^{- 1}, ab 〉 = 〈 a, b 〉 = G$ .

L9(v) $〈 a 〉$ is a cyclic normal subgroup of G of order 6, and the subgroups A and C also have order 6. These three subgroups are maximal in G. Hence $〈 A, 〈 a 〉〉 = 〈 C, 〈 a 〉〉 = G$ . Assume for a contradiction that A = C. Then $b \in C = {1, a^{2}, a^{4}, ab, a^{3} b, a^{5} b}$ , which is false. Since $〈 a^{2} 〉$ is the unique subgroup of order 3 of G, we conclude that $A \cap 〈 a 〉 = A \cap C = 〈 a 〉 \cap C = 〈 a^{2} 〉$ .

Altogether it follows, with Lemma 3.5, that L is isomorphic to L₉, and then G is not L₉-free. □

The next lemma shows how we can construct an entire class of groups that are not L₉-free.

Lemma 3.7.

Suppose that $p \neq q$ and that G = PQ, where P is an elementary abelian normal Sylow p-subgroup of G and Q is a cyclic Sylow q-subgroup of G. Suppose that Q acts irreducibly on $[P, Q] \neq 1$ and that $| C_{P} (Q) | \geq 3$ . Then G is not L₉-free.

Proof.

Since Q is abelian, we see that $E : = C_{Q} (P) E G$ . We claim that $G / C_{Q} (P)$ is not L₉-free. Therefore we may suppose that E = 1.

Our hypotheses imply that $[P, Q]$ is not centralized by Q, and in particular $| [P, Q] | \geq 3$ . Moreover $| C_{P} (Q) | \geq 3$ by hypothesis. Since P is elementary abelian, Lemma 1.1 gives that $P = [P, Q] \times C_{P} (Q)$ .

We let $V \leq [P, Q]$ and $D \leq C_{P} (Q)$ be subgroups of minimal order such that $| V | \geq 3$ and $| D | \geq 3$ and we set $A : = V \times D$ . If p is odd, then A has order p², and if p = 2, then $| A | = 2^{4} = 16$ . In the first case A has $\frac{p^{2} - 1}{p - 1} = p + 1 \geq 4$ subgroups isomorphic to V. In the second case A has $\frac{15 \cdot 14}{3} = 70$ subgroups isomorphic to V, where $3 \cdot \frac{14}{2} - 2 = 19$ of these subgroups intersect V non-trivially and 19 of them intersect D non-trivially. In both cases, we find subgroups T and S of A isomorphic to V such that $| {D, T, S, V} | = 4$ and $T \cap V = T \cap D = T \cap S = S \cap D = S \cap V = 1 = E$ .

We recall that A is elementary abelian, and then it follows that L9 (i) and L9 (ii) hold and that $A = 〈 S, V 〉 = 〈 T, V 〉$ (*).

We further set $U : = Q$ . Then $U \cap D \leq Q \cap P = E = 1$ and $〈 U, D 〉 = UD = : C$ , which implies L9 (iii).

If $X \in {T, S}$ , then $X ≰ C_{P} (Q)$ and then the irreducible action of Q on $[P, Q]$ and Lemma 1.2 yield that $V \leq [P, Q] \leq 〈 X, U 〉$ . Using (*) it follows that $D \leq A \leq 〈 V, X, U 〉 = 〈 X, U 〉$ . Combining all this information gives that $〈 X, U 〉 = [P, Q] DQ$ . Now if we set $F : = [P, Q] DQ$ , then we have L9 (iv).

To prove our claim, it remains to show that property L9 (v) of Lemma 3.5 is satisfied.

We set $B : = D Q^{x}$ for some $x \in {[P, Q]}^{#}$ .

If $y \in {1, x}$ , then $D \leq A \cap D Q^{y} = D (A \cap Q^{y}) = D$ and hence $A \cap B = A \cap C = D$ . We further have that $D \leq B \cap C = D (Q \cap D Q^{x}) = D C_{Q} (x) = D$ , by Lemma 1.4 (b) and (c), because Q acts irreducibly on $[P, Q]$ .

In addition, the irreducible action of Q^x on $[P, Q]$ and Lemma 1.2 yield that $[P, Q] \leq 〈 A, Q^{x} 〉$ . It follows that $〈 A, B 〉 = [P, Q] D Q^{x} = F$ . Finally we deduce from Part (d) of Lemma 1.4 that $〈 B, C 〉 = D 〈 Q, Q^{x} 〉 = D 〈 {[x, Q]}^{Q^{x}} 〉 Q^{x} = D [P, Q] Q^{x} = F .$

□

4 Group orders with few prime divisors

Much of our analysis will focus on non-nilpotent groups with a small number of primes dividing their orders. The next lemma sheds some light on why this situation naturally occurs.

Lemma 4.1.

Suppose that G is L₉-free. Then G possesses a normal Sylow subgroup.

Proof.

Assume that this is false. Since G is L₉-free and hence L₁₀-free, [Citation9, Corollary C] is applicable. Then G is metacyclic because it does not have any normal Sylow subgroup, and it follows that G is supersoluble. Then Satz VI.9.1(c) in [Citation3] gives a contradiction. □

Lemma 4.2.

Suppose that G is a p-group. Then the following statements are equivalent:

L(G) is modular.
G is L₅-free.
G is L₉-free.
G is L₁₀-free.

Proof.

This lemma follows from Theorem 2.1.2 in [Citation7] and Lemma 2.1 in [Citation9], since L₉ is a sublattice of L₁₀ containing L₅. □

Lemma 4.3.

Suppose that $p \neq q$ and that G is an L₉-free {p, q}-group. Let P be a normal Sylow p-subgroup of G and let $Q \in {Syl}_{q} (G)$ . If G is not nilpotent, then Q is cyclic or $Q ≅ Q_{8}$ .

Proof.

First we note that all subgroups and sections of G are L₉-free and that G is L₁₀-free.

Let G be non-nilpotent and assume for a contradiction that Q is neither cyclic nor isomorphic to Q₈. Given that G is L₁₀-free, we may apply Theorem B of [Citation9] and we see that neither (a), (b) nor (c) hold. Therefore p = 3 and q = 2. Now there are $a, b \in Q$ such that $〈 a, b 〉$ is not cyclic and b is an involution. If, for all choices of b, we have that $C_{P} (b) = C_{P} (Q)$ , then $Ω_{1} (Q)$ acts element-wise fixed-point-freely on $P / C_{P} (Q)$ , contradicting 8.3.4 (b) of [Citation5]. Therefore we may choose b such that a does not centralize $C_{P} (b)$ , and we also choose a of minimal order under these constraints. Then a² centralizes P and a inverts an element $x \in C_{P} (b)$ by a result of Baer (e.g. 6.7.7 of [Citation5]). If follows that a inverts $Ω_{1} (〈 x 〉)$ and we may suppose that x has order 3. Now $〈 x, a, b 〉 / C_{〈 a 〉} (x)$ is isomorphic to D₁₂, contrary to Lemma 3.6. □

Lemma 4.4.

Suppose that $p \neq q$ and that G is an L₉-free {p, q}-group. Furthermore, let P be a normal Sylow p-subgroup of G and let $Q \in {Syl}_{q} (G)$ be cyclic such that $1 \neq [P, Q]$ is elementary abelian.

Then every subgroup of Q acts irreducibly or by inducing (possibly trivial) power automorphism on $[P, Q]$ . Moreover, $C_{P} (Q)$ is a cyclic 2-group and P is abelian.

Proof.

First we note that all subgroups and sections of G are L₉-free and that all subgroups and sections of $[P, Q]$ are elementary abelian, by hypothesis. In addition Lemma 2.2 of [Citation9] yields that $P = C_{P} (Q) \times [P, Q]$ , as G is also L₁₀-free. In particular P is abelian if $C_{P} (Q)$ is.

Assume that the lemma is false and let G be a minimal counterexample.

Since $[P, Q]$ is elementary abelian, we introduce the following notation with Maschke’s theorem:

Let $n \in N$ and let $M_{1}, \dots, M_{n} \leq [P, Q]$ be Q-invariant and such that $[P, Q] = M_{1} \times \dots \times M_{n}$ and that Q acts irreducibly on $M_{1}, \dots, M_{n}$ , respectively. Lemmas 2.3.5 of [Citation7] and 4.2 yield that $Ω_{1} (C_{P} (Q))$ is elementary abelian. Now there are $r \in N$ and cyclic subgroups $M_{n + 1}, \dots, M_{n + r}$ of $Ω_{1} (C_{P} (Q))$ such that $Ω_{1} (C_{P} (Q)) = M_{n + 1} \times \dots \times M_{n + r}$ .

We set $H_{1} : = (M_{1} \times \dots \times M_{n + r - 1}) Q$ and $H_{2} : = (M_{2} \times \dots \times M_{n + r}) Q$ .

Then for every $i \in {1, 2}$ the group $O_{p} (H_{i})$ is elementary abelian. Moreover H_i is a proper subgroup of G and then the minimal choice of G implies that every subgroup of Q either induces (possibly trivial) power automorphisms on $[O_{p} (H_{i}), Q]$ or acts irreducibly on it.

(1) $C_{P} (Q) = 1$ and $n \leq 2$ .

Proof. We assume for a contradiction that $n + r \geq 3$ .

Then Q does not act irreducibly on both $O_{p} (H_{1})$ and $O_{p} (H_{2})$ , and it follows that Q induces (possibly trivial) power automorphisms on $[O_{p} (H_{i}), Q]$ .

We suppose first that $C_{P} (Q) = 1$ . Then $[O_{p} (H_{i}), Q] = O_{p} (H_{i})$ for both $i \in {1, 2}$ . Therefore Lemma 1.5.4 of [Citation7], together with the fact that $1 \neq M_{2} \leq O_{p} (H_{1}) \cap O_{p} (H_{2})$ , provides some $k \in N$ such that $a^{y} = a^{k}$ for every $a \in O_{p} (H_{1}) O_{p} (H_{2}) = [P, Q] Ω_{1} (C_{P} (Q))$ . But this means that Q, and hence every subgroup of Q, induces (possibly trivial) power automorphism on $O_{p} (H_{1}) O_{p} (H_{2}) = [P, Q]$ in this case. Thus G is not a counterexample, which is a contradiction.

We conclude that $C_{P} (Q) \neq 1$ and now there is some $i \in {1, 2}$ such that $1 \neq [O_{p} (H_{i}), Q]$ and $C_{O_{p} (H_{i})} (Q) \neq 1$ . Then H_i satisfies the hypotheses of our lemma and it follows that $C_{O_{p} (H_{i})} (Q)$ is an non-trivial 2-group. In particular p = 2. But then Lemma 1.3 provides the contradiction that $[O_{p} (H_{i}), Q] = 1$ .

For the proof of (1), we assume for a further contradiction that $r = n = 1$ . Then Q acts irreducibly on the elementary abelian group $[P, Q]$ and Lemma 3.7, applied to $([P, Q] \times Ω_{1} (C_{P} (Q))) Q$ , gives that $| Ω_{1} (C_{P} (Q)) | = 2$ . In particular we have that p = 2. Thus the minimal choice of G and Lemma 1.3 yield, for every proper subgroup U of Q, that U centralizes P or acts irreducibly on $[P, Q] = [P, U]$ .

Since G is a counterexample, it follows that $C_{P} (Q)$ is not cyclic. But $| Ω_{1} (C_{P} (Q)) | = 2$ and therefore $C_{P} (Q)$ is a generalized quaternion group. It follows that $C_{P} (Q) ≅ Q_{8}$ by Lemma 4.2. In this case $1 \neq Z : = Z (C_{P} (Q)) E G$ and G/Z satisfies the hypotheses of our lemma, but not the conclusion. Thus G is not a minimal counterexample, contrary to our choice. □

(2) Q acts irreducibly on $P = [P, Q]$ .

Proof. Assume for a contradiction that n = 2. By hypothesis Q is cyclic, and then we may suppose that $C_{Q} (M_{1}) \leq C_{Q} (M_{2}) = : Q_{0}$ . If $C_{Q} (M_{1}) = C_{Q} (M_{2})$ , then Lemma 2.8 of [Citation9] implies that Q induces power automorphisms on P. Thus G is not a counterexample, which is a contradiction.

Therefore $C_{Q} (M_{1}) ≨ Q_{0}$ and $1 \neq [M_{1}, Q_{0}] \leq [P, Q_{0}]$ . The minimal choice of G yields that Q₀ acts irreducibly or by inducing power automorphisms on $[P, Q_{0}]$ and that $C_{P} (Q_{0})$ is a cyclic 2-group. Now we notice that $M_{2} \leq C_{P} (Q_{0})$ , but $M_{2} ≰ 1 = C_{P} (Q)$ , whence we deduce a contradiction from 2.2.5 of [Citation5]. □

Since G = PQ is a counterexample to the lemma, Q has a proper subgroup U that does not act irreducibly on $[P, Q] = P$ and it also does not induce power automorphisms on $[P, Q]$ . In particular it does not act trivially. Since PU is a proper subgroup of our minimal counterexample G, it follows that $1 \neq C_{P} (U) \neq P$ . But $C_{P} (U)$ is Q-invariant, because Q is abelian. This is a final contradiction with regard to (2). □

Lemma 4.5.

Suppose that q is odd and that G is an L₉-free ${2, q}$ -group. Suppose further that P is a normal Sylow 2-subgroup of G such that $[P, Q]$ is hamiltonian and let $Q \in {Syl}_{q} (G)$ . Then one of the following holds:

G is nilpotent or
$[P, Q] ≅ Q_{8}$ and there exists a group I of order at most 2 such that $P = [P, Q] \times I$ and Q is a cyclic 3-group. Moreover $[P, Q] Q / Z ([P, Q] Q) ≅ {Alt}_{4}$ .

Proof.

We suppose that G is not nilpotent.

Then Q is not normal in G and Lemma 4.3 implies that Q is cyclic. Furthermore, P is L₉-free and hence it is modular by Lemma 4.2. Since $[P, Q]$ is hamiltonian, Theorem 2.3.1 of [Citation7] provides subgroups $P_{0}, I \leq P$ such that $P_{0} ≅ Q_{8}$ , I is elementary abelian and $P = P_{0} \times I$ .

We recall that the automorphism group of Q₈ is isomorphic to ${Sym}_{4}$ . Thus, if $Q_{8} ≅ P_{1} \leq P$ is Q-invariant, but not centralized by Q, then $Q_{8} ≅ P_{1} \leq [P, Q]$ and $1 \neq | Q / C_{Q} (P_{1}) | = 3$ . It follows that Q is a cyclic 3-group and $[P_{1}, Q] Q / Z ([P_{1}, Q] Q) ≅ {Alt}_{4}$ .

We conclude that our assertion holds if I = 1. Now suppose that $I \neq 1$ . We recall that $P E G$ and therefore $Φ (P_{0}) = Φ (P) ⊴ G$ . Since $| Φ (P_{0}) | = 2$ , it follows that $Φ (P_{0}) \leq Z (G)$ , and then $\bar{G} : = G / Φ (P_{0})$ is not nilpotent because G is not. Furthermore, $\bar{G}$ is L₉-free, $\bar{I} ≅ I \neq 1, \bar{Q} ≅ Q$ and ${\bar{P}}_{0}$ is elementary abelian of order 4. In particular $\bar{P}$ is elementary abelian, hence it is a non-hamiltonian 2-group of order at least 8. Lemma 4.4 states that $\bar{Q}$ acts irreducibly on $[\bar{P}, \bar{Q}] \neq 1$ or induces power automorphisms on it. The second case is not possible by Lemma 1.3.

Hence $Q ≅ \bar{Q}$ acts irreducibly on $[\bar{P}, \bar{Q}] = \bar{[P, Q]}$ and, by Lemma 4.4, we see that $C_{\bar{P}} (\bar{Q})$ is a cyclic 2-group. Since $I \neq 1$ and $Ω_{1} (P) = Φ (P_{0}) \times I$ , we have that $1 \neq \bar{I} = \bar{Ω_{1} (P)}$ is $\bar{Q}$ -invariant and ${\bar{P}}_{0}$ is a non-cyclic complement of $\bar{I}$ in $\bar{P}$ . This implies that $\bar{I} = C_{\bar{P}} (\bar{Q})$ is cyclic and elementary abelian at the same time. Thus $I ≅ \bar{I}$ has order 2 and with Lemma 1.1 we deduce that $C_{P} (Q) Φ (P_{0}) = I Φ (P_{0}) = Ω_{1} (P)$ is elementary abelian of order 4. Then we deduce that $C_{P} (Q) = Ω_{1} (P)$ and then $[P, Q] \cap C_{P} (Q) \neq 1$ . Moreover, since $[P, Q] C_{P} (Q) = P = P_{1} \times I \leq P_{1} C_{Q} (P)$ , it follows that $| [P, Q] | \leq | P_{1} | \leq 8$ . In conclusion, $[P, Q]$ is a subgroup of order at most 8 admitting an automorphism of odd order that centralizes $Ω_{1} ([P, Q])$ . It follows that $[P, Q] ≅ Q_{8}$ and then, together with the fact that $I \leq Z (G)$ , our assertions follow. □

Definition 4.6.

Suppose that Q is a cyclic q-group that acts coprimely on the p-group P. We say that the action of Q on P avoids L₉ (and we indicate more technical details by writing “of type ( $\cdot$ )”) if and only if one of the following is true:

(std)Every subgroup of Q acts irreducibly or by inducing (possibly trivial) power automorphisms on the elementary abelian group $[P, Q] = P$ .

(cent)Every subgroup of Q acts irreducibly or trivially on the elementary abelian group $[P, Q]$ , P is abelian and $C_{P} (Q)$ is a nontrivial cyclic 2-group.

(hamil) $[P, Q] ≅ Q_{8}, P = [P, Q] \times I$ , where I is a group of order at most 2, and Q is a cyclic 3-group such that $[P, Q] Q / Z ([P, Q] Q) ≅ {Alt}_{4}$ .

Lemma 4.7.

Suppose that p is an odd prime and that G is an L₉-free ${2, p}$ -group. Let further P be a normal Sylow p-subgroup of G and let $Q \in {Syl}_{2} (G)$ be isomorphic to Q₈ and such that $1 \neq [P, Q]$ is elementary abelian.

Then $p \equiv 3 mod 4, | P | = p^{2}$ and Q acts faithfully on P.

Proof.

We set $Z : = Ω_{1} (Q)$ . If $Z ⊴ G$ , then $Z \leq Z (G)$ and we consider $\bar{G} : = G / Z$ . Then $\bar{G}$ is an L₉-free ${2, p}$ -group, $\bar{P}$ is a normal Sylow p-subgroup of $\bar{G}$ and $\bar{Q} \in {Syl}_{2} (\bar{G})$ . Since $\bar{Q}$ is neither cyclic nor isomorphic to Q₈, Lemma 4.3 is applicable and we see that $\bar{G}$ is nilpotent. But then G is also nilpotent, contrary to our hypothesis that $[P, Q] \neq 1$ . Thus $Ω_{1} (Q)$ is not normal in G and Q acts faithfully on P. Now, for all $y \in Q$ of order 4, we apply Lemma 4.4 on $[P, Q] 〈 y 〉$ to deduce that $〈 y 〉$ either induces power automorphisms on $[P, Q]$ or acts irreducibly on it. Theorem 1.5.1 of [Citation7] states that ${Pot}_{G} (P)$ is abelian, but $Q ≅ Q_{8}$ is not, which means that we may choose y such that y does not induce power automorphisms on P. In particular P is not cyclic of prime order. Moreover p is odd and therefore 4 divides $(p + 1) (p - 1) = p^{2} - 1$ , and Satz II 3.10 of [Citation3] yields that $| P | \leq p^{2}$ . It follows that $| P | = p^{2}$ . More precisely, as $| P | \neq p$ , the result implies that $p \equiv 3 (mod 4)$ , and then the proof is complete. □

Definition 4.8.

Suppose that $Q ≅ Q_{8}$ acts coprimely on the p-group P. We say that the action of Q on P avoids L₉ if and only if $p \equiv 3 mod 4, | P | = p^{2}$ and Q acts faithfully on P.

Lemma 4.9.

Suppose $Q ≅ Q_{8}$ and that P is a p-group on which Q acts avoiding L₉. Then P is elementary abelian, $Ω_{1} (Q)$ inverts P, and every subgroup of Q of order at least 4 acts irreducibly on P.

Proof.

Since a cyclic group of order p² has an abelian automorphism group by 2.2.3 of [Citation5], it follows that P is elementary abelian. If $1 \neq R$ is a cyclic subgroup of P, then $| Aut (R) | = p - 1$ and therefore R does not admit an automorphism of order 4. Additionally, $[P, Ω_{1} (Q)]$ is Q-invariant and, since Q acts faithfully on P, we see that $[P, Ω_{1} (Q)] \neq 1$ . Furthermore, Lemma 1.1 gives that $[P, Ω_{1} (Q)] \cap C_{P} (Ω_{1} (Q)) = 1$ because P is abelian. Moreover, Q has rank 1, and then it follows that Q acts faithfully on $[P, Ω_{1} (Q)]$ . This implies that $| [P, Ω_{1} (Q)] | \neq p$ and consequently $[P, Ω_{1} (Q)] = P$ . Hence 8.1.8 of [Citation5] states that $Ω_{1} (Q)$ inverts P. In particular $Ω_{1} (Q)$ inverts every cyclic subgroup R of P.

In addition, these arguments show that every subgroup U of order 4 of Q does not normalize any nontrivial proper subgroup of the elementary abelian group P. This means that U acts irreducibly on P. □

Corollary 4.10.

Suppose that $p \neq q$ and that G is an L₉-free {p, q}-group such that $P \in {Syl}_{p} (G)$ is normal in G and $Q \in {Syl}_{q} (G)$ .

Then either G is nilpotent and P and Q are modular or Q is a batten and it acts on P avoiding L₉.

In particular, if G is not nilpotent, then Q is isomorphic to Q₈ or cyclic and $[P, Q]$ is elementary abelian or isomorphic to Q₈, where in the second case q = 3.

Proof.

By hypothesis G is L₉-free, hence P and Q are, too. Then Lemma 4.2 implies that P and Q are modular.

Suppose that G is not nilpotent. Then Lemma 4.3 applies: Q is cyclic or isomorphic to Q₈ and hence it is a batten. Moreover Lemma 2.2 of [Citation9] states that $[P, Q]$ is a hamiltonian 2-group or elementary abelian. In the first case Lemma 4.5 gives the assertion. In the second case our statement follows from Lemmas 4.4 and 4.7. □

Lemma 4.11.

Let Q be a nilpotent batten that acts on the p-group P avoiding L₉, and suppose that U is a subgroup of Q.

Then U induces power automorphisms on P or it acts irreducibly on $[P, Q] / Φ ([P, Q])$ .

Proof.

First suppose that $Q ≅ Q_{8}$ . Then Lemma 4.9 implies that every subgroup of order at least 4 of Q, and in particular Q itself, acts irreducibly on $P = [P, Q] / Φ ([P, Q])$ . Moreover, the involution of Q inverts P by Lemma 4.9, and then the statement holds.

Next we suppose that Q is cyclic. Then Definition 4.6 gives the assertion unless the action of Q on P avoids L₉ of type (hamil). In this case every proper subgroup of Q centralizes P, while Q acts irreducibly on $[P, Q] / Z ([P, Q]) = [P, Q] / Φ ([P, Q])$ . □

Next we investigate groups of order divisible by more than two primes. This needs some preparation.

Lemma 4.12.

Suppose that P and R are distinct Sylow subgroups of G, that $Q \in {Syl}_{q} (G)$ is cyclic and that it normalizes P and R, but does not centralize them. Suppose further that R normalizes every Q-invariant subgroup of P. If $C_{Q} (P) = C_{Q} (R)$ , then G is not L₉-free.

Proof.

We suppose that $C_{Q} (P) = C_{Q} (R) = : E$ . Then E is a normal subgroup of G because Q is abelian.

We claim that G/E is not L₉-free, and for this we may suppose that E = 1. Then $Q \neq 1$ acts faithfully on P and R. Now we need a technical step before we move on:

There are a Q-invariant subgroup D of P and elements $x, y \in D$ such that $C_{Q} (x) = C_{Q} (y) = C_{Q} (x y^{- 1}) = 1$ and $D = [x, Q] = [y, Q] = [x y^{- 1}, Q]$ . (*)

Since Q is cyclic, there is some $u \in Q$ such that $〈 u 〉 = Q$ . Let $x_{0} \in [P, Q]$ be such that $Ω_{1} (Q)$ does not centralize x₀.

Then $[x_{0}, Q] = [x_{0}, 〈 u 〉, Q] = [〈 [x_{0}, u] 〉, Q] = [[x_{0}, u], Q]$ by Lemma 1.1, and for all integers n we have the following: ${(x_{0}^{- 1} x_{0}^{u})}^{u^{n}} = 1$ iff $(HTML translation failed)$ iff $x_{0}^{u} = 1$ . It follows that $[x_{0}, u] \in [[x_{0}, u], Q] = [x, Q]$ and $C_{Q} ([x_{0}, u]) = C_{Q} (x_{0}) = 1$ .

Now we set $x : = [x_{0}, u], y : = x^{u}$ and $D : = [x, Q]$ . Then D is Q-invariant and we have that $x, y \in D, C_{Q} (x) = C_{Q} (y) = 1$ and $D = [x, Q] = [y, Q]$ . If we set $z : = x y^{- 1}$ , then $z = [x^{- 1}, u]$ and we can use the information from the end of the previous paragraph:

$[z, Q] = [x^{- 1}, Q] = D$ and $C_{Q} (z) = C_{Q} (x^{- 1}) = C_{Q} (x) = 1$ . This concludes the proof of (*).

We use (*) and its notation and, similarly, we find a Q-invariant subgroup R₀ of R and an element $h \in R_{0}$ such that $C_{Q} (h) = 1$ and $R_{0} = [h, Q]$ .

We set $S : = Q^{x}, T = Q^{y}$ , $A : = DQ, B : = D R_{0}, U : = Q^{h}, C : = D Q^{h}$ and $F : = D R_{0} Q$ , and we claim that ${A, B, C, D, E, F, S, T, U}$ is isomorphic to L₉. The properties L9 (i) and L9 (iii) of Lemma 3.5 follow from the choice of D, since h and Q normalize D.

For L9 (ii) we first note that $D \cap Q^{x} = D \cap Q^{y} = 1 = E$ and $〈 D, Q^{x} 〉 = 〈 D, Q^{y} 〉 = DQ$ , since $x \in D$ and hence $y = x^{u} \in D$ . Next, Lemma 1.4 (b) yields that $T \cap S = Q^{x} \cap Q^{y} \leq C_{Q} {(x y^{- 1})}^{y} = 1$ . Part (d) of the same lemma shows that $〈 T, S 〉 = 〈 Q^{x}, Q^{y} 〉 = 〈 {[x y^{- 1}, Q]}^{Q^{x y^{- 1}}} 〉 Q^{x y^{- 1}} = D Q^{x y^{- 1}} = DQ = A$ , as $x y^{- 1} \in D$ .

For all $z \in {x, y}$ we calculate that $〈 Q^{z}, Q^{h} 〉 = 〈 {[z h^{- 1}, Q]}^{Q^{z h^{- 1}}} 〉 Q^{z h^{- 1}} =$ $〈 {({[z, Q]}^{h^{- 1}} [h^{- 1}, Q])}^{Q^{z h^{- 1}}} 〉 Q^{z h^{- 1}}$ $= 〈 D, R_{0} 〉 Q^{z h^{- 1}} = F$ by Lemma 1.4 (d). Thus L9 (iv) of Lemma 3.5 is true.

We moreover have that $〈 A, B 〉 = 〈 D, Q, R_{0} 〉 = F = 〈 D, Q^{h}, R_{0} 〉$ and $A \cap B = DQ \cap D R_{0} = D (Q \cap D R_{0}) = D = D (Q^{h} \cap D R_{0}) = C \cap B$ . Finally $A \cap C = DQ \cap D Q^{h} = D (Q \cap D Q^{h}) \leq D C_{Q} (h) = D$ by Lemma 1.4 (b).

Using Lemma 3.5 we conclude that G/E is not L₉-free, and hence G is not L₉-free. □

Corollary 4.13.

Suppose that $p, q$ and r are pairwise distinct primes and that G is a directly indecomposable L₉-free {p, q, r}-group. Suppose further that $P \in {Syl}_{p} (G)$ and $R \in {Syl}_{r} (G)$ are normal in G and let $Q \in {Syl}_{q} (G)$ .

Then Q is cyclic and $C_{Q} (P) \neq C_{Q} (R)$ .

Proof.

Since G is directly indecomposable, we see that Q acts non-trivially on both P and R. Moreover, PQ and RQ are L₉-free by hypothesis, and then we conclude that Q is cyclic or isomorphic to Q₈.

In the first case, our assertion follows from Lemma 4.12, and in the second case, we choose a maximal subgroup Q₁ of Q. Then Q₁ acts irreducibly on P and R by Lemma 4.9, and the same Lemma shows that $Φ (Q)$ inverts P and R. Thus Q₁ acts on P and on R avoiding L₉, respectively, and it acts faithfully. This contradicts Lemma 4.12. □

We explain another example where a subgroup lattice contains L₉.

Lemma 4.14.

Suppose that p, q and r are pairwise distinct primes and that G is a {p, q, r}-group. Suppose further that $P \in {Syl}_{p} (G)$ is normal in G and that $Q \in {Syl}_{q} (G)$ and $R \in {Syl}_{r} (G)$ are cyclic groups such that $R E RQ$ . Suppose that $| R | = r$ and $C_{Q} (R) = 1$ .

If R acts irreducibly on P, but non-trivially, and if $1 \neq [P, Q]$ is elementary abelian, then G is not L₉-free.

Proof.

We first remark that G is soluble, because $P E PR E PRQ = G$ . We will construct the lattice L₉ in L(G) using Lemma 3.5. For this we set $E : = 1$ and $D : = P$ . Then $D \neq E$ and we see that L9 (i) is true.

Next, we recall that $1 \neq [P, Q]$ by hypothesis. Assume that $| [P, Q] | = 2$ . Then Q, which normalizes $[P, Q]$ , must centralize it, and then Lemma 1.1 gives a contradiction.

Therefore $| [P, Q] | ≩ 2$ . As a consequence, we find $a, b \in {[P, Q]}^{#}$ such that $a \neq b$ , and then we set $S : = R^{a}$ and $T : = R^{b}$ . Now $D \cap S = 1 = E = D \cap T$ and $〈 D, T 〉 = P R^{b} = PR = P R^{a} = 〈 D, S 〉$ . We set $A : = PR$ . In addition, since R acts irreducibly, but non-trivially on P, it follows that $C_{R} (b a^{- 1}) ≨ R$ . Then the fact that $| R | = r$ gives that $C_{R} (b a^{- 1}) = 1 = E$ .

Lemma 1.4 (b) shows that $S \cap T = {(R \cap R^{b a^{- 1}})}^{a} \leq C_{R} {(b a^{- 1})}^{a} = E$ , and now we recall that $[b a^{- 1}, R] \neq 1$ . Moreover, R acts irreducibly on P, and then Part (e) of the same lemma yields the following:

$〈 T, S 〉 = {〈 R, R^{b a^{- 1}} 〉}^{a} = {(〈 {[b a^{- 1}, R]}^{R^{b a^{- 1}}} 〉 R^{b a^{- 1}})}^{a} = PR = A$ . We conclude that L9 (ii) holds.

For L9 (iii) we set $U : = Q$ and $C : = 〈 D, Q 〉 = PQ$ . Then we note that $U \cap D = Q \cap P = 1 = E$ .

Assume for a contradiction that $X : = 〈 R^{c}, Q 〉$ has odd order for some $c \in {a, b}$ .

In both cases X is a $p'$ -Hall subgroup of the soluble group G = PRQ and therefore $R^{c} = O_{r} (X)$ . It follows that Q normalizes R^c and then that $Q^{c^{- 1}}$ and Q normalize R.

Since $N_{P} (R)$ is R-invariant and R acts irreducibly, but non-trivially on P, we conclude that $N_{G} (R) = RQ$ . Thus Sylow’s theorem provides some $y \in R$ such that $Q^{c^{- 1}} = Q^{y}$ . Now $[yc, Q] \leq PR \cap Q = 1$ . In addition $| R | = r$ and $[R, Q] \neq 1$ by hypothesis. Together this gives that $yc \in C_{G} (Q) \cap PR \leq PQ \cap PR = P (Q \cap PR) = P$ , by Dedekind’s modular law. Altogether we have that $yc \in C_{P} (Q)$ . We recall that $c \in {a, b} \subseteq P$ , and then $y = yc c^{- 1} \in P$ . But we chose $y \in R$ and now $y \in R \cap P = 1$ , whence $Q^{c^{- 1}} = Q$ . In other words, $c \in N_{P} (Q)$ , and this means that $[Q, c] \leq Q \cap P = 1$ and $c \in C_{P} (Q)$ . We recall that $c \in {[P, Q]}^{#}$ and that $[P, Q]$ is elementary abelian by hypothesis. Then Lemma 1.1 implies that $[P, Q] = [P, Q, Q] \times C_{[P, Q]} (Q)$ , and this contradicts the fact that $c \in C_{P} (Q) \cap [P, Q]$ .

It follows that X has even order and since R^c acts irreducibly on P, we conclude that $P \leq X$ . This implies that $〈 S, U 〉 = 〈 R^{a}, Q 〉 = PRQ = G = 〈 R^{b}, Q 〉 = 〈 T, Q 〉$ and then L9 (iv) holds for $F : = G$ .

We finally set $B : = P Q^{z}$ for some $z \in R^{#}$ . Then $R = 〈 z 〉$ and Lemma 1.4 (b) and (c), together with our hypothesis, show that $P \leq PQ \cap P Q^{z} = P (Q \cap P Q^{z}) \leq P C_{Q} (z) = P C_{Q} (R) = P$ . Thus we have that $B \cap C = P = D$ . We further see that $A \cap C = PR \cap PQ = P (R \cap PQ) = P = D$ and $A \cap B = PR \cap P Q^{z} = P (R \cap P Q^{z}) = P = D$ . Since $〈 A, B 〉 = 〈 PR, P Q^{z} 〉 = PQR = G$ and $〈 B, C 〉 = 〈 PQ, P Q^{z} 〉 = P 〈 {[Q, z]}^{Q^{z}} 〉 Q^{z} = P [Q, R] Q = PRQ = G$ by Lemma 1.4 (d), we finally obtain L9 (v).

Altogether Lemma 3.5 gives the assertion. □

Proposition 4.15.

Suppose that $p, q$ , and r are pairwise distinct primes and that G is a non-nilpotent L₉-free {p, q, r}-group with normal Sylow p-subgroup P. Suppose further that $R \in {Syl}_{r} (G)$ and $Q \in {Syl}_{q} (G)$ are not normal in G, that $R E RQ$ and $[R, Q] \neq 1$ .

Then RQ is a batten, P is elementary abelian of order p^r, R and Q act irreducibly on P and $Φ (Q)$ induces non-trivial power automorphisms on P.

Proof.

We proceed in a series of steps.

(1) The groups PQ and PR are not nilpotent, Q is cyclic and $R ≅ Q_{8}$ or $| R | = r$ . In addition $R = [R, Q]$ and $[R, C_{Q} (P)] \leq C_{R} (P)$ .

Proof. By hypothesis R is not normal in G, but Q normalizes R. Hence $P ≰ N_{G} (R)$ and in particular PR is not nilpotent. But PR is L₉-free, because G is. Moreover, RQ is non-nilpotent and L₉-free, again by hypothesis. Then Corollary 4.10 implies that R and Q are battens, that R acts on P avoiding L₉ and that Q acts on R avoiding L₉. More specifically, R and Q are cyclic or isomorphic to Q₈, and $[P, R]$ as well as $[R, Q]$ are elementary abelian or isomorphic to Q₈.

It follows that $R ≅ Q_{8}$ or that R is cyclic of order r. In both cases Lemma 1.1 yields $R = [R, Q]$ and the avoiding L₉ action of Q on R gives that Q is cyclic.

In addition $[P, C_{Q} (P), R] = 1, [P, R, C_{Q} (P)] \leq [P, C_{Q} (P)] = 1$ and then the Three Subgroups Lemma (see for example 1.5.6 of [Citation5]) implies that $1 = [R, C_{Q} (P), P]$ . Thus $[R, C_{Q} (P)] \leq C_{R} (P)$ .

If it was true that $[P, Q] = 1$ , then $R = [R, Q] = [R, C_{Q} (P)]$ would centralize P. But this is a contradiction. □

(2) $C_{R} (P) = 1$ and $C_{Q} (P) \leq Z (G)$ .

Proof. Since $[P, R] \neq 1$ by (1), we can apply Corollary 4.10 to PR, and this shows that R acts on P avoiding L₉. Otherwise (1) implies that $R ≅ Q_{8}$ and then Definition 4.8 gives that R acts faithfully on P. If $| R | = r$ , then R acts faithfully of P because $[P, R] \neq 1$ . In both cases we see that $C_{R} (P) = 1$ , and then the last statement of (1) implies that $[R, C_{Q} (P)] \leq C_{R} (P) = 1$ . Then $C_{Q} (P)$ centralizes P and R, and Q is cyclic by (1), and therefore it follows that $C_{Q} (P) \leq Z (G)$ . □

(3) Z(G) = 1 or $p \neq 2$ .

Proof. We suppose that p = 2. Let $- : G \to G / Z (G)$ be the natural homomorphism. We show that $\bar{G}$ satisfies the hypotheses of our lemma.

From (1) we see that none of the groups P, Q or R is contained in Z(G). We even have that $R \cap Z (G) = 1$ by (1). In particular $p, q, r \in π (\bar{G})$ and $\bar{G}$ is L₉-free. We see that $\bar{P}$ is a normal Sylow p-subgroup of G and that $\bar{Q} \in {Syl}_{q} (G)$ and $\bar{R} \in {Syl}_{r} (G)$ are such that $\bar{R} E \bar{R} \bar{Q} \leq G$ . Let $X \in {R, Q}$ . If $\bar{X} E \bar{G}$ , then $XZ (G) E G$ and X is a characteristic Sylow subgroup of XZ(G) and hence normal in G. This is a contradiction.

We deduce that all hypotheses of the lemma hold for $\bar{G}$ and that $[\bar{R}, \bar{Q}] = \bar{[R, Q]} = \bar{R} \neq 1$ . Therefore, if $Z (G) \neq 1$ , then the minimal choice of G implies that $Φ (\bar{Q})$ induces non-trivial power automorphisms on the elementary abelian group $\bar{P}$ . Then Lemma 1.3 yields that $p \neq 2$ . □

(4) $C_{P} (R) = 1$ , and the groups $[P, Q]$ and $P = [P, R]$ are elementary abelian.

Proof. As PR and PQ are not nilpotent by (1), Corollary 4.10 implies that X acts on P avoiding L₉ and that $[P, X]$ is elementary abelian or isomorphic to Q₈ for both $X \in {Q, R}$ .

Assume for a contradiction that $[P, X]$ is isomorphic to Q₈ for some $X \in {Q, R}$ . Then X acts on P of type (hamil). Hence we obtain a group I of order 1 or 2 such that $P ≅ Q_{8} \times I$ . It follows that $Aut (P)$ is a {2, 3}-group. But this is impossible because Q and R both act coprimely and non-trivially on P by (1), and p = 2, q and r are pairwise distinct.

We conclude that $[P, Q]$ and $[P, R]$ are elementary abelian, and then it follows that P is abelian, by Lemma 4.4, applied to PR.

Assume for a further contradiction that $C_{P} (R) \neq 1$ . As R avoids L₉ in its action on P, it follows from Lemma 4.9 that R is not isomorphic to Q₈. Now (1) yields that R is cyclic and we may apply Lemma 4.4 to PR, because $[P, R]$ is elementary abelian. It follows that $C_{P} (R)$ is a cyclic 2-group and thus q and r are odd.

Furthermore $C_{P} (R)$ is normalized by Q, because Q normalizes P and R. Since q is odd, it follows that $C_{P} (R)$ is centralized by Q. We recall that P is abelian, and then (3) yields that $C_{P} (R) \leq Z (G) = 1$ . This is a contradiction.

Altogether $C_{P} (R) = 1$ and Lemma 1.1 gives that $P = [P, R]$ is elementary abelian. □

(5) $C_{P} (Q) = 1$ and Q acts on P avoiding L₉ of type (std).

Proof. Since PQ is not nilpotent and Q is cyclic by (1), Corollary 4.10 implies that the action of Q on P avoids L₉. But P is elementary abelian by (4), and therefore the action is not of type (hamil).

We assume for a contradiction that $C_{P} (Q) \neq 1$ . Then it follows that PQ is not of type (std). We consequently have type (cent) and we see that $C_{P} (Q)$ is a cyclic 2-group. In particular p = 2, and thus q and r are odd. Then $| R | = r$ by (1).

Next we claim that G satisfies the hypotheses of Lemma 4.14.

First, q and r are pairwise distinct odd primes and G is a finite ${2, q, r}$ -group. From above, (1) and our assumption we see that $P \in {Syl}_{2} (G)$ is normal in G and that $Q \in {Syl}_{q} (G)$ and $R \in {Syl}_{r} (G)$ are cyclic groups such that $R E RQ$ . We have shown that R has order r.

We recall that $C_{P} (Q)$ is a cyclic 2-group (first paragraph). As r is odd and $C_{P} (R) = 1$ by (4), we see that $C_{P} (Q)$ is not R-invariant. But $C_{P} (C_{Q} (R))$ is R-invariant, and now the irreducible action of R on P and the fact that $1 \neq C_{P} (Q) \leq C_{P} (C_{Q} (R))$ show that $P = C_{P} (C_{Q} (R))$ . Then (2) and (4) imply that $C_{Q} (R) \leq C_{Q} (P) \leq Z (G) = 1$ .

By (4) P is an elementary abelian 2-group, and $P = [P, R] \neq 1$ by (1) and (4). In particular R does not induce power automorphism on P by Lemma 1.3. As $C_{P} (R) = 1$ by (4), we deduce that R acts irreducibly on P (using Lemma 4.4). Finally, (1) and (4) yield that $1 \neq [P, Q]$ is elementary abelian.

All hypotheses of Lemma 4.14 are satisfied now, and we infer that G is not L₉-free. This is a contradiction.

Thus $C_{P} (Q) = 1$ and we deduce that the action of Q on P is not of type (cen). It remains that Q acts on P avoiding L₉ of type (std). □

(6) R and Q act irreducibly on P. If $X \leq Q$ induces power automorphisms on P, then X centralizes R.

Proof. We recall from (2) that $C_{Q} (P) \leq Z (G)$ and $C_{R} (P) = 1$ . Consequently $C_{RQ} (P) = C_{Q} (P) \leq Z (RQ)$ and $RQ / C_{Q} (P)$ is isomorphic to a subgroup of $Out (P)$ . Since RQ is not nilpotent, we see that $RQ / C_{Q} (P)$ is not nilpotent.

The group P is an elementary abelian p-group by (4) and hence, if $X \leq RQ$ induces power automorphisms on it, then it follows that $X C_{Q} (P) / C_{Q} (P) \leq Z (RQ / C_{Q} (P))$ , see page 177 of [Citation3]. We denote this fact by (*). Then we deduce that $[R, X] \leq C_{Q} (P) \leq Z (RQ)$ (see above) and therefore $[R, X] = [X, R] = [X, R, R] = 1$ by Lemma 1.1.

In addition, the fact (*) shows that neither R nor Q induces power automorphisms on P. It follows from (5) and Definition 4.6 (std) that Q acts irreducibly on P. Furthermore (1), together with Corollary 4.10, yields that R acts on P avoiding L₉. Since $C_{P} (R) = 1$ by (4), this action has type (std) or (hamil). In the first case, the irreducible action follows from Definition 4.6 (std), and in the second case, it follows from Lemma 4.9. □

(7) $C_{Q} (R)$ induces nontrivial power automorphisms on P.

Proof. We set $Q_{0} : = C_{Q} (R)$ and we assume that Q₀ acts irreducibly on P. Then II 3.11 of [Citation3] implies that $RQ = C_{RQ} (Q_{0})$ is isomorphic to a subgroup of the multiplicative group of some field of order $| P |$ , and it follows that RQ is cyclic. This is a contradiction.

Thus (6) and Lemma 4.11 show that Q₀ induces power automorphisms on P. Since R normalizes every Q-invariant subgroup of P by (7), we see from Lemma 4.12 and (2) that $C_{Q} (R) ≩ C_{Q} (P)$ . □

(8) $| P | = p^{q}$ and $Φ (Q) = C_{Q} (R)$ .

Proof. We recall that Q is cyclic, by (1). Moreover $[R, Q] \neq 1$ , whence $C_{Q} (R) < Q$ and therefore we may choose $y \in Q$ such that $C_{Q} (R) < 〈 y 〉$ .

By (5) and Lemma 4.11, it follows that every subgroup of $〈 y 〉$ either acts irreducibly on P or induces power automorphism on it (in particular normalizing every subgroup of P). Then $P 〈 y 〉$ satisfies (b) of Lemma 3.1 in [Citation8], which implies that it satisfies one of the possibilities 3.1 (i)–3.1 (iii). By (5) and the choice of y, we see that 3.1 (i) is not true. Further (7) provides some $x \in C_{Q} (R)$ that induces a power automorphism of order q on P. This implies that q divides p – 1 and therefore $P 〈 y 〉$ satisfies (ii) of Lemma 3.1 (b) in [Citation8]. It follows that $| P | = p^{q}$ and that, if k is the largest positive integer such that q^k divides p – 1, then y induces an automorphism of order $q^{k + 1}$ on P. We conclude that $q^{k + 1} = | 〈 y 〉 : C_{〈 y 〉} (P) | = | 〈 y 〉 : C_{R} (P) |$ , because Q is cyclic. Finally, we deduce that o(y) is uniquely determined, that $Q = 〈 y 〉$ and that $C_{Q} (P) = Φ (Q)$ . □

By (6), we see that R and Q act irreducibly on P, and (1) gives that Q is cyclic. Then (8) and (7) say that $Φ (Q)$ induces nontrivial power automorphisms on P. In addition P is elementary abelian by (4), and it has order p^r by (8). If $| R | = r$ , then R is a cyclic group of order r and RQ is a batten. In particular G satisfies the assertion of our lemma.

But G is a counterexample, and then it follows that $R ≅ Q_{8}$ and r = 2. Then (4) yields that PR fulfills the hypothesis of Lemma 4.7, and consequently $p^{r} = p^{2} = | P | = p^{q}$ by (8). This is our final contradiction, because $q \neq r$ . □

Definition 4.16.

Suppose that B is a non-nilpotent batten that acts coprimely on the p-group P. We say that the action of B on P avoids L₉ if and only if $[P, Z (B)] \neq 1$ and if one of the following occurs:

(Cy) $[P, B (B)] = 1$ and Q acts on P avoiding L₉ for every Sylow subgroup Q of B different from $B (B)$ or

(NN)P is elementary abelian of order $p^{| B : B (B) Z (B) |}$ and the Sylow subgroups of B act irreducibly on P, while Z(B) induces power automorphisms on P.

As in Definition 4.6, we specify the type of the L₉-avoiding action by writing that “B acts on P avoiding L₉ of type ( $\cdot$ )”.

Lemma 4.17.

Let B be a batten that acts non-trivially and avoiding L₉ on the p-group P. Then the following hold:

If $C_{P} (B) \neq 1$ , then p = 2.
Either $P = [P, B] \times C_{P} (B)$ , where $[P, B]$ is elementary abelian and $C_{P} (B)$ is cyclic, or $P = [P, B] \times I$ , where I is a group of order at most 2 and $[P, B] ≅ Q_{8}$ .
$C_{P} (B)$ is centralized by every automorphism of P of order coprime to p that leaves $C_{P} (B)$ invariant.

Proof.

If $B ≅ Q_{8}$ , then Definition 4.8 and Lemma 4.9 imply that P is elementary abelian and that B acts irreducibly on it. We conclude that $P = [P, B]$ is elementary abelian and we deduce from Lemma 1.1 that $C_{P} (B) = 1$ . Hence, in this case, all statements of our lemma hold.

Now suppose that B is not nilpotent and that it acts of type (NN). Then Definition 4.16 states that, once more, P is elementary abelian and B acts irreducibly on it. Again we see that $P = [P, B]$ is elementary abelian, and as before all statements hold.

Next we suppose B is not nilpotent and that it acts of type (Cy), or that B is cyclic. In the first case B has a cyclic Sylow subgroup Q that acts on P avoiding L₉ such that $C_{P} (Q) = C_{P} (B)$ and $[P, B] = [P, Q]$ by Definition 4.16. In the second case we set $Q : = B$ .

Then, in both cases, Q is a cyclic group that acts on P avoiding L₉ such that $C_{P} (Q) = C_{P} (B)$ and $[P, B] = [P, Q]$ . If Q acts of type (std), then $P = [P, Q]$ is elementary abelian by Definition 4.6. Again we deduce the statements of our lemma.

Suppose that Q acts of type (cent). Then Definition 4.6 yields that $[P, Q] = [P, B]$ is elementary abelian, that P is abelian and that $C_{P} (Q) = C_{P} (B)$ is a cyclic 2-group. In particular $P = [P, B] \times C_{P} (B)$ by Lemma 1.1. It also follows that $C_{P} (B)$ is centralized by every automorphisms of P of odd order that leaves $C_{P} (B)$ invariant. These are the statements of our lemma.

Finally, suppose that Q acts of type (hamil). Then Definition 4.6 yields that $[P, Q] ≅ Q_{8}$ and $P = [P, Q] \times I$ , where I is a group of order at most 2. In particular statement (a) is true. Moreover, we deduce that $C_{P} (Q) \leq Ω_{1} (P) = Φ ([P, Q]) \times I$ , where $Φ ([P, Q])$ is cyclic of order 2. In particular $Φ ([P, Q])$ is centralized by B and by every automorphisms of P. We conclude that $C_{P} (Q)$ is elementary abelian of order at most 4 and that every automorphism of P centralizes a cyclic subgroup of order 2. This implies (b). □

5 Avoiding L₉

We now work toward a classification of arbitrary L₉-free groups, and therefore we need to understand in more detail the group structures that appear when “L₉ is avoided” in the sense of the previous section.

Definition 5.1.

Suppose that K is a batten group that acts coprimely on the p-group P. We say that the action of K on P avoids L₉ if and only if $[P, K] \neq 1$ and every batten of K either centralizes P or avoids L₉ in its action on P.

Lemma 5.2.

Let K be a batten group that acts coprimely on the p-group P avoiding L₉. Suppose further that $L \leq K$ and $L_{0} E L$ such that $[P, L] \neq 1 = [P, L_{0}]$ . Then $L / L_{0}$ acts on P avoiding L₉. In particular $L / L_{0}$ is a batten group.

Proof.

By induction we may suppose that K is a batten and that either $L_{0} = 1$ and L is a maximal subgroup of K or that L₀ is a minimal normal subgroup of K = L. Thus either $| L_{0} |$ has order q or $| K : L | = q$ . Since $L_{0} \leq C_{K} (P)$ , we first remark that $L / L_{0}$ induces automorphisms on P.

If $K ≅ Q_{8}$ , then K acts faithfully on P by Definition 4.8. Thus $L_{0} \leq C_{K} (P) = 1$ and it follows that L is a cyclic group of order 4. Thus Lemma 4.9 yields that $Ω_{1} (L)$ inverts P and that L acts irreducibly on the elementary abelian group $P = [P, L]$ . Then we see that $L ≅ L / L_{0}$ acts on P avoiding L₉ of type (std).

Next suppose that K is cyclic. Then $L / L_{0}$ is cyclic. If K acts of type (std) on P, then L and every subgroup of L act irreducibly or via inducing power automorphisms on the elementary abelian group $P = [P, K]$ . Since $[P, L] \neq 1$ and power automorphisms are universal, by Lemma 1.5.4 of [Citation7], it follows that $P = [P, L]$ . Moreover, the action of L on P is equivalent to that of $L / L_{0}$ , and then it follows that $L / L_{0}$ acts on P avoiding L₉ of type (std).

If K acts on P of type (cent), then L and all its subgroups act irreducibly or trivially on the elementary abelian group $[P, K]$ . Again the fact that $[P, L] \neq 1$ implies that $P = [P, L]$ , and then $C_{P} (L) = C_{P} (K)$ by Lemma 1.1. Since the action of L on P is equivalent to that of $L / L_{0}$ , it follows that $L / L_{0}$ acts on P avoiding L₉ of type (cent).

We suppose now that K acts of type (hamil). Then K is a cyclic 3-group and $K / C_{K} (P)$ has order 3. It follows that L = K. But again, the action of $L / L_{0} = K / L_{0}$ on P is equivalent to the action of K on P, which means that it has type (hamil).

We finally suppose that K is a non-nilpotent batten. Let R be a Sylow subgroup of K such that $K = B (K) \cdot R$ . Suppose first that $L / L_{0}$ is a q-group. Then our choice of L and L₀ implies that $L / L_{0} ≅ R$ . If K acts on P of type (Cy) in this case, then it follows that $L / L_{0} ≅ R$ is cyclic and that it acts on P avoiding L₉, according to Definition 4.16. Otherwise, if K acts of type (NN), then $B (K) ≰ C_{K} (P)$ and then $L_{0} = 1$ . It follows that L = R acts irreducibly on the elementary abelian group P, whence $P = [P, L] = [P, L / L_{0}]$ . In addition $Φ (L)$ and all of its subgroups induce power automorphism on P. Altogether the cyclic group $L / L_{0} ≅ L$ acts on P of type (std).

Now we suppose that $L / L_{0}$ does not have prime power order. Then $L_{0} \leq Φ (R) = Z (K)$ . Now if $L / L_{0}$ is nilpotent, then $L \neq K$ and therefore $L_{0} = 1$ . It follows from Lemma 2.5 that $L = Z (K) \times B (K)$ . We have already proven that R acts on P avoiding L₉, and then Z(K) also acts on P avoiding L₉. In addition $B (K)$ either centralizes P or it acts irreducibly on the elementary abelian group $P = [P, B (K)]$ . Since $B (K)$ has prime order, it follows that the cyclic group $B (K)$ acts on P avoiding L₉ of type (std). Altogether $L / L_{0} ≅ L = Z (K) \times B (K)$ acts on P avoiding L₉.

Finally, suppose that $L / L_{0}$ is not nilpotent. Then L is not nilpotent and hence Lemma 2.7 implies that L = K. We conclude that $L_{0} \neq 1$ . Since $[P, Z (K)] \neq 1$ by Definition 4.16, it follows that L₀ is a proper subgroup of Z(K) and that $L / L_{0} = K / L_{0} ≅ B (K) ⋊ R / L_{0}$ is a non-nilpotent batten. If $[P, B (K)] = 1$ , then our investigation above imply that the cyclic group $R / L_{0}$ acts on P avoiding L₉, and then $K / L_{0}$ acts on P avoiding L₉. Otherwise $1 \neq [P, B (K)]$ is elementary abelian of order $p^{| K : B (K) Z (K) |} = p^{| K / L_{0} : B (K) Z (K) / L_{0} |}$ , moreover $B (K) ≅ B (K) L_{)} / L_{0}$ and $R / L_{0}$ act irreducibly on P. At the same time $Z (K / L_{0}) = Z (K) / L_{0}$ induces power automorphisms on P. Altogether $K / L_{0}$ acts on P avoiding L₉ of type (NN). □

Lemma 5.3.

Let K be a batten group that acts coprimely on the p-group P avoiding L₉. Then the following assertions are true:

If $L E K$ , then $[P, K] = [P, L]$ or $[P, L] = 1$ .
$[P, K]$ is elementary abelian or isomorphic to Q₈.

Proof.

Let $L E K$ be such that $[P, L] \neq 1$ . Then L is a batten group by Lemma 2.7 and therefore there is a batten B of L such that $[P, B] \neq 1$ . Assume for a contradiction that $[P, B] \leq [P, L] ≨ [P, K] \leq P$ . Then the fact that $P \neq [P, B]$ implies that $C_{P} (B) \neq 1$ by Lemma 1.1. In addition B avoids L₉ in its action on P, by Lemma 5.2. Since B is a batten of L, it is characteristic in L, and therefore $B E K$ . In particular $C_{P} (B)$ is K-invariant and hence it is centralized by K by Lemma 4.17 (c). This implies that $C_{P} (B) = C_{P} (K)$ . Finally $[P, K] = [[P, B] C_{P} (B), K] = [[P, B] C_{P} (K), K] = [[P, B], K] \leq [P, B] ≨ [P, K]$ , which is a contradiction. In particular (a) is true.

Together with Lemma 4.17 (b), the statement in (b) follows from (a). □

Lemma 5.4.

Let K be a batten group that acts on the p-group P avoiding L₉, and suppose that H is a subgroup of K. Then H centralizes P or $[P, H] = [P, K]$ .

Moreover, H induces power automorphisms on P or it acts irreducibly on $[P, K] / Φ ([P, K])$ .

Proof.

Let $H \leq K$ . If H centralizes P, then it induces power automorphisms on P. We may suppose that $[P, H] \neq 1$ . Then H has a q-subgroup Q such that $[P, Q] \neq 1$ . Therefore, if $Q E K$ , then we have that $[P, K] = [P, Q]$ by Lemma 5.3 (a). Then the fact that $[P, Q] \leq [P, H] \leq [P, K]$ yields that $[P, H] = [P, K]$ .

Assume for a contradiction that $[P, K] \neq [P, H]$ . Then Lemma 2.8 implies that K has a non-nilpotent batten B such that $B = B (B) Q$ . We moreover deduce that $[P, Q] ≨ [P, K]$ and $[P, B] = [P, K]$ by Lemma 5.3 (a), because $B E K$ . Since the action of K on P avoids L₉, the action of B also does. If B acts of type (Cy), then we obtain the contradiction that $[P, Q] = [P, B]$ . Thus B acts of type (NN) and in particular Q acts irreducibly on P. But this is impossible as well. It follows that $[P, H] = [P, K]$ .

Assume for a further contradiction that $H \leq K$ neither induces power automorphisms on P nor does it act irreducibly on $[P, K] / Φ ([P, K]) = [P, H] / Φ ([P, H])$ . Then there is a batten B of H that neither induces power automorphisms on P nor does it act irreducibly on $[P, H] / Φ ([P, H])$ . Similarly to the arguments above, we deduce that $[P, K] = [P, H] = [P, B]$ , and Lemma 5.2 gives that B avoids L₉ in its action on P. Therefore Lemma 4.11 yields that B is not nilpotent. From Definition 4.16 we further see that B does not act of type (NN), and thus B acts of type (Cy) on P. Consequently $[P, B (B)] = 1$ and B has a cyclic Sylow subgroup Q such that $B = B (B) Q$ and Q acts on P avoiding L₉. Again we have $[P, K] = [P, B] = [P, Q]$ and Lemma 4.11 gives that Q induces power automorphisms on P or acts irreducibly on $[P, Q] / Φ ([P, Q]) = [P, K] / Φ ([P, K])$ . In the first case $B = C_{B} (P) Q$ induces power automorphism on P and in the second case B acts irreducibly on $[P, K] / Φ ([P, K])$ . This is a contradiction. □

Lemma 5.5.

Let K be a batten group that acts on the p-group P avoiding L₉, and suppose that H is a subgroup of K that acts non-trivially on $R \leq P$ .

Then $C_{H} (R) = C_{H} (P), C_{P} (H) = C_{P} (K)$ and $[P, H] = [P, K]$ .

Proof.

From Lemma 5.4 we see that $[P, H] = [P, K]$ . In addition $C_{P} (K) \leq C_{P} (H) \leq C_{P} (Q)$ for every q-subgroup Q of H and every prime q. Let Q be a q-subgroup of H for some prime q such that $C_{P} (Q) \neq P$ . Then Q is a batten by Lemma 2.7, and it acts on P avoiding L₉ by Lemma 5.2. If $Q E K$ , then K centralizes $C_{P} (Q)$ by Lemma 4.17(c). Thus $C_{P} (K) \leq C_{P} (H) \leq C_{P} (Q) \leq C_{P} (K)$ , and this gives that $C_{P} (K) = C_{P} (H)$ . If Q is not a normal subgroup of K, then Lemma 2.8 provides a non-nilpotent batten B of K such that $B = B (B) Q$ . Since the action of K on P avoids L₉, the action of B also does. If B acts of type (NN), then Q acts irreducibly on P and therefore $C_{P} (Q) = 1 \leq C_{P} (K)$ . Again we deduce that $C_{P} (K) = C_{P} (H)$ . If B acts of type (Cy), then $[P, B (B)] = 1$ and hence $C_{P} (B) = C_{P} (Q)$ . But now B is a normal subgroup of K, and then Lemma 4.17(c) gives that $C_{P} (B) = C_{P} (Q) \leq C_{P} (K)$ . As above we deduce that $C_{P} (K) = C_{P} (H)$ .

Finally, suppose that $R \leq P$ is H-invariant, but not centralized by H, and set $H_{0} : = C_{H} (R) \geq C_{H} (P)$ . Assume for a contradiction that H₀ does not centralize P. Then we deduce, as above, that $C_{P} (K) = C_{P} (H_{0}) \geq R$ . This is a contradiction, because H does not centralize R. □

Corollary 5.6.

Let K be a batten group that acts non-trivially and avoiding L₉ on the p-group P. Then the following hold:

If $C_{P} (K) \neq 1$ , then p = 2.
If $C_{P} (K) = 1$ , then $P = [P, K]$ is elementary abelian.
If K induces power automorphisms on P, then $P = [P, K]$ is elementary abelian of odd order. In particular $C_{P} (K) = 1$ in this case.

Proof.

Let B be a batten of K that does not centralize P. Then Lemma 5.5 implies that $C_{P} (K) = C_{P} (B)$ . Thus Part (a) and (b) of Lemma 4.17 yield the statements (a) and (b) of our lemma. For Part (c) we suppose that K induces power automorphisms on P. Then P is not an elementary abelian 2-group by Lemma 1.3. If $C_{P} (K) = 1$ , then our assertion holds by (b). Otherwise p = 2 by (a), and then Lemma 1.3 implies that $[P, K]$ is neither elementary abelian nor isomorphic to Q₈, contradicting Part (b) of Lemma 5.3.

For the final comment we just use that p is odd and then apply (a). □

Lemma 5.7.

Let B be a batten that acts on the p-group P avoiding L₉. Let $R \leq P$ be B-invariant and $R_{0} \leq C_{P} (B)$ .

Then B avoids L₉ in its action on $R / R_{0}$ .

Proof.

Since B centralizes R₀, the action of B on $R / R_{0}$ is well-defined.

We first suppose that $B ≅ Q_{8}$ . Then Lemma 4.9 yields that B acts irreducibly on P, and then it follows that R = P and $R_{0} = 1$ . Thus our assertion is true in this case.

Next suppose that B is cyclic. If B acts of type (std) on P, then $R_{0} \leq C_{P} (B) = 1$ . If B acts irreducibly on P, then again P = R and there is nothing left to prove. Otherwise B and all of its subgroups induce power automorphisms on P, and hence on $R = [R, B]$ as well. It follows that B also acts of type (std) on $R ≅ R / R_{0}$ .

Suppose now that B acts of type (cent). Then, since B does not centralize R and B acts irreducibly on $[P, B]$ , it follows that $[P, B] \leq R$ . Moreover P is abelian and then we use the fact that $R_{0} \leq C_{P} (B)$ . This gives that $R / R_{0} = [R / R_{0}, B] \times C_{R / R_{0}} (B) ≅ [R, B] \times C_{R} (B) / R_{0}$ , where $C_{R} (B) / R_{0}$ is a cyclic 2-group. Since every subgroup of B that does not centralize P acts irreducibly on $[P, B] ≅ [R / R_{0}, B]$ in this case, there are two possibilities for the action of B on $R ≅ R / R_{0}$ : If $R_{0} \neq C_{R} (B)$ , then B acts of type (cent), and otherwise it acts of type (std).

Suppose now that B acts of type (hamil). Then the cyclic 3-group B acts irreducibly on $[P, B] / Φ ([P, B])$ , by Lemma 5.4, and we see again that $[P, B] \leq R$ . It follows that $R ≅ Q_{8} \times I$ , where I is a group of order at most 2, and then $R_{0} \leq C_{R} (B) \leq Φ ([R, B]) \times I$ . We remark that $[R / R_{0}, B] B / Z ([R / R_{0}, B] B) ≅ [R, B] B / Z ([R, B] B) ≅ [P, B] B / Z ([P, B] B) ≅ {Alt}_{4}$ .

If $R_{0} \cap [P, B] = 1$ , then $R / R_{0} ≅ Q_{8} \times \tilde{J}$ for some group J of order $\frac{| I |}{| R_{0} |}$ . Thus B acts on $R / R_{0}$ of type (hamil) in this case.

Otherwise we have that $R_{0} \geq Φ ([P, B])$ and therefore $R / R_{0}$ is elementary abelian of order 4 or 8. Moreover B acts irreducibly on $[R / R_{0}, B]$ , which is a group of order 4. In addition every proper subgroup of B centralizes $R / R_{0}$ . Consequently, if $R / R_{0} = [R / R_{0}, B]$ , then B acts on $R / R_{0}$ of type (std) or of type (cent).

The final case is that B is not nilpotent, and we suppose that B acts of type (NN) on P. Then Definition 4.16 yields that B acts irreducibly on P. Hence there is nothing left to prove.

Suppose that B acts of type (Cy). Then we choose a Sylow subgroup Q of B such that $B = B (B) Q$ . Then $[R / R_{0}, B (B)] = [R, B (B)] = [P, B (B)] = 1$ and Q acts on P avoiding L₉ in such a way that $Φ (Q) = Z (B)$ does not centralize P. Then Lemma 5.5 yields that $Φ (Q)$ does not centralize R. In particular, we have that $[R / R_{0}, Z (B)] \neq 1$ . In addition Q acts on $R / R_{0}$ avoiding L₉, by our arguments above. Altogether B acts on $R / R_{0}$ avoiding L₉ of type (Cy) in this final case. □

6 The first implication

We now investigate the general case.

Proposition 6.1.

Let G be a finite L₉-free group. Then G = NK, where N is a nilpotent normal Hall-subgroup of G with modular Sylow subgroups and K is a batten group. Moreover, for all $p \in π (N)$ , every batten of K acts on $O_{p} (N)$ avoiding L₉ or it centralizes $O_{p} (N)$ .

Proof.

We first remark that G is L₁₀-free, whence Theorem A of [Citation9] implies that G is soluble. Furthermore, Corollary C of [Citation9] provides normal Hall-subgroups N and M of G such that $N \leq M$ and such that N is nilpotent, M/N is a 2-group and G/M is metacyclic. We choose N as large as possible with these constraints. From Lemma 4.1 we see that $N \neq 1$ . We also have that every Sylow subgroup of N is L₉-free, and hence it is modular by Lemma 4.2. In addition the Schur-Zassenhaus Theorem (see for example 3.3.1. of [Citation5]) provides a complement K of N in G.

(1) If RQ is a non-nilpotent Hall {r, q}-subgroup of K, where R is a normal Sylow r-subgroup of RQ and $Q \in {Syl}_{q} (RQ)$ , then RQ is a batten. For all $p \in π (N)$ , the group RQ centralizes $O_{p} (N)$ or acts on it avoiding L₉.

Proof. If there is some $p \in π (N)$ such that $[O_{p} (N), R] \neq 1$ , then we set $P : = O_{p} (N)$ . Otherwise the maximal choice of N implies that R is not a normal subgroup of K. Then, using the solubility of G, we find a prime $s \in π (K) ∖ {r}$ and a normal s-subgroup T of K such that $[T, R] \neq 1$ (see 5.2.2 of [Citation5]). Then $s \neq q$ because $1 \neq [R, Q] \leq R$ and $1 \neq [T, R] \leq T$ . In this case we set $P : = T$ .

In both cases $p, r$ and q are pairwise different primes and PRQ is a non-nilpotent {p, r, q}-subgroup that satisfies the hypothesis of Proposition 4.15. For this we note that $P E PRQ, P ≰ N_{G} (R)$ and $R ≰ N_{G} (Q)$ . Since $[R, Q] \neq 1$ , the assertion in (1) follows. □

(2) Every Sylow subgroup S of K is a batten, and for all $p \in π (N)$ it is true that S centralizes $O_{p} (N)$ or acts on $O_{p} (N)$ avoiding L₉.

Proof. Let S be a Sylow subgroup of K. If S centralizes N, then the choice of N provides some Sylow subgroup R of K such that RS is not nilpotent. Then (1) implies that RS is a batten, then that S is cyclic and hence that S is a batten.

Let $p \in π (N / C_{N} (S))$ . Then $O_{p} (N) S$ is an L₉-free {p, q}-group for some prime q. Hence Corollary 4.10 implies the assertion. □

(3) K is a batten group.

Proof. Let $1 \neq B \leq K$ be such that there is some $K_{1} \leq K$ such that $K = K_{1} \times B$ , where $(| K_{1} |, | B |) = 1$ and B is not a direct product of nontrivial subgroups of coprime order. In particular B is a Hall subgroup of K. If B is nilpotent, then B is a Sylow q-subgroup of K for some prime q. In this case (2) implies that B is a batten.

Assume for a contradiction that B is not a batten of K. Then B is not nilpotent and therefore (1) yields that $| B |$ is divisible by at least three different primes. Since $B \leq G$ is L₉-free, Lemma 4.1 provides a normal Sylow r-subgroup R of B for some prime $r \in π (B)$ . We remark that R is a normal subgroup of $K = K_{1} \times B$ . In addition B is is not a direct product of non-trivial subgroups with coprime order and hence there are a prime $q \in π (B)$ and some $Q \in {Syl}_{q} (B)$ such that RQ is not nilpotent. Now B is a Hall subgroup of K and thus RQ is a Hall subgroup of K. In particular (1) implies that RQ is a batten and it follows that $| R | = r$ and $1 \neq C_{Q} (R) = Φ (Q)$ . We further see, from Definition 4.16 and (1), that for every $p \in π (N)$ with the property $[O_{p} (N), R] \neq 1$ we have that $| O_{p} (N) | = p^{q}$ . Since R is a normal Sylow subgroup of K, the maximal choice of N provides some $p \in π (N)$ such that R does not centralize $P : = O_{p} (N)$ . In particular we have that $| P | = p^{q}$ .

Let $s \in π (B) ∖ {q, r}$ and let S be a Sylow s-subgroup of B such that QS = SQ. Such a subgroup exists by Satz VI. 2.3 in [Citation3]. If S does not centralize R, then RS is not nilpotent and therefore our arguments above show that $p^{s} = | P | = p^{q}$ . This is impossible because $r \neq s$ . Consequently $[R, S] = 1$ .

Since B is directly indecomposable, we conclude that SQ is not nilpotent. But SQ is a Hall subgroup of B and then it is a Hall subgroup of K. In particular (1) yields that SQ is a batten. From the fact that $1 \neq C_{Q} (R) = Φ (Q)$ we conclude that $| Q | \neq q$ , and then $| S | = s$ and $S E SQ$ . In addition $C_{Q} (S) = Φ (Q) = C_{Q} (R)$ . But $(R \times S) Q$ is an L₉-free group, and this situation contradicts Corollary 4.13. □

We conclude that, by construction, G = NK, where N is a nilpotent normal subgroup of G with modular Sylow subgroups and K is a batten group, by (3), such that for all $p \in π (N)$ it is true that every batten of K centralizes $O_{p} (N)$ or acts on $O_{p} (N)$ avoiding L₉, by (1) and (2). □

The converse of Proposition 6.1 is false, as can be seen in the following example and subsequent lemma.

Example 6.2.

Let $H = C_{19} \times C_{19}, J = C_{5} \times C_{5}$ and let $x, y \in GL (2, 19) \times GL (2, 5)$ be such that $x = ((\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}), (\begin{matrix} 0 & 3 \\ 1 & 0 \end{matrix})) and y = ((\begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix}), (\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix})) .$

Then $xy = ((\begin{matrix} - 4 & 0 \\ 0 & - 4 \end{matrix}), (\begin{matrix} 3 & 1 \\ 2 & 3 \end{matrix})) = yx$ . Hence $G : = (H \times J) ⋊ (〈 x 〉 \times 〈 y 〉)$ is a group.

Moreover, $N : = H \times J$ is a nilpotent normal subgroup of G with modular Sylow subgroups. Since $x^{8} = {(x^{2})}^{4} = {((\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}))}^{4} = {((\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}))}^{2} = 1$

and

$y^{9} = {((\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}), (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}))}^{3}$ , it follows that x and y have coprime order.

Thus $K : = 〈 x 〉 \times 〈 y 〉$ is cyclic, which means that it is a batten group.

We see that x and y induce non-trivial power automorphisms on H. Thus every batten of K acts on $H = O_{19} (N)$ avoiding L₁₉ of type (std). In addition x² and y³ induce power automorphisms on J. Since x and y act irreducibly on J, it follows that every batten of K acts on $J = O_{5} (N)$ avoiding L₁₉ of type (std), too.

Altogether G satisfies the conclusion of Proposition 6.1.

On the other hand we observe that $π (K) = {2, 3} = π (〈 x, y 〉 / 〈 x^{2} 〉) = π (K / C_{K} (H))$ and that $C_{H} (C_{K} (J)) = C_{H} (〈 y^{3} 〉) = 1$ .

If $g \in HJ$ centralizes $〈 x 〉$ or $〈 y 〉$ , then g = 1. Thus the following lemma yields that G is not L₉-free.

Lemma 6.3.

Let H be a non-trivial abelian group where all non-trivial Sylow subgroups are non-cyclic elementary abelian, let L be a cyclic group inducing power automorphism on H such that $π (L) = π (L / C_{L} (H))$ and let $1 \neq J$ be an abelian group admitting L as a group of automorphisms such that the action of L on $O_{p} (J)$ avoids L₉ for every $p \in π (J)$ and such that $(| H |, | J |) = 1$ .

Let $π : = {q \in π (L) | C_{O_{q} (L)} (H) < C_{O_{q} (L)} (J)}$ . Suppose that $C_{H} (C_{L} (J)) = 1$ and, for all $g \in {(H \times J)}^{#}$ , suppose that g centralizes neither $O_{π} (L)$ nor $O_{π'} (L)$ .

Then $(H \times J) ⋊ L$ is not L₉-free.

Proof.

We first set $L_{1} : = O_{π} (L)$ and $L_{2} : = O_{π'} (L)$ . Since none of the groups L₁ nor L₂ is centralized by any element of $HJ ∖ {1} \neq \emptyset$ , it follows that L₁ and L₂ are both non-trivial.

For every odd prime $p \in π (H)$ there is an elementary abelian subgroup H_p of H that has order p². In particular there are elements a_p and b_p of H such that $H_{p} = 〈 a_{p} 〉 \times 〈 b_{p} 〉$ .

We set $a : = \prod_{p \in π (H)} a_{p}$ and $b : = \prod_{p \in π (H)} b_{p}$ .

These elements are well-defined (in the sense that the ordering of the primes does not matter) because H is abelian. For every $p \in π (J)$ , we further see that L acts on $O_{p} (J)$ avoiding L₉. Since J is abelian, we deduce from Lemma 5.3(b) that $[O_{p} (J), L]$ is elementary abelian. In addition L acts irreducibly on $[O_{p} (J), L]$ or it induces power automorphisms on $O_{p} (L)$ , by Lemma 5.4. We choose $x_{p} \in {[O_{p} (J), L]}^{#}$ . Then L acts irreducibly on $P : = 〈 x_{p}^{L} 〉$ . Next we choose $l \in L$ such that $L = 〈 l 〉$ . Then $1 \neq x_{p}^{l} \in P \leq HJ$ and $1 \neq [l, x_{p}] \in P \leq HJ$ . Thus our hypothesis implies that $1 \neq [[l, x_{p}], L_{1}] \leq P$ and $1 \neq [x_{p}^{l}, L_{2}] \leq P$ . Altogether we have that $P = 〈 x_{p}^{L} 〉 = 〈 {(x_{p}^{l})}^{L} 〉 = 〈 {[[l, x_{p}], L_{1}]}^{L} 〉 = 〈 {[x_{p}^{l}, L_{2}]}^{L} 〉$ . Moreover, Lemma 5.5 yields that $C_{L} (x_{p}) = C_{L} (P) = C_{L} (O_{p} (J))$ .

For every $p \in π (L)$ we choose $1 \neq x_{p} \in [O_{p} (J), L]$ , and then we set $x : = \prod_{p \in π (J)} x_{p}$ .

Since J is abelian, it follows that $C_{L} (J) = \underset{p \in π (J)}{\cap} C_{L} (O_{p} (J)) = \underset{p \in π (J)}{\cap} C_{L} (x_{p}) = C_{L} (x)$ . Next we set $y : = x^{l}$ . Then our previous arguments show that $J_{0} : = 〈 x^{L} 〉 = 〈 y^{L} 〉 = 〈 {[[l, x], L_{1}]}^{L} 〉 = 〈 {[y, L_{2}]}^{L} 〉$ (*).

We will construct a subgroup lattice L₉ using Lemma 3.5. For this we set $E : = C_{L} (HJ) E HJL$ and $D : = C_{L} (J)$ .

For every $q \in π'$ we have $C_{O_{q} (L)} (H) \geq C_{O_{q} (L)} (J)$ and so $C_{O_{q} (L)} (J) \leq E$ . This implies that $C_{L_{2}} (J) = C_{L_{2}} (HJ) \leq E$ ( $* *$ ) and that $D = C_{L} (J) = C_{L_{1}} (J)$ . We conclude that $C_{L_{1}} (x) E = (L_{1} \cap C_{L} (x)) E = (L_{1} \cap C_{L} (J)) E = C_{L_{1}} (J) E = DE = D$ .

If $q \in π = π (L_{1})$ , then $C_{O_{q} (L)} (H) < C_{O_{q} (L)} (J) \leq C_{L_{1}} (x)$ and therefore $C_{L_{1}} (H) = C_{L_{1}} (HJ) \leq E$ ( $* * *$ ). Then it follows that $E \cap L_{1} = C_{L_{1}} (HJ) = C_{L_{1}} (H) < C_{L_{1}} (x) \leq D \cap L_{1}$ . In particular $E \neq D$ and hence (L9(i)) of Lemma 3.5 holds.

Next we set $A : = 〈 a 〉 L_{1}^{x} E, S : = L_{1}^{ax} E$ and $T : = L_{1}^{a^{- 1} x} E$ . Then we have that A contains the subgroups S, T and D ( $= C_{L_{1}} (x) E$ ). In addition, if $c \in {a, a^{- 1}, a^{2}}$ , then we see that $〈 c 〉 = 〈 a 〉$ , because o(a) is odd by construction. Since $C_{〈 a 〉} (D) \leq C_{H} (D) = C_{H} (C_{L} (J)) = 1$ , it follows that $〈 a 〉 = 〈 c 〉 = [c, D] \times C_{〈 a 〉} (D) = [c, D]$ , by Lemma 1.1. The group L₁ induces power automorphisms on H, which means that it normalizes $[c, D]$ . Together with Part (d) of Lemma 1.4 we conclude that $〈 D, L_{1}^{c} E 〉 = 〈 {[c, D]}^{L_{1}^{c} E} 〉 L_{1}^{c} E = [c, D] L_{1}^{c} E = 〈 c 〉 L_{1} E = A^{x^{- 1}} .$

In particular, since D centralizes x, it follows that $〈 D, T 〉 = 〈 D, S 〉 = A$ . Moreover, we have that $A \geq 〈 T, S 〉 = {〈 L_{1}, L_{1}^{a^{2}} E 〉}^{a^{- 1} x} \geq {〈 D, L_{1}^{a^{2}} E 〉}^{a^{- 1} x} = {(A^{x^{- 1}})}^{a^{- 1} x} = A$ and therefore we conclude that $〈 S, T 〉 = A$ as well. Next Lemma 5.5 gives that $C_{L_{1}} (a) = C_{L_{1}} (H) \leq E$ by ( $* * *$ ). Furthermore $L_{1}^{c}$ is a π-group for all $c \in {a, a^{- 1}, a^{2}}$ , whence $L_{1} E \cap L_{1}^{c} \leq O_{π} (L_{1} E) \cap L_{1}^{c} = L_{1} \cap L_{1}^{c} = C_{L_{1}} (c) = C_{L_{1}} (a) \leq E$ by Part (b) of Lemma 1.4. Altogether Dedekind’s modular law gives that $L_{1} E \cap L_{1}^{c} E = (L_{1} E \cap L_{1}^{c}) E \leq E$ for all $c \in {a, a^{- 1}, a^{2}}$ . We conclude that $T \cap S = L_{1}^{ax} E \cap L_{1}^{a^{- 1} x} E \leq E^{x} = E$ , that $D \cap T \leq {(L_{1} E \cap L_{1}^{a^{- 1}} E)}^{x} \leq E^{x} = E$ and that $D \cap S \leq {(L_{1} E \cap L_{1}^{a} E)}^{x} \leq E$ . With all these properties, we see that (L9(ii)) of Lemma 3.5 is true.

Now we set $C : = 〈 b 〉 D L_{2}$ and $U : = 〈 b 〉 L_{2} E$ . Then $C = 〈 D, U 〉$ and $D \cap U = C_{L} (J) \cap 〈 b 〉 L_{2} E = (C_{L} (J) \cap 〈 b 〉 L_{2}) E = C_{L_{2}} (J) E$ by Dedekind’s modular law and by Part (b) of Lemma 1.4. Hence ( $* *$ ) implies that (L9(iii)) of Lemma 3.5 holds.

Let $c \in {a, a^{- 1}}$ and let $X : = 〈 L_{1}^{cx}, L_{2} 〉$ . Then X contains a π-Hall subgroup as well as a $π (L_{2})$ -Hall subgroup of HJL. Since HJL is soluble, there is a $π (L)$ -Hall subgroup K of X such that $L_{2} \leq K$ and some $g \in HJ$ such that $L^{g} = K$ . It follows that $L_{2} \leq K \cap L = L^{g} \cap L = C_{L} (g)$ by Lemma 1.4(b). The hypothesis of our lemma yields that g = 1 and therefore $L = K \leq X$ . From there we obtain some $h \in X$ such that $L_{1}^{h} = L_{1}^{cx}$ and hence $L_{1} = L_{1} \cap L_{1}^{h x^{- 1} c^{- 1}} \leq C_{L} (h x^{- 1} c^{- 1})$ by Lemma 1.4(b). This forces $cx = h \in X$ , and then $c, x \in X$ , because H and J have coprime order and centralize each other. Altogether $〈 c 〉 = 〈 a 〉$ and $〈 x^{L} 〉 = J_{0}$ are subgroups of X, and we conclude that $X = 〈 a 〉 J_{0} L$ . Thus $〈 U, T 〉 = 〈 L_{1}^{a^{- 1} x}, 〈 b 〉 L_{2} E 〉 = 〈 b 〉〈 L_{1}^{a^{- 1} x}, L_{2}, E 〉 = 〈 b 〉 XE = 〈 a, b 〉 J_{0} L = 〈 b 〉〈 L_{1}^{ax}, L_{2}, E 〉 = 〈 U, L 〉 .$

We set $F : = 〈 a, b 〉 J_{0} L$ in order to obtain Part (L9(iv)) of Lemma 3.5. Moreover, Dedekind’s law and Part (a) of Lemma 1.4 gives that $\begin{matrix} A \cap C = 〈 a 〉 L_{1}^{x} E \cap 〈 b 〉 D L_{2} = (〈 a 〉 L_{1}^{x} E \cap 〈 b 〉 L_{2}) D \\ = (〈 a 〉 {(L_{1} E \cap L_{2})}^{x} \cap 〈 b 〉 (L_{1} E \cap L_{2})) D = (〈 a 〉 C_{L_{2}} {(HJ)}^{x} \cap 〈 b 〉 C_{L_{2}} (HJ)) D \\ = (〈 a 〉 C_{L_{2}} (HJ) \cap 〈 b 〉 C_{L_{2}} (HJ)) D = (〈 a 〉 \cap 〈 b 〉 C_{L_{2}} (HJ)) C_{L_{2}} (HJ) D = C_{L_{2}} (HJ) D = D . \end{matrix}$

We set $B : = 〈 ab 〉 L^{y}$ . Then $A \cap B = {(〈 a 〉 L_{1} E \cap 〈 ab 〉 L^{y x^{- 1}})}^{x} \leq {(〈 a 〉 (L_{1} E \cap H L^{y x^{- 1}}))}^{x} \leq {(〈 a 〉 C_{L_{1} E} (y x^{- 1}))}^{x} \leq {(〈 a 〉 C_{L} (J))}^{x} = 〈 a 〉 D$ by Lemma 1.4(b), because $〈 {(y x^{- 1})}^{L} 〉 \cap H \leq J \cap H = 1$ .

In a similar way we obtain that $A \cap B \leq 〈 ab 〉 D$ and therefore $D \leq A \cap B \leq 〈 a 〉 D \cap 〈 ab 〉 D = (〈 a 〉 \cap 〈 ab 〉 D) D = D$ .

We further calculate that $B \cap C = 〈 ab 〉 L^{y} \cap 〈 b 〉 D L_{2} \leq 〈 ab 〉 (L^{y} \cap {HDL}_{2}) \leq 〈 ab 〉 C_{D L_{2}} (y) \leq 〈 ab 〉 C_{L} (J) = 〈 ab 〉 D$ and similarly $B \cap C \leq 〈 b 〉 D$ . Therefore $D \leq B \cap C \leq 〈 ab 〉 D \cap 〈 b 〉 D = (〈 ab 〉 \cap 〈 b 〉 D) D = D$ .

Finally (*) and Part (d) of Lemma 1.4 yield that $\begin{matrix} 〈 A, B 〉 = 〈〈 a 〉 L_{1}^{x} E, 〈 ab 〉 L^{y} 〉 = 〈 a, b 〉〈 L_{1}^{x} E, L^{y} 〉 = 〈 a, b 〉〈 {[y x^{- 1}, L_{1}]}^{L} 〉 L = 〈 a, b 〉〈 {[[l, x], L_{1}]}^{L} 〉 L \\ = 〈 a, b 〉 J_{0} L = F = 〈 a, b 〉〈 {[y, L_{2}]}^{L} 〉 L = 〈 a, b 〉〈 L^{y}, D L_{2} 〉 = 〈〈 ab 〉 L^{y}, 〈 b 〉 D L_{2} 〉 = 〈 B, C 〉 . \end{matrix}$

Altogether ${A, B, C, D, E, F, S, T, U}$ satisfies every condition of Lemma 3.5, which means that it is isomorphic to L₉. □

The previous lemma and Lemma 4.12 motivate the following definition:

Definition 6.4.

Here we define a class $L$ of finite groups, and each group in $L$ has a type.

We say that $G \in L$ has type (N, K) if and only if the following hold:

( $L$ 1) $G = N ⋊ K$ , where N is a normal nilpotent Hall subgroup of G with modular Sylow subgroups and K is a batten group.

( $L$ 2)If $p \in π (N)$ , then every batten of K centralizes $O_{p} (N)$ or it acts on it avoiding L₉.

( $L$ 3)For all Sylow subgroups Q of K and all distinct Sylow subgroups P and R of N that are not centralized by Q, we have that $C_{Q} (P) \neq C_{Q} (R)$ .

( $L$ 4)Suppose that $H \leq N$ is abelian, that its nontrivial Sylow subgroups are not cyclic and that $L \leq {Pot}_{K} (H)$ is cyclic and such that $π (L) = π (L / C_{L} (H))$ . Let $1 \neq J \leq N$ be L-invariant and abelian, suppose that $(| H |, | J |) = 1, [H, J] = 1$ and $C_{H} (C_{L} (J)) = 1$ , and set $π : = {q \in π (L) | C_{O_{q} (L)} (H) < C_{O_{q} (L)} (J)}$ .

Then there is some $g \in {(HJ)}^{#}$ that centralizes $O_{π} (L)$ or $O_{π'} (L)$ .

Theorem 6.5.

Let G be a finite L₉-free group. Then $G \in L$ .

Proof.

From Lemma 6.1 we see that G = NK and that ( $L$ 1) and ( $L$ 2) are satisfied.

For ( $L$ 3) we let Q be a Sylow subgroup of K and we let P and R be distinct Sylow subgroups of N that are not centralized by Q. Since N is a nilpotent normal Hall subgroup of G, it follows that $[P, R] = 1$ and that Q normalizes P and R. Then $(P \times R) Q$ is directly indecomposable and L₉-free, whence Corollary 4.13 gives that $C_{Q} (P) \neq C_{Q} (R)$ .

Finally, we look at ( $L$ 4) and we assume that it is not true. Then there is an abelian subgroup H of N such that the nontrivial Sylow subgroups are not cyclic, and we find a cyclic group $L \leq {Pot}_{K} (H)$ such that $π (L) = π (L / C_{L} (H))$ and a nontrivial L-invariant abelian subgroup J of N such that $(| H |, | J |) = 1, [H, J] = 1$ and $C_{H} (C_{L} (J)) = 1$ . Let $π : = {q \in π (L) | C_{O_{q} (L)} (H) < C_{O_{q} (L)} (J)}$ . Then we have, for all $g \in {(HJ)}^{#}$ , that g centralizes neither $O_{π} (L)$ nor $O_{π'} (L)$ for $π : = {q \in π (L) | C_{O_{q} (L)} (H) < C_{O_{q} (L)} (J)}$ .

We note that P does not centralize $O_{π} (L) \leq L$ . Then we find a prime $q \in π (L)$ such that a Sylow q-subgroup Q of L does not centralize P. Using Lemma 5.2, we see that Q acts on $O_{p} (N)$ avoiding L₉ and then Lemma 5.7 yields that Q acts non-trivially on P, and avoiding L₉. Now we may apply Corollary 5.6: Since L induces power automorphisms on P, Part (c) shows that P is elementary abelian. Then the hypotheses of Lemma 6.3 are satisfied. It says that $(H \times J) ⋊ L$ is not L₉-free, which is false. We conclude that ( $L$ 4) holds. □

7 The class $L$

Lemma 7.1.

All groups in the class $L$ are soluble.

Proof.

Let $G \in L$ be of type (N, K). Then N is nilpotent normal Hall subgroup of G and $G / N ≅ K$ is a direct product of p-groups or of groups whose order is divisible by exactly two primes. Thus G/N is soluble as well, and it follows that G is soluble. □

Lemma 7.2.

Let $G \in L$ be of type (N, K) and $π : = π ([N, K])$ . Then every subgroup of $O_{π} (N)$ is normal in N.

Proof.

Let U be a subgroup of $O_{π} (N)$ . Then $U E N$ if and only if $O_{p} (U) E O_{p} (N)$ for all $p \in π$ , since N is nilpotent. Let $p \in π$ . We note that this implies that $O_{p} (N)$ is not centralized by K. In particular there is a batten of K that acts non-trivially on $O_{p} (N)$ and avoiding L₉. If $O_{p} (N)$ is abelian, then $O_{p} (U) E O_{p} (N)$ . If $O_{p} (N)$ is not abelian, then we apply Lemma 4.17 (b) to a batten B of K that acts non-trivially on $O_{p} (N)$ . The first possibility described there implies that $O_{p} (N)$ is abelian, which is not the case here. Thus the second possibility holds, and then $O_{p} (N) ≅ Q_{8} \times I$ , where I is cyclic of order at most 2. We conclude that $O_{p} (N)$ is hamiltonian and it follows that $O_{p} (U) \ unlhd O_{p} (N)$ . □

Lemma 7.3.

Let $G \in L$ be of type (N, K) and $X \leq G$ . Then there is some $x \in [N, K]$ such that $X = (N \cap X) (K^{x} \cap X)$ .

Proof.

We set $M : = N \cap X$ . Then M is a normal Hall subgroup of X, because N is one of G. Then the Schur-Zassenhaus Theorem provides a complement C of M in X, and we notice that C and M have coprime orders. Therefore $π (C) = π (X) ∖ π (M) \subseteq π (G) ∖ π (N) = π (K)$ . It follows that C is contained in a complement for N in G. Since G is soluble by Lemma 7.1, such a complement is conjugate to K, and thus we find $g \in G$ such that $C \leq K^{g}$ . The coprime action of K on N yields, together with Lemma 1.1, that $N = C_{N} (K) [N, K]$ , and therefore $G = KN = K C_{N} (K) [N, K]$ . We notice that $[N, K] E G$ and we let $x \in [N, K]$ and $y \in K C_{N} (K) \leq N_{G} (K)$ be such that g = yx. Then $C = K^{g} \cap X = K^{x} \cap X$ and hence $X = MC = (N \cap C) (K^{x} \cap X)$ . □

Lemma 7.4.

Let $G \in L$ be of type (N, K) and suppose that $U \leq G$ . Then U is a group in class $L$ of type $(U \cap N, U \cap K^{g})$ for some $g \in [N, K]$ .

Proof.

Lemma 7.3 provides some $g \in [N, K]$ such that $U = (U \cap N) \cdot (U \cap K^{g})$ . By conjugation we may suppose that g = 1 and we set $K_{1} : = U \cap K$ . Then Lemma 2.7 yields that $K_{1} = U \cap K^{g} \leq K^{g} ≅ K$ is a batten group. Moreover, $M : = U \cap N \leq N$ is a normal nilpotent Hall subgroup of U with modular Sylow subgroups, by ( $L$ 1). This means that ( $L$ 1) holds for U, and now we turn to ( $L$ 2) and let $p \in π (M)$ . Suppose that B is a batten of K₁ that does not centralize $O_{p} (M)$ . Then it does not centralize $O_{p} (N)$ and therefore Lemma 5.2 implies that $B ≅ B / 1$ acts on $O_{p} (N)$ avoiding L₉. Then we may apply Lemma 5.7 to see that B also acts on $O_{p} (M) / 1 ≅ O_{p} (M)$ avoiding L₉.

This gives property $(L 2)$ of Definition 6.4 for U, and $(L 4)$ follows because $M \leq N$ and $K_{1} \leq K$ .

For ( $L$ 3) we let Q₁ be a Sylow subgroup of K₁ and we let $p, r \in π (M)$ be different primes such that $[O_{p} (M), Q_{1}] \neq 1 \neq [O_{r} (M), Q_{1}]$ . We need to prove that $C_{Q_{1}} (O_{p} (M)) \neq C_{Q_{1}} (O_{r} (M))$ .

First we let Q be a Sylow subgroup of K that contains Q₁. Then $[O_{p} (N), Q] \neq 1 \neq [O_{r} (N), Q]$ and therefore $C_{Q} (O_{p} (N)) \neq C_{Q} (O_{r} (N))$ , using Property ( $L$ 3) for G. In particular, these centralizers cannot both be trivial, and we may suppose that $C_{Q} (O_{p} (N)) \neq 1$ . Then Q does not act faithfully on $O_{p} (N)$ , but the action of Q on $O_{p} (N)$ avoids L₉. Definition 4.8 immediately gives that $Q ≅ Q_{8}$ . Then it follows that Q is cyclic, which means that the subgroup lattice L(Q) of Q is a chain, and Q₁ is also cyclic.

We assume for a contradiction that $C_{Q_{1}} (O_{p} (M)) = C_{Q_{1}} (O_{r} (M))$ . Then Lemma 5.5, with Q₁ in the role of H, $O_{p} (M)$ in the role of R and $O_{p} (N)$ in the role of P, gives that $C_{Q_{1}} (O_{p} (M)) = C_{Q_{1}} (O_{p} (N))$ . Similarly $C_{Q_{1}} (O_{r} (M)) = C_{Q_{1}} (O_{r} (N))$ , and then it follows that $C_{Q_{1}} (O_{p} (N)) = C_{Q_{1}} (O_{r} (N))$ . We recall that $C_{Q} (O_{p} (N)) \neq C_{Q} (O_{r} (N))$ and that Q is cyclic, and now we may suppose that $C_{Q} (O_{p} (N)) ≨ C_{Q} (O_{r} (N))$ . This forces $C_{Q_{1}} (O_{p} (N)) ≨ C_{Q} (O_{p} (N))$ , which is impossible. Thus $C_{Q_{1}} (O_{p} (M)) \neq C_{Q_{1}} (O_{r} (M))$ and ( $L$ 3) holds for U as well. □

Lemma 7.5.

Let $G \in L$ be of type (N, K) and suppose that S is a normal Sylow q-subgroup of K that centralizes N for some prime q. Let K₁ be a Hall $q'$ -subgroup of K.

Then G is also of type $(N \times S, K_{1})$ . In particular, if we choose (N, K) such that $| N |$ is as large as possible, then $π (K) = π (K / C_{K} (N))$ .

Proof.

We show that $G = (N \times S) K_{1}$ satisfies ( $L 1$ )–( $L 4$ ) of Definition 6.4, and we first note that K₁ is a batten group by Lemma 2.7. The structure of K forces all Sylow subgroups of K to be cyclic or quaternion, more specifically S is cyclic or isomorphic to Q₈. This means that S is modular. Since S is a normal Sylow q-subgroup of K and N is a Hall subgroup of G, by hypothesis, it follows that N × S is a Hall subgroup of G where all Sylow subgroups are modular.

By hypothesis $[N, S] = 1$ and N is nilpotent, hence N × S is nilpotent, too. This is ( $L 1$ ).

For ( $L 2$ ) we let B be a batten of K₁ and $p \in π (N \times S)$ . We keep in mind that B is not necessarily a batten of K – if it is, then it centralizes $O_{p} (N \times S)$ or it acts on it avoiding L₉, because of ( $L 2$ ) for G.

Now we suppose that B is not a batten of K and that $[O_{p} (N \times S), B] \neq 1$ . Then SB is a non-nilpotent batten of K. If $p \neq q$ , then SB acts on $O_{p} (N)$ avoiding L₉, and $[O_{p} (N), S] = 1$ . Then Definition 4.16 implies that SB acts of type (Cy) and it follows that B acts on $O_{p} (N)$ avoiding L₉. Finally suppose that q = p. Then $Φ (B)$ centralizes $S = O_{p} (NS)$ , while B induces power automorphisms on the cyclic group S of order p. Thus SB satisfies (std) of Definition 4.6, and we deduce that B acts on S avoiding L₉.

We turn to ( $L 3$ ). Let Q be a Sylow subgroup of K₁ and let P and R be distinct Sylow subgroups of N × S that are not centralized by Q. First we note that Q is a Sylow subgroup of K because K₁ is a Hall subgroup of K by hypothesis. Therefore, if $PR \leq N$ , then we immediately have that $C_{Q} (R) \neq C_{Q} (P)$ , by ( $L 3$ ) in G.

Without loss suppose that $R ≰ N$ , i.e. R = S. Then we recall that Q was chosen not to centralize P and R = S, which means that Q and S cannot come from distinct battens of K, but their product must be a non-nilpotent batten of K. Moreover, $[P, QS] \neq 1$ . We obtain from Lemma 2.5 and Definition 4.16 that $Φ (Q) = Z (Q) = C_{Q} (S)$ does not centralize P and therefore $C_{Q} (R) \neq C_{Q} (P)$ . This is ( $L 3$ ).

Finally, let $H \leq N \times S$ be such that its nontrivial Sylow subgroups are not cyclic, let $L \leq {Pot}_{K_{1}} (H)$ be cyclic and such that $π (L) = π (L / C_{L} (H))$ and let $1 \neq J$ be an L-invariant abelian subgroup of M such that $(| H |, | J |) = 1$ , $[H, J] = 1$ and $C_{H} (C_{L} (J)) = 1$ .

As K is a batten group, the set $π (K / C_{K} (S))$ contains at most one element. We recall that $L \leq K_{1} \leq K$ , and then it follows that, for every set of primes π, the group S centralizes $O_{π} (L)$ or $O_{π'} (L)$ . In order to prove ( $L 4$ ) of Definition 6.4, we may thus suppose that H and J are subgroups of N.

Then $L \leq K_{1} \leq K$ shows that $L \leq {Pot}_{K} (H)$ . Hence we apply ( $L 4$ ) of Definition 6.4 to G, i.e. to the type (N, K). If $π : = {q \in π (L) | C_{O_{q} (L)} (H) < C_{O_{q} (L)} (J)}$ , then we find some $g \in {(HJ)}^{#}$ that centralizes $O_{π} (L)$ or $O_{π'} (L)$ .

Altogether, $G = (N \times S) K_{1}$ satisfies Definition 6.4.

We now suppose that $| N |$ is as large as possible and we assume for a contradiction that $S \leq C_{K} (N)$ is a Sylow subgroup of K. Then S is not normal in K, hence there is a non-nilpotent batten B of K such that $B = B (B) S$ . For all $p \in π (N)$ it follows that $[O_{p} (N), Z (B)] \leq [O_{p} (N), S] = 1$ , by Lemma 2.5. Then Definition 4.16 yields that B does not act on $O_{p} (N)$ avoiding L₉, whence B centralizes $O_{p} (N)$ . But now $B (B) \leq C_{K} (N)$ and $B (B)$ is a normal Sylow subgroup of K. This contradicts the maximal choice of N. □

Lemma 7.6.

Let $G \in L$ and suppose that $M E G$ . Then $G / M \in L$ .

Proof.

By induction we may suppose that M is a minimal normal subgroup of G. We recall that G is soluble by Lemma 7.1, and we let r be prime such that M is an elementary abelian r-group. If M has a complement C in G, then $G / M ≅ C$ and Lemma 7.4 gives that $C \in L$ and hence $G / M \in L$ .

Consequently we may suppose that M does not have a complement in G. We choose $N, K \leq G$ such that G has type (N, K) and such that $| N |$ is as large as possible. Then $π (K) = π (K / C_{K} (N))$ by Lemma 7.5.

First suppose that $r \in π (K)$ . Then $M \leq K$ and we see that $[N, M] \leq N \cap M \leq N \cap K = 1$ , because N is a Hall subgroup of G. Hence $M \leq C_{K} (N)$ . Next we let B be a batten of K that contains M. Then $M \leq C_{B} (N)$ , which means that for all $p \in π (N)$ , B does not act faithfully on $O_{p} (N)$ .

Since there is some $p \in π (N)$ such that B acts on $O_{p} (N)$ avoiding L₉, it follows from Definition 4.8 that B is not isomorphic to Q₈. In addition $M ≨ Z (B)$ , if B is not nilpotent, by Definition 5.1. Therefore, in this case, the section B/M is a non-nilpotent batten as well. We conclude that K/M is a batten group.

Assume for a contradiction that $r \in π (N)$ , but that $M ≰ C_{N} (K)$ . Then we note that $C_{M} (K) E G$ by Lemma 7.2, and we deduce that $C_{M} (K) = 1$ , because M is a minimal normal subgroup of G. This forces $M \leq [O_{r} (N), K]$ . Then $[O_{r} (N), K]$ is not elementary abelian, because otherwise M would have a complement in this commutator and hence in G. But we are working under the hypothesis that it does not.

Now Lemmas 5.4 and 4.17(b) imply that $M \leq [O_{r} (N), K] ≅ Q_{8}$ . But now $Z ([O_{r} (N), K])$ is the unique subgroup of order 2 of $[O_{r} (N), K]$ , which means that it must be centralized by K and contained in M. But this is a contradiction. Thus $M \leq C_{N} (K)$ and Corollary 5.6 implies that p = 2. We summarise that $M \leq C_{K} (N)$ and that K/M is a batten group or $M \leq C_{N} (K)$ and p = 2.

Let $- : G \to G / M$ be the natural homomorphism. Then $\bar{G} = \bar{N} \cdot \bar{K}$ , where $\bar{N}$ is a normal nilpotent Hall subgroup of G with modular Sylow subgroups, since sections of modular p-subgroups are modular. Moreover $\bar{K}$ is a batten group. Thus $\bar{G}$ satisfies $(L 1)$ of Definition 6.4. We further deduce $(L 2)$ from Lemmas 5.2 and 5.7.

For $(L 3)$ we let Q be a Sylow subgroup of K and we let $p, s \in π (\bar{N})$ be distinct primes such that $[O_{s} (\bar{N}), \bar{Q}] \neq [O_{p} (\bar{N}), \bar{Q}]$ . Then $[O_{s} (N), Q] \neq [O_{p} (N), Q]$ and therefore $C_{Q} (O_{s} (N)) \neq C_{Q} (O_{p} (N))$ . Since $M \leq C_{N} (Q)$ or $M \leq C_{K} (O_{s} (N)) \cap C_{K} (O_{p} (N))$ , it follows that $C_{\bar{Q}} (O_{s} (\bar{N})) \neq C_{\bar{Q}} (O_{p} (\bar{N}))$ .

We finally let $1 \neq \bar{H} \leq \bar{N}$ be abelian with non-cyclic Sylow subgroups and $\bar{L} \leq {Pot}_{\bar{K}} (\bar{H})$ be cyclic with $π (\bar{L}) = π (\bar{L} / C_{\bar{L}} (\bar{H}))$ and we let $\bar{J}$ be an abelian $\bar{L}$ -invariant subgroup of $\bar{N}$ such that $(| \bar{H} |, | \bar{J} |) = 1$ and $C_{\bar{H}} (C_{\bar{L}} (\bar{J})) = 1$ . We set $\bar{π} : = {q \in π (\bar{L}) | \forall \bar{Q} \in {Syl}_{q} (\bar{L}) : C_{\bar{Q}} (\bar{H}) < C_{\bar{Q}} (\bar{J})}$ .

Then we assume for a contradiction that every nontrivial element $\bar{g} \in \bar{H} \bar{J}$ centralizes neither $O_{π} (\bar{L})$ nor $O_{π'} (\bar{L})$ . Then Lemma 1.1 yields that $[\bar{H} \bar{J}, \bar{L}] = \bar{H} \bar{J}$ . We choose pre-images H, L and J in G of smallest possible order. Then $HJ = [HJ, L]$ and $π (\bar{X}) = π (X)$ for all $X \in {H, L, J}$ , because G is soluble by Lemma 7.1. In particular we have that $(| H |, | J |) = 1$ .

If $r \in π (X)$ for some $X \in {H, J}$ , then r = 2. Then our assumption implies that $C_{O_{2} (X)} (L) \leq M$ . It follows that $X ≅ \bar{X}$ or that $M \leq Φ (X) = Φ ([X, L])$ . In the second case, we apply Lemmas 5.4 and 4.17. Together they show that $[O_{2} (N), K] = [O_{2} (N), L] ≅ Q_{8}$ and thus $π (L / C_{O_{2} (N)} (L)) = {3}$ . This means that $O_{2} (N)$ centralizes $O_{π} (L)$ or $O_{π'} (L)$ . But then we also have that $[O_{2} (\bar{X}), O_{σ} (\bar{L})] = 1$ for some $σ \in {π, π'}$ , which is a contradiction. We deduce that $\bar{H} ≅ H$ and $\bar{J} ≅ J$ . In particular $H \leq N$ is abelian, with non-cyclic Sylow subgroups, and $J \neq 1$ is an abelian L-invariant subgroup of N. Since $\bar{L}$ is cyclic, it follows from our arguments above that L is also cyclic. Moreover $π (L) = π (\bar{L}) = π (\bar{L} / C_{\bar{L}} (\bar{H})) = π (L / C_{L} (H))$ , because $M \leq C_{K} (N)$ or $M \cap L = 1$ .

We now investigate the action of L on H. Since $\bar{H} ≅ H$ and $M \cap L = 1$ or $M \leq C_{L} (H)$ , we see that L induces power automorphisms on H. In addition Lemma 1.1 shows that $\bar{C_{H} (C_{L} (J))} ≅ C_{\bar{H}} (\bar{C_{L} (J)}) ≅ C_{\bar{H}} (C_{\bar{L}} (\bar{J})) = 1$ and then the fact that $H \cap M = 1$ yields that $C_{H} (C_{L} (J)) = 1$ . Altogether we obtain, by applying ( $L 4$ ) to G, some $g \in H J^{#}$ such that g centralizes $O_{π} (L)$ or $O_{π'} (L)$ , where $π : = {q \in π (L) | \forall Q \in {Syl}_{q} (L) : C_{Q} (H) < C_{Q} (J)}$ . Since $\bar{H} \bar{J} ≅ HJ$ , it follows that $\bar{g} \neq 1$ and $[O_{π} (\bar{L}), \bar{g}] = 1$ or $[O_{π'} (\bar{L}), \bar{g}] = 1$ . Again we use that $\bar{L}$ acts on $\bar{H} ≅ H$ and $\bar{J} ≅ L$ equivalently to L, because $M \cap L \leq C_{L} (N)$ . Then we see that $\bar{π} : = {q \in π (\bar{L}) | \forall \bar{Q} \in {Syl}_{q} (\bar{L}) : C_{\bar{Q}} (\bar{H}) < C_{\bar{Q}} (\bar{J})} = π .$

This is a contradiction. □

Lemma 7.7.

Let $G \in L$ be of type (N, K) such that $C_{K} (N) = 1$ , let $q \in π (K)$ and let $Q \in {Syl}_{q} (K)$ . Then $1 \neq [N, Ω_{1} (Q)]$ has prime power order.

Proof.

We apply Lemma 7.4 and we see that $N Ω_{1} (Q) \in L$ has type $(N, Ω_{1} (Q))$ . Since $Ω_{1} (Q)$ does not centralize N, there is a prime $p \in π (N)$ such that $[O_{p} (N), Ω_{1} (Q)] \neq 1$ . It follows that $C_{Ω_{1} (Q)} (O_{p} (N)) = 1$ . Now $(L 3)$ implies that $Ω_{1} (Q)$ centralizes $O_{r} (N)$ for every $r \in π (N) ∖ {p}$ , and this shows that $1 \neq [N, Ω_{1} (Q)] \leq [O_{p} (N), Ω_{1} (Q)] \leq O_{p} (N)$ . □

Lemma 7.8.

Let $G \in L$ be of type (N, K) such that $C_{K} (N) = 1$ , let $q \in π (K)$ and let $Q \in {Syl}_{q} (K)$ . Let $p \in π (N)$ be such that $Ω_{1} (Q)$ does not centralize $P : = O_{p} (N)$ . Then the following hold:

$N_{G} (Ω_{1} (Q)) = (O_{p'} (N) K) \times C_{P} (K)$ .
$G = [P, Ω_{1} (Q)] N_{G} (Ω_{1} (Q))$ .
$[P, Ω_{1} (Q)] = [P, K]$ acts transitively on $Ω_{1} {(Q)}^{G} = {Q_{0} \leq G | | Q_{0} | = q}$ .
If $X \leq G$ , then there is some $x \in [P, Ω_{1} (Q)]$ such that $X = (X \cap P) N_{X} (Ω_{1} {(Q)}^{x})$ .
Suppose that $X \leq G$ , that q divides $| X |$ and that $x \in P$ . Then $X = (X \cap P) N_{X} (Ω_{1} {(Q)}^{x})$ if and only if $Ω_{1} {(Q)}^{x} \leq X$ .

Proof.

We set $P_{0} : = [P, Ω_{1} (Q)]$ . Then Lemma 7.7 implies that $P_{0} = [N, Ω_{1} (Q)]$ and so $O_{p'} (N) \leq C_{G} (Ω_{1} (Q)) \leq N_{G} (Ω_{1} (Q))$ . Furthermore K acts on P avoiding L₉ and then we have that $P_{0} = [P, Q] = [P, K]$ by Lemma 5.4. Next $K \leq N_{G} (Ω_{1} (Q))$ from Lemma 2.6.

Since N is nilpotent, we conclude that $N_{G} (Ω_{1} (Q)) = (O_{p'} (N) \times N_{P} (Ω_{1} (Q))) K$ . But we also have that $[N_{P} (Ω_{1} (Q)), Ω_{1} (Q)] \leq P \cap Ω_{1} (Q) = 1$ , whence $N_{P} (Ω_{1} (Q)) \leq C_{P} (Ω_{1} (Q)) = C_{P} (K)$ by Lemma 5.5. Consequently $N_{P} (Ω_{1} (Q)) = C_{P} (Ω_{1} (Q))$ and it follows that $N_{G} (Ω_{1} (Q)) = (O_{p'} (N) K) \times C_{P} (K)$ , as stated in (a).

For (b) we recall that, by (a), the subgroups K and $O_{p'} (N)$ normalize $Ω_{1} (Q)$ . Then $G = P O_{p'} (N) K \leq P N_{G} (Q) \leq G$ . Moreover $P E G$ and Lemma 1.1 implies that $P = C_{P} (Ω_{1} (Q)) P_{0}$ , where $C_{P} (Ω_{1} (Q)) \leq N_{G} (Ω_{1} (Q))$ and therefore $G = P_{0} N_{G} (Ω_{1} (Q))$ as stated in (b).

From there we deduce that P₀ acts transitively on $Ω_{1} {(Q)}^{G}$ by conjugation. The second statement of (c) follows because G is soluble (Lemma 7.1), together with the fact that $Ω_{1} (Q)$ is the unique subgroup of its order in the Hall subgroup K of G (Lemma 2.6). This means that every subgroup of order q of G is conjugate to $Ω_{1} (Q)$ , completing (c).

For (d) and (e) we let $X \leq G$ . Lemma 7.3 provides some $g \in G$ such that $X = (N \cap X) (K^{g} \cap X)$ . Moreover, (a) implies that K^g and $O_{p'} (N) (= O_{p'} {(N)}^{g})$ normalize $Ω_{1} {(Q)}^{g}$ , and then we summarise: $X = (N \cap X) (K^{g} \cap X) \leq (P \cap X) (O_{p'} (N) \cap X) (K^{g} \cap X) \leq (P \cap X) N_{X} (Ω_{1} {(Q)}^{g}) \leq X .$

Using (b) we see that $G = N_{G} (Ω_{1} (Q)) P_{0}$ , and then we take $y \in N_{G} (Ω_{1} (Q))$ and $x \in P_{0}$ such that g = yx. Now we deduce that $X = (P \cap X) N_{X} (Ω_{1} {(Q)}^{yx}) = (P \cap X) N_{X} (Ω_{1} {(Q)}^{x})$ , as stated in (d).

Finally, suppose that q divides $| X |$ and that $x \in P$ . Suppose first that $X = (X \cap P) N_{X} (Q^{x})$ . Then q divides $| N_{X} (Ω_{1} {(Q)}^{x}) |$ , which provides a subgroup Q₀ of order q in $N_{X} (Ω_{1} {(Q)}^{x}) \leq N_{G} (Ω_{1} {(Q)}^{x})$ and, by (d), there is some $y \in P_{0}$ such that $Ω_{1} {(Q)}^{y} = Q_{0} \leq N_{G} (Ω_{1} {(Q)}^{x}) = N_{G} {(Ω_{1} (Q))}^{x}$ . Then $Ω_{1} (Q)$ and $Ω_{1} {(Q)}^{y x^{- 1}}$ are subgroups of $N_{G} (Ω_{1} (Q))$ and therefore $\begin{matrix} [y x^{- 1}, Ω_{1} (Q)] = [[y x^{- 1}, Ω_{1} (Q)], Ω_{1} (Q)] \leq [[P, Ω_{1} (Q)] \cap N_{G} (Ω_{1} (Q)), Ω_{1} (Q)] \\ \leq [C_{P} (Ω_{1} (Q)), Ω_{1} (Q)] = 1 \end{matrix}$ by Lemma 1.1. We conclude that $Ω_{1} {(Q)}^{x} = Ω_{1} {(Q)}^{y} = Q_{0} \leq N_{X} (Ω_{1} {(Q)}^{x}) \leq X$ .

Now, conversely, suppose that $Ω_{1} {(Q)}^{x} \leq X$ . Then (d) provides some $z \in P_{0}$ such that $X = (P \cap X) (N_{X} (Ω_{1} {(Q)}^{z}))$ . In the paragraph above we have shown that $Ω_{1} {(Q)}^{z} \leq X$ . We apply (c) to $X = (X \cap N) (X \cap K^{g})$ , which is a group in $L$ by Lemma 7.4, and we obtain some $y \in P \cap X$ such that $Ω_{1} {(Q)}^{zy} = Ω_{1} {(Q)}^{x}$ . We conclude that $X = X^{y} = {(P \cap X)}^{y} (N_{X^{y}} (Ω_{1} {(Q)}^{zy})) = (P \cap X) (N_{X} (Ω_{1} {(Q)}^{x})),$ because $P \cap X E X$ . □

Lemma 7.9.

Let $G \in L$ be of type (N, K) and let $p \in π (N)$ such that K induces non-trivial power automorphisms on $P : = O_{p} (N)$ .

Then for all $X, Y \leq G$ , there is some $i \in {0, 1}$ such that $| 〈 X, Y 〉 \cap P | = | (P \cap X) (P \cap Y) | \cdot p^{i}$ .

In addition i = 0 if and only if there is some $g \in P$ such that for both $Z \in {X, Y}$ we have $Z \leq (Z \cap P) O_{p'} (N) K^{g}$ .

Proof.

We first remark that Lemma 7.2 gives that every subgroup of P is normal in N, and hence in G, because K normalizes every subgroup of P as well. In addition $P = [P, K]$ is elementary abelian by Corollary 5.6 (c).

Let $X, Y \leq G$ . Then Lemma 7.3 provides $x, y \in P$ such that $X \leq (X \cap P) O_{p'} (N) K^{x}$ and $Y \leq (Y \cap P) O_{p'} (N) K^{y}$ .

This implies that $〈 X, Y 〉 \leq (X \cap P) (Y \cap P) 〈 x^{- 1} y 〉 O_{p'} (N) K^{x}$ , bearing in mind that $X \cap P$ and $Y \cap P$ are normal subgroups of G, and therefore $〈 X, Y 〉 \cap P = (P \cap X) (P \cap Y) 〈 x^{- 1} y 〉 .$

Since P is elementary abelian, we see that $o (x y^{- 1}) \in {1, p}$ and we deduce the first assertion.

If it is possible to choose x = y, then $〈 X, Y 〉 \cap P = (P \cap X) (P \cap Y) 〈 x^{- 1} y 〉 = (P \cap X) (P \cap Y)$ and in particular i = 0 in the statement of the lemma.

For the converse we suppose that i = 0, i.e. $| 〈 X, Y 〉 \cap P | = | (P \cap X) (P \cap Y) |$ . Then $x^{- 1} y \in 〈 X, Y 〉 \cap P = (P \cap X) (P \cap Y)$ and thus we find $x_{0} \in X \cap P$ and $y_{0} \in Y \cap P$ such that $x^{- 1} y = x_{0} y_{0}$ . Then $g : = y y_{0}^{- 1} = x x_{0} \in P \cap X \cap Y$ . We note that x₀ normalizes X, centralizes $P \cap X$ and normalizes $O_{p'} (N)$ , which implies that $X = X^{x_{0}} \leq {((X \cap P) O_{p'} (N) K^{x})}^{x_{0}} = (X \cap P) O_{p'} (N) K^{x x_{0}}$ . Similarly $Y \leq (Y \cap P) O_{p'} (N) K^{y y_{0}^{- 1}}$ . □

Lemma 7.10.

Let $G \in L$ be of type (N, K). Suppose that X and Y are subgroups of G such that $〈 X, Y 〉 = G$ and let B is a batten of K. Suppose that K has a normal q-complement H. Then one of the following hold:

$q ∤ | G : X |$ ,
$q ∤ | G : Y |$ , or
q = 2, K has a section isomorphic to Q₈ and 4 divides $(| X |, | Y |)$ .

Proof.

Let $Q \in {Syl}_{q} (K)$ and suppose that q divides neither $| G : X |$ nor $q | | G : Y |$ . Then Lemma 7.3 gives maximal subgroups M_X and M_Y of Q such that $X \leq {NHM}_{X}$ and $Y \leq {NHM}_{Y}$ .

If Q is cyclic, then $M_{Y} = M_{X} = Φ (Q)$ and it follows that $G = 〈 X, Y 〉 \leq NH Φ (Q) \neq NHQ = G$ . This is impossible. We conclude that Q is not cyclic and then, since K is a batten group, it follows that $Q ≅ Q_{8}$ . We assume for a contradiction that $4 ∤ | X |$ . Then $X \leq NH Φ (Q)$ and hence $G = 〈 X, Y 〉 \leq {NHM}_{Y} \neq NHQ = G$ , which is again a contradiction. □

Lemma 7.11.

Let $G \in L$ be of type (N, K) such that K acts irreducibly on $[O_{p} (N), K] / Φ ([O_{p} (N), K])$ for some prime $p \in π (N)$ . If $X, Y \leq G$ are such that $〈 X, Y 〉 = G$ , then X or Y acts irreducibly on $[O_{p} (N), K] / Φ ([O_{p} (N), K])$ .

Proof.

Let $X, Y \leq G$ be such that $〈 X, Y 〉 = G$ and let $P : = O_{p} (N)$ . We assume for a contradiction that neither X nor Y act irreducibly on $[P, K] / Φ ([P, K])$ . Lemma 7.2 implies that N normalizes every subgroup of P (*). Moreover, by Lemma 7.3, there are $x, y \in N$ such that $X = (X \cap N) (X \cap K^{x})$ and $Y = (X \cap N) (Y \cap K^{y})$ and, by assumption, neither $X \cap K^{x}$ nor $Y \cap K^{y}$ act irreducibly on $[P, K] / Φ ([P, K])$ . It follows from Lemma 5.4 that $X \cap K^{x}$ and $Y \cap K^{y}$ both induce power automorphisms on P and that $| P | \neq p$ . Thus (*) yields that X and Y normalize every subgroup of P. Then also $G = 〈 X, Y 〉$ normalizes every subgroup of P, which contradicts the irreducible action of K. □

8 The main result

Theorem 8.1.

If $G \in L$ , then G is L₉-free.

Proof.

Assume for a contradiction that the statement is false and let G be a minimal counterexample. Then there is a sublattice $L = {E, S, T, D, U, A, B, C, F}$ of L(G) isomorphic to L₉, and in particular $L$ satisfies the relations in Definition 3.1.

We let G be of type (N, K) where, among the minimal counterexamples, we choose G such that $| N |$ is as large as possible and we set $π {(K)}^{*} : = π (K) ∖ {| B (H) | | H is a non‐nilpotent batten of K} .$

Then K has a normal q-complement for every $q \in π {(K)}^{*}$ and Lemma 7.10 is applicable.

We will first analyze how $L$ fits into the subgroup lattice of G.

(1) F = G, $C_{K} (N) = 1$ and every subgroup of N is normal in N.

Proof. The group F is a subgroup of G that is not L₉-free, and Lemma 7.4 yields that $F \in L$ . Hence the minimal choice of G implies that F = G.

Similarly, it follows from Lemma 3.4 that G is not a direct product of two non-trivial groups of coprime order. Let $p \in π (N)$ . Then $N = O_{p'} (N) \times O_{p} (N)$ because N is nilpotent. If K centralizes $O_{p} (N)$ , then $G = O_{p'} (N) K \times O_{p} (N)$ , where the direct factors have coprime order by ( $L 1$ ). But we just saw above that such a direct decomposition of G is not possible. Therefore $[O_{p} (N), K] \neq 1$ and p divides $| [O_{p} (N), K] |$ , which divides $| [N, K] |$ . We conclude from Lemma 7.2 that every subgroup of $O_{p} (N)$ is normal in N. This implies that every subgroup of N is a normal subgroup of N, because N is nilpotent.

Since we have chosen N as large as possible, Lemma 7.5 implies that $π (K) = π (K / C_{K} (N))$ . Let $q \in π (C_{K} (N))$ . Then the previous equation forces $q \in π (K / C_{K} (N))$ , and therefore a Sylow q-subgroup of K has order at least q². In particular, for all non-nilpotent battens V of K, we have that $B (V) ≰ C_{K} (N)$ . It follows that $q \in π {(K)}^{*}$ . Now Lemma 7.10 provides $X, Y \in {T, U, B} \subseteq L$ such that $X \neq Y$ and $O_{q} (C_{K} (N)) \leq X \cap Y = E$ . Since $O_{q} (C_{K} (N))$ is characteristic in $C_{K} (N) E NK = G$ , it follows that $G / O_{q} (C_{K} (N))$ is not L₁₀-free. Moreover, $G / O_{q} (C_{K} (N)) \in L$ by Lemma 7.6. Since G is a minimal counterexample, we conclude that $O_{q} (C_{K} (N)) = 1$ . We recall that G is soluble, by Lemma 7.1, hence $C_{K} (N)$ is soluble, and then there must exist a prime $q \in π (C_{K} (N))$ such that $O_{q} (C_{K} (N)) \neq 1$ . This gives a contradiction, and therefore $C_{K} (N) = 1$ . □

We remark that, by (1), we may apply Lemmas 7.7 and 7.8.

(2) For all $p \in π (N)$ we have that $O_{p} (N) \cap D E G$ . In particular $N \cap D \ unlhd G$ and $N \cap E = 1$ .

Proof. Let $H \in {E, D}$ , let $p \in π (N)$ and set $P : = O_{p} (N)$ . If H = E, then we set $M : = {U, T, B}$ and otherwise we set $M : = {A, B, C}$ . Then for all distinct $X, Y \in M$ , we have that $X \cap Y = H$ and $〈 X, Y 〉 = F$ .

Assume for a first contradiction that $H \cap P$ is not a normal subgroup of G. Since every subgroup of N is normal in N by (1), it follows that K does not induce power automorphism on P. Then Lemma 5.4 implies that K acts irreducibly on $\tilde{P} : = [P, K] / Φ ([P, K])$ . We apply Lemma 7.11 twice to find some $X \in M$ and some $Y \in M ∖ {X}$ such that X and Y act irreducibly on $\tilde{P}$ . In particular $[P, K] = [H \cap P, Y] = [H \cap P, X] \leq X \cap Y = H$ . It follows that $[P, K] \leq H \cap P \leq [P, K] C_{P} (K)$ , which yields that $H \cap P$ is normalized by K and hence $H \cap P E NK = G$ . This is a contradiction. We deduce that $P \cap D E G$ as stated, in particular $N \cap D E G$ and also $N \cap E E G$ .

For the final statement in (2) we use that $G / (N \cap E)$ is not L₉-free. Then the minimality of G and Lemma 7.6 give that $N \cap E = 1$ . □

(3) For every $q \in π (K)$ such that $1 \neq Q \in {Syl}_{q} (D)$ , one of the following holds:

$[N, Ω_{1} (Q)] \leq D \leq A$ or K induces power automorphisms on $[N, Ω_{1} (Q)]$ and $[N, Ω_{1} (Q)] \cap A \neq 1$ .

Moreover E = 1.

Proof. We adopt the same notation as in the previous step, which means that $H \in {E, D}, p \in π (N)$ and $P : = O_{p} (N)$ . If H = E, then $M : = {U, T, B}$ , and otherwise $M : = {A, B, C}$ . Whenever $X, Y \in M$ are distinct, then $X \cap Y = H$ and $〈 X, Y 〉 = F$ .

Let $q \in π (K) \cap π (H)$ and $Q \in {Syl}_{q} (D)$ . By conjugation we may suppose that $Q \leq K$ . Then $Ω_{1} (Q) = Ω_{1} (Q_{0})$ for some Sylow q-subgroup Q₀ of K by Lemmas 2.6 and 7.7 provides some $p \in π (N)$ such that $1 \neq [N, Ω_{1} (Q)]$ is a p-group. Now Lemma 7.8 (e) states that $X = (X \cap P) N_{X} (Ω_{1} (Q))$ for all subgroups $X \in M$ (*).

Let X and Y be distinct elements of $M$ and assume for a contradiction that $X \cap P$ and $Y \cap P$ are subgroups of $C_{P} (K)$ . Then $G \overset{(1)}{=} F = 〈 X, Y 〉 \overset{(*)}{\leq} 〈 C_{P} (K), N_{X} (Ω_{1} (Q)), N_{Y} (Ω_{1} (Q)) 〉 \leq C_{P} (K) N_{G} (Ω_{1} (Q)) = N_{G} (Ω_{1} (Q)),$ which contradicts (1).

For the remainder of this proof we let X and Y in $M$ be such that their intersection with P is not contained in $C_{P} (K)$ . We note that $C_{P} (K) = C_{P} (Ω_{1} (Q))$ by Lemma 5.5. Then it follows that $1 \neq [P \cap X, Ω_{1} (Q)] = [[P \cap X, Ω_{1} (Q)], Ω_{1} (Q)]$ and hence $[P, K] \cap X ≰ C_{P} (K)$ . In a similar way we observe that $[P, K] \cap Y ≰ C_{P} (K)$ .

If K acts irreducibly on $[P, K] / Φ ([P, K])$ , then Lemma 7.11 yields that X or Y, say X, acts irreducibly on $[P, K] / Φ ([P, K])$ . It follows that $[P, K] \leq X$ and $[P, K] \cap Y \leq X \cap Y = H$ . Let $Z \in M ∖ {X, Y}$ . Then Lemma 7.11 yields that Z or Y, say Y, acts irreducibly on $[P, K] / Φ ([P, K])$ and contains $[P, K] \cap Y$ . Since $[P, K] \cap Y ≰ C_{P} (K)$ , it follows that $[P, K] \leq V \cap X = H$ . By (1), this is only possible if H = D, in other words $π (K) \cap π (E) = \emptyset$ . Then the fact that $E \cap N = 1$ (see (1) once more) forces E = 1.

The previous paragraph also gives that $[N, Ω_{1} (Q)] = [P, Ω_{1} (Q)] = [P, K] \leq D \leq A$ , by Lemma 5.4, in the case where K acts irreducibly on $[P, K] / Φ ([P, K])$ .

Otherwise Lemma 5.4 gives that K and hence $Ω_{1} (Q)$ induce power automorphisms on P. Together with (1) this means that every subgroup of P is normal in G. Now, if $V, W \in M$ are distinct, then $P N_{G} (P) = G = 〈 V, W 〉 \overset{(*)}{\leq} (V \cap P) (W \cap P) N_{G} (Ω_{1} (Q))$ and it follows that $P = (V \cap P) (W \cap P)$ . We deduce that $| P | = \frac{| V \cap P | \cdot | W \cap P |}{| H \cap P |}$ for all $W, V \in M$ . In particular $| W \cap P | = | X \cap P | \neq 1$ for all $W \in M$ . This implies all our claims about D, because $P = [P, K] = [P, Ω_{1} (Q)] = [N, Ω_{1} (Q)]$ by Lemma 5.4 and $A \in M$ .

If, still in the power automorphism case, we have that H = E, then we recall that $E \cap P = 1$ by (1). Therefore $| (U \cap P) | \cdot | (T \cap P) | = \frac{| (U \cap P) | \cdot | (T \cap P) |}{| E \cap P |} = | (U \cap P) (T \cap P) | = | P | = | (U \cap P) (A \cap P) |$ $= \frac{| U \cap P | \cdot | A \cap P |}{| P \cap E |} = | U \cap P | \cdot | A \cap P | .$

This implies that $A \cap P = T \cap P \neq 1$ . But in this case we may interchange T by S in $M$ . Since $1 \neq T \cap P = A \cap P = S \cap P$ , we arrive at the contradiction that $1 \neq S \cap T \cap P = E \cap P \leq E \cap N = 1$ (by (1)). Thus, we have that $π (K) \cap π (E) = \emptyset$ in this case as well. Again it follows that E = 1. □

The remainder of the proof is dedicated to constructing a subgroup of G that violates Property ( $L 4$ ).

(4) If $p \in π (N)$ , then $A \cap K$ does not centralize $O_{p} (N)$ . In particular $A \cap K \neq 1$ .

Proof. We assume for a contradiction that $A \cap K$ centralizes $P : = O_{p} (N)$ for some $p \in π (N)$ .

It follows from Lemma 7.3 that, for every subgroup X of $G = P O_{p'} (N) K$ , there is some $x \in P$ such that $X \leq (X \cap P) O_{p'} (N) K^{x}$ . Since $[P, A \cap K] = 1$ , we further have THAT $X \leq (X \cap P) O_{p'} (N) K^{y}$ for all $X \leq A$ and $y \in P$ (*).

Assume that $A \cap P = 1$ . Then for both $X \in {U, B}$ , we see that $P \leq G \overset{(1)}{=} F = 〈 U, A 〉 \overset{(*)}{\leq} (P \cap X) O_{p'} (N) K^{x}$ and therefore $P = P \cap U = P \cap B \leq U \cap B = E$ . But this contradicts (3).

Thus $A \cap P \neq 1$ and then (1), together with our assumption at the beginning of the proof, imply that every subgroup of $A \cap P$ is normal in A. Suppose that $X, Y \in {S, T, D}$ are distinct. We recall that $A \cap P \leq A = 〈 X, Y 〉 \leq (X \cap P) (Y \cap P) O_{p'} (N) K$ , and then it follows that $A \cap P = (X \cap P) (Y \cap P)$ . Since $X \cap Y = E \overset{(2)}{=} 1$ , we know more: $T \cap P ≅ S \cap P ≅ D \cap P$ and $| T \cap P |^{2} = | A \cap P |$ .

If K induces power automorphisms on P and if $X \in {A, T, S}$ , then $(X \cap P)$ and $(U \cap P)$ are normal subgroups of G by (1). In addition there is some $u \in P$ such that $U \leq (U \cap P) O_{p'} (N) K^{u}$ and then $X \leq (X \cap P) O_{p'} (N) K^{u}$ by (*). We apply Lemma 7.9 to see that $| P \cap A | = | P : P \cap U | = | P \cap S | = \sqrt{| P \cap A |}$ . Now it follows that $P \cap A = 1$ , which is a contradiction.

We conclude that K does not induce power automorphisms on P. Then Lemma 5.4 gives that K acts irreducibly on $[P, K] / Φ ([P, K])$ . Since $P \cap D E G$ by (2), we see that either $[P, K] \leq D$ or that $1 \neq P \cap D \leq C_{P} (K)$ .

In the first case $T \cap P ≅ D \cap P \geq [P, K]$ and $T \cap D = E \overset{(2)}{=} 1$ . This implies, together with Lemma 1.1, that $C_{P} (K)$ has a subgroup isomorphic to $[P, K]$ . We apply Lemma 5.4, in combination with Part (b) of Lemma 4.17, and we deduce that $[P, K]$ is cyclic of order 2 and that, therefore, K centralizes it. This is impossible.

It follows that the second case holds, i.e. $1 \neq P \cap D \leq C_{P} (K)$ . Then p = 2 by Corollary 5.6 (a). If $[P, K]$ is not elementary abelian, then Part (b) of Lemmas 4.17 and 5.4 give that $[P, K] ≅ Q_{8}$ and $P = [P, Q] \times I$ , where I is a subgroup of P of order at most 2. We note that $T \cap P ≅ D \cap P$ and $T \cap D = E \overset{(3)}{=} 1$ , and then we conclude that $T \cap P$ and $D \cap P$ are cyclic of order 2. Consequently $A \cap P = (T \cap P) \cdot (D \cap P) = Ω_{1} (P) E G$ . Now $1 \overset{(3)}{=} E = U \cap A \geq U \cap Ω_{1} (P)$ and therefore $U \leq O_{π'} (N) K^{u}$ . We arrive at a contradiction: $P \leq G = 〈 U, A 〉 \leq (P \cap A) O_{π'} (N) K^{u}$ , because $[P, K] ≅ Q_{8}$ is not elementary abelian.

So we finally have that $[P, K]$ is elementary abelian. Then Lemma 4.17(b) gives that $C_{P} (K)$ is cyclic and Lemma 1.1 shows that $T \cap P ≅ D \cap P \leq C_{P} (K)$ .

Together with the fact that $T \cap D = E \overset{(3)}{=} 1$ , we obtain that $D \cap P$ is cyclic of order 2. It follows that $P \cap D, T \cap P$ and $P \cap S$ have order 2 and hence $P \cap A$ is elementary abelian of order 4. Moreover $A \cap [P, K]$ is cyclic of order 2, and therefore it equals one of the subgroups $T \cap P, S \cap P$ or $D \cap P$ . The last case is not possible because $D \cap P \leq C_{P} (K)$ . By symmetry between S and T in the lattice we may suppose that $T \cap P \leq [P, K]$ . Recall that K acts irreducibly on $[P, K]$ while $A \cap K$ centralizes P. In particular (1) implies that A does not act irreducibly on $[P, K]$ . Thus Lemma 7.11, together with the fact that $G \overset{(1)}{=} F = 〈 A, U 〉$ , gives that U acts irreducibly on $[P, K]$ . We also know that $T \cap U = E \overset{(3)}{=} 1$ , and this implies that $[P, K] ≰ U$ . Then Lemma 1.2 yields that $U \cap P \leq C_{P} (K)$ . But we recall that $C_{P} (K)$ is cyclic, and its unique involution is contained in D. Then the fact that $U \cap D = E \overset{(3)}{=} 1$ implies that $U \cap P = 1$ . Finally, we see that $G \overset{(1)}{=} F = 〈 U, T 〉 \overset{(*)}{\leq} [P, K] O_{p'} (N) K^{u}$ , which gives a contradiction. □

By conjugation and by Lemma 7.3 we may suppose that $A = (A \cap N) (A \cap K)$ .

We set $π : = π {(K)}^{*} \cap π (A)$ and we let L₁ be a Hall π-subgroup of $A \cap K$ .

Then we let $σ : = {p \in π (N) | [O_{p} (N), Q] \neq 1 for some Q \leq L_{1}, where | Q | \in π} .$

(5) If $p \in π (N)$ and $[O_{p} (A), L_{1}] = 1$ , then $O_{p} (N) \leq D$ . Moreover $π \neq \emptyset \neq σ$ .

Proof. First suppose that L₁ centralizes $P : = O_{p} (N)$ for some $p \in π (N)$ . Then (4) implies that $A \cap K \neq L_{1}$ and then there is a non-nilpotent batten V of K such that $Q : = B (V) \leq A$ and $[P, Q] \neq 1$ . Now $| Q | = q$ is a prime and $[N, Q] = [P, Q]$ by Lemma 7.7. In addition we see, from Definition 4.16, that Q does not induce power automorphisms on P and that $P = [P, Q]$ . If q divides $| D |$ , then (3) implies that $P = [N, Q] \leq D$ , as stated.

Now we suppose that q does not divide $| D |$ . Let $R \leq K$ be such that V = QR. Up to conjugation we may suppose that $R \cap L_{1}$ is a Sylow subgroup of L₁. Then R does not centralize P by Definition 5.1 and hence $R ≰ L_{1}$ . It follows that $R ≰ A$ , from the definition of L₁. Since $Q \cdot Φ (R) = Q \cdot Z (V)$ is nilpotent by Lemma 2.5, we deduce that A has a normal q-complement. Moreover $A \in L$ , by Lemma 7.4, whence we may apply Lemma 7.10 to A. Then we see that S and T have orders divisible by q, because $q \notin π (D)$ . Let $t, s \in [P, Q]$ be such that $Q^{s} \leq S$ and $Q^{t} \leq T$ . The irreducible action of Q on P, together with the fact that $P \cap T \cap S \leq E \overset{(3)}{=} 1$ , implies that $P \cap T = 1$ or $P \cap S = 1$ . Without loss $P \cap T = 1$ . Then $T \leq N_{G} (Q^{t})$ by Lemma 7.8 (e). Assume that A normalizes Q^t. Then $Q^{s} \leq N_{G} (Q^{t})$ , which implies that $[t s^{- 1}, Q] Q^{s} = 〈 Q^{s}, Q^{t} 〉 \leq N_{G} (Q^{t})$ by Lemma 4.1.1 (b) of [Citation7]. Then the irreducible action of Q^t on P forces t = s, contradicting the fact that $T \cap S = E \overset{(3)}{=} 1$ .

Thus A does not normalize Q^t and $〈 T, D 〉 = A ≰ N_{G} (Q^{t})$ , which yields that $D ≰ N_{G} (Q^{t}) = (O_{p'} (N) K) \times C_{P} (K) = O_{p'} (N) K$ by Part (a) of Lemma 7.8. It follows that $D \cap P \neq 1$ and therefore $P \leq D$ , because $P \cap D E G$ by (2) and K acts irreducibly on P.

We turn to the second statement and assume for a contradiction that $π = π (A) \cap π (K) = \emptyset$ . Then $A \cap K = 1$ , contrary to (4). Hence if $π = π (A) \cap π {(K)}^{*} = \emptyset$ , then we can draw two conclusions: First $L_{1} = 1$ and the statement we just proved gives that $N \leq D$ . Second, there must be a prime in $π (K) ∖ π {(K)}^{*}$ dividing $| A |$ . By definition of $π {(K)}^{*}$ , such a prime is $| B (V) |$ for some non-nilpotent batten V of K. We choose such a non-nilpotent batten V and set $Q : = B (V)$ . Then there are some prime r and an r-subgroup $R \leq K$ such that QR = V, and $r \in π {(K)}^{*}$ because of the structure of non-nilpotent battens (Definition 2.1). In particular $r ∤ | A |$ by assumption and Lemma 7.10 yields that B and U contain a conjugate of R. We recall that $N \leq D \leq B$ by the first consequence of our assumption and because of the structure of the lattice L₉. Then Lemma 7.8(c) gives that $Ω_{1} {(R)}^{G} \subseteq B$ . Thus $B \cap U = E \overset{(3)}{=} 1$ , which is a contradiction. This proves that $π \neq \emptyset$ .

If $σ = \emptyset$ , then for all $p \in π (N)$ , all $q \in π$ and all q-subgroups Q of L₁, we have that $[O_{p} (N), Q] = 1$ . Then $[N, L_{1}] = 1$ by definition of L₁, and $L_{1} \neq 1$ because $π \neq \emptyset$ . But then $1 \neq L_{1} \leq C_{K} (N)$ , contrary to (1). □

Next we set $H : = O_{σ} (N)$ and we prove that H is a candidate for the desired properties in ( $L 4$ ).

(6) The non-trivial Sylow subgroups of H are elementary abelian (in particular N is abelian), but not cyclic, and $K \leq {Pot}_{K} (H)$ .

Proof. By definition $H \leq N$ is nilpotent. If, for all $p \in σ$ , the group K does not act irreducibly on $[O_{p} (N), K] / Φ ([O_{p} (N), K])$ , then Lemma 5.4 gives that $K \leq {Pot}_{K} (H)$ . In particular $O_{p} (N)$ is not cyclic. Moreover, Corollary 5.6 yields that $O_{p} (N) = [O_{p} (N), K]$ is elementary abelian. Therefore, our claim is satisfied in this case.

Let us assume for a contradiction that there is some $p \in σ$ such that K acts irreducibly on the group $[O_{p} (N), K] / Φ ([O_{p} (N), K])$ . We set $P : = O_{p} (N)$ and we choose $Q \leq L_{1}$ of order $q \in π$ such that $[P, Q] \neq 1$ , by the definition of σ.

Case 1: $[P, Q] \leq A$ .

Then Lemma 7.8(c) implies that $Q^{g} \leq A$ for every $g \in G$ . We recall that $U \cap A = E \overset{(3)}{=} 1$ , and then it follows first that $q \notin π (U)$ and then that $q | | B |$ , by Lemma 7.10. Here we use that $G = F = 〈 B, U 〉$ by (1). Thus we find some $g \in G$ such that $Q^{g} \leq D$ , because $D = A \cap B$ . We apply (3) to observe that $[P, Q] = [N, Q] \leq D$ and then D contains every conjugate of Q by Lemma 7.8(c). Recall that $q \notin π (U)$ , which implies that $q \in π (T)$ by Lemma 7.10. Again we use that $G = F = 〈 T, U 〉$ . But this contradicts the fact that $T \cap D = E = 1$ by (3).

Case 2: $[P, Q] ≰ A$ .

Then $[P, Q] ≰ D$ , in particular $P ≰ D$ , and $P \cap D E G$ by (2). This means that K stabilizes the subgroup $[P \cap D, K] /$ of $[P, K] / Φ ([P, K])$ , while acting irreducibly. This forces $[P \cap D, K] \leq Φ ([P, K])$ , and together with coprime action (Lemma 1.1) we see that K centralizes $P \cap D$ .

By Lemma 7.3 we know that $A \in L$ , with type $(A \cap N, A \cap K)$ . Additionally, since $q \in π {(K)}^{*}$ , the group K has a normal q-complement. Then $A \cap K$ also has a normal q-complement. We may apply Lemma 7.10 to $A = 〈 T, S 〉 = 〈 T, D 〉 = 〈 S, D 〉$ . It yields that at least two of the groups D, T, S have a subgroup of order $| Q |$ . As $| Q | ∤ | E |$ by (2), there is some $g \in G$ such that $Q^{g} \neq Q$ and $Q^{g} \leq A$ . Then Part (e) of Lemma 7.8 implies that $P \cap A ≰ C_{P} (Q) = C_{P} (K)$ by Lemma 5.5. In the present case we have that $[P, A] ≰ A$ , and then Lemma 1.2 gives that A does not act irreducibly on $[P, K] / Φ ([P, K])$ . Thus $A \cap K$ induces power automorphism on P by Lemma 5.4, and these automorphisms are not trivial because $Q \leq A \cap K$ . Corollary 5.6 (c) implies that $P = [P, A \cap K] \overset{5.4}{=} [P, K]$ is elementary abelian and in particular that $D \cap P \leq C_{P} (K) \overset{5.5}{=} C_{P} (A \cap K) = 1$ . Hence there is some $d \in [P, K]$ such that $D = N_{D} (Q^{d})$ , by Lemma 7.8(b). In addition we see from Lemma 7.11 that B and C act irreducibly on P. If it was true that $P \leq B$ , then it would follow that $1 \neq P \cap A \leq P \cap B \cap A \leq P \cap D \leq C_{P} (K) = 1$ , which is a contradiction. We conclude that $P ≰ B$ and hence $P \cap B = 1$ because of the irreducible action of B on P, and similarly $P \cap C = 1$ .

Then Lemma 7.8(b) provides $b, c \in P$ such that $B = N_{B} (Q^{b})$ and $C = N_{C} (Q^{c})$ . If $Q^{b} = Q^{c}$ , then $G = F = 〈 B, C 〉 \leq N_{G} (Q^{b})$ , which is false.

Consequently $Q^{b} \neq Q^{c}$ , and we can use Lemma 7.8(a), Dedekind’s modular law and Lemma 1.4(c). Together this shows that $\begin{matrix} D \leq N_{B} (Q^{b}) \cap N_{C} (Q^{c}) = O_{p'} (N) K^{b} \cap O_{p'} (N) K^{c} = O_{p'} (N) (K^{b} \cap O_{p'} (N) K^{c}) \\ \leq O_{p'} (N) C_{K} (b c^{- 1}) \leq C_{G} (b c^{- 1}), \end{matrix}$ because N is nilpotent and because $〈 {(b c^{- 1})}^{K} 〉 \cap O_{p'} (N) \leq P \cap O_{p'} (N) = 1$ .

We recall that $P = [P, K]$ is abelian, hence it is contained in Z(N), and then it follows that D centralizes $1 \neq b c^{- 1} \in [P, K]$ . Here we also use Lemma 5.4, i.e. that D centralizes P. We conclude that $D = N_{D} (Q^{h})$ for all $h \in P$ . Again Lemma 7.8(b) provides $t, s \in P$ such that $T = (T \cap P) N_{T} (Q^{t})$ and $S = (T \cap S) N_{C} (Q^{s})$ . Let $X \in {T, S}$ . Then $P \cap A \leq A = 〈 D, X 〉 \leq (X \cap P) N_{G} (Q^{x})$ , which implies that $P \cap X = P \cap A$ . But now $P \cap A \leq P \cap S \cap T = P \cap E = 1$ , by (3), which gives a final contradiction. □

(7) For all $p \in σ$ we have that $O_{p} (N) \cap A \neq 1$ .

Proof. We set $P : = O_{p} (N)$ for some $p \in σ$ . Then there is some $Q \leq L_{1}$ of order $q \in π$ such that $[P, Q] \neq 1$ , by the definition of σ. Then $q \in π {(K)}^{*}$ and we see that K as well as $K \cap A$ have a normal q-complement. Since $A \in L$ by Lemma 7.4, we may apply Lemma 7.10. We notice that $A = 〈 D, T 〉 = 〈 D, S 〉$ , and then it follows that $q \in π (D)$ or that $q \in π (T) \cap π (S)$ . In the first case (3) implies our assertion. In the second case we use Lemma 7.8(c). It gives that $Q^{s} \leq S$ and $Q^{t} \leq T$ for some $t, s \in [P, K]$ . We deduce that $[t^{- 1} s, Q] \leq 〈 Q^{s}, Q^{t} 〉 \leq A$ by Lemma 4.1.1 of [Citation7]. In conclusion, our assertion is true or $t^{- 1} s$ centralizes Q. But in the second case, we see that $Q^{t} = Q^{s} \leq T \cap S = E \overset{(3)}{=} 1$ , and this gives a contradiction. □

(8) For all $q \in π$ we have that q divides $| S |, | T |$ , and $| B |$ .

Furthermore, if $Q \leq L_{1}$ and $| Q | = q$ , then $[N, Q] \cap T = [N, Q] \cap S = 1$ , but $[N, Q] \cap U \neq 1$ .

Proof. Suppose that $Q \leq L_{1}$ has order q. Then $[N, Q]$ is a p-group by Lemma 7.7, for some prime $p \in σ$ . We set $P = O_{p} (N)$ . Then for every $X \in {T, S}$ there is some $x \in P$ such that $X = (X \cap P) N_{X} (Q^{x})$ , by Lemma 7.8(b). Using (6) we see that K induces power automorphisms on P. Thus $P = [P, K]$ is elementary abelian by Corollary 5.6(c), and then Lemma 7.9 is applicable.

Assume for a contradiction that $P ≩ (P \cap A) (P \cap U)$ . Then, for all $X \in {T, S, A}$ , we have that $P \cap 〈 X, U 〉 = P \cap F \overset{(1)}{=} P > (P \cap A) (P \cap U) \geq (P \cap X) (P \cap U) .$

Hence Lemma 7.9 yields that $| P | = | (P \cap X) (P \cap U) | p \overset{(2)}{=} | (P \cap X) | | (P \cap U) | p$ and, for all $X \in {T, S, A}$ , we see that $p \cdot | P \cap X | = | P : P \cap U |$ . In particular, we have that $| P \cap A | = | P \cap S | = | P \cap T |$ . Therefore, the fact that $T, S \leq A$ implies that $P \cap A = P \cap T = P \cap S \leq P \cap (T \cap S) = P \cap E \overset{(3)}{=} 1$ . This contradicts (7).

We conclude that $P = (P \cap A) (P \cap U)$ and hence Lemma 7.9 provides some element $g \in P$ such that, for both $X \in {A, U}$ , it is true that $X \leq (X \cap P) O_{p'} (N) K^{g} = (X \cap P) N_{G} {(Q)}^{g}$ .

Assume for a contradiction that $q \in π (U)$ . Then Part (e) of Lemma 7.8 yields that $Q^{g} \in A \cap U = E \overset{(3)}{=} 1$ . This is impossible. Thus $q \notin π (U)$ and we deduce that $q \in π (X)$ for all $X \in {T, S, B}$ . Here we use Lemma 7.10 and the fact that $G \overset{(1)}{=} F = 〈 U, X 〉$ . Next we apply Lemma 7.9, which gives that $(P \cap A) \neq (P \cap S) (P \cap T)$ . Then the same lemma yields, for both combinations of ${X, Y} \in {T, S}$ , that p divides $| P \cap Y | | P \cap X | | P \cap U | = | P \cap A | | P \cap U | = | P | \leq | P \cap X | | P \cap U | \cdot p .$

It follows that $p \cdot | P \cap Y | \leq p$ and then that $S \cap P = T \cap P = 1$ . Finally (6) and Lemma 7.9 yield that $p^{2} \leq | P | \leq | P \cap T | | P \cap U | \cdot p = | P \cap U | \cdot p$ . We conclude that $P \cap U \neq 1$ . □

(9) For every $p \in σ$ there is some $r \in π {(K)}^{*}$ such that a Sylow r-subgroup of U does not centralize $O_{p} (N)$ .

Proof. We set $P : = O_{p} (N)$ . Since $p \in σ$ , there is some $Q \leq L_{1}$ such that $| Q | = q \in π$ . In addition (6) implies that P is elementary abelian, and then $P \leq Z (N)$ because N is nilpotent. By Lemma 7.8(b) and (8) there is some $s \in P$ such that $S = N_{S} (Q^{s})$ .

Assume for a contradiction that U centralizes P. Then $U = U^{s} = (U \cap P) N_{U} (Q^{s})$ by Lemma 7.8(d) and hence $G \overset{(1)}{=} F = 〈 U, S 〉 \leq (U \cap P) N_{G} (Q^{s})$ implies that $U \cap P = P$ . With (7) we obtain the contradiction that $1 \neq P \cap A \leq U \cap A = E \overset{(3)}{=} 1$ .

It follows that U does not centralize P. We recall that $P \leq Z (N)$ and then we obtain a prime $r \in π (K)$ such that a Sylow r-subgroup of U does not centralize P. Since K induces power automorphisms on P by (6), Definition 5.1 gives, for every non-nilpotent batten V of K, that $B (V)$ centralizes P. This implies that $r \in π {(K)}^{*}$ . □

We are now able to define L and J. Let $u \in N$ be such that $U = (U \cap N) (U \cap K^{u})$ .

We set $\tilde{π} : = {r \in π {(K)}^{*} | [H, R] \neq 1 for some R \in {Syl}_{r} (U)}$ and we let L₂ be a Hall $\tilde{π}$ -subgroup of $(U^{u^{- 1}} \cap K)$ . Then $L_{2} \leq K$ and $L = 〈 L_{1}, L_{2} 〉$ is a subgroup of K. In addition (5) and (9) show that $\tilde{π} \neq \emptyset$ .

Next we set $ρ : = {p \in π (N) | [O_{p} (N), R] \neq 1 for some R \leq L_{2} with | R | \in \tilde{π}}$ and $J : = [O_{ρ} (N), K]$ . Then $ρ \neq \emptyset$ by Lemma 7.7, because $\tilde{π} \neq \emptyset$ , and hence $J \neq 1$ . Finally, we note that J is L-invariant by construction.

(10) $σ \cap ρ = \emptyset$ , i.e. $| H |$ and $| J |$ are coprime.

Proof. We assume for a contradiction that the prime p divides $| H |$ and $| J |$ . Then by definition there are $q \in π$ and $r \in \tilde{π}$ and subgroups $Q \leq L_{1}$ and $R \leq L_{2}$ such that the following hold:

$| Q | = q, | R | = r, 1 \neq [N, Q] \leq O_{p} (N) = : P$ and $1 \neq [N, R] \leq P$ .

In particular (6) shows that K induces power automorphisms on P. It follows from Corollary 5.6 (c) and Lemma 5.4 that $P = [P, K] = [P, Q] = [N, Q] = [N, R]$ , and then (8) shows that $P \cap S = 1$ . Moreover $P \cap A \neq 1$ by (7). We apply Lemma 7.9 to obtain some $i, j \in {1, 0}$ such that $p^{i} | P \cap U | = | (P \cap U) (P \cap S) | p^{i} = | P | = | (P \cap U) (P \cap A) | p^{j} \overset{(2)}{=} | (P \cap U) | | (P \cap A) | p^{j} .$

This implies that $| P \cap A | p^{j} = p^{i}$ , and we obtain that i = 1 and j = 0.

In particular we see that $P = (U \cap P) (A \cap P)$ . Then Lemma 7.9 and the fact that $A \cap U = E \overset{(3)}{=} 1$ give some element $g \in P$ such that $X \leq (P \cap X) K^{g} = (P \cap X) N_{G} {(R)}^{g}$ , where $X \in {A, U}$ . Assume for a contradiction that $r \in π (A)$ . Then 7.8(d) forces $R^{g} \leq U \cap A = E \overset{(3)}{=} 1$ . This is impossible, hence $r \notin π (A)$ and we apply Lemma 7.10 to $G \overset{(1)}{=} F = 〈 A, B 〉$ . Then it follows that $r \in π (B)$ .

Moreover $q \in π (B) \cap π (T)$ by (8). Since $B \cap T = E \overset{(3)}{=} 1$ , this is not possible, hence Lemma 7.9 and (8) give that $| P | = | P \cap B | \cdot | P \cap T | \cdot p = | P \cap B | \cdot p$ . On the other hand we have that $r \in π (B) \cap π (U)$ , which is also impossible because $B \cap U = E \overset{(3)}{=} 1$ . Now Lemma 7.9 implies that $| P \cap U | \cdot | P \cap B | \cdot p = | P | = | P \cap B | \cdot p$ . In particular we see that $P \cap U = 1$ , and this contradicts (8). □

We summarize:

The definitions of σ and H imply that $H \leq N$ , and (6) shows that the non-trivial Sylow subgroups of H are not cyclic. In addition $L \leq K$ induces power automorphisms on H.

From (10) we deduce that $π \cap \tilde{π} = \emptyset$ . A Hall $π {(K)}^{*}$ -subgroup of K is nilpotent by Lemma 2.7, which means that L is nilpotent. In particular we see that $L = L_{1} \times L_{2}$ . Let $r \in π (L) = π \cup \tilde{π}$ and $R = O_{r} (L)$ . If $r \in π$ , then $[H, R] \geq [[N, Ω_{1} (R)], R] \neq 1$ , and if $r \in \tilde{p}$ , then $[H, R] \neq 1$ by (9). These arguments show that $π (L) = π (L / C_{L} (H))$ (*).

We assume for a contradiction that L is not cyclic. Then, since L is nilpotent and a batten group by Lemma 2.7, we deduce that $O_{2} (L) ≅ Q_{8}$ . Definition 4.8 implies that, for all $p \in π (N)$ , the group $O_{2} (L)$ does not induce non-trivial power automorphisms on $O_{p} (N)$ . In particular $O_{2} (L)$ centralizes H, which contradicts (*). Thus L is cyclic.

Furthermore, we already saw that J is L-invariant and that $1 \neq J$ . Then (10) gives that $(| H |, | J |) = 1$ , and since N is nilpotent, this forces $[H, J] = 1$ .

Assume for a contradiction that J is not abelian. Then Lemma 4.17(b) and Lemma 5.4 show that $O_{2} (J) = [O_{2} (N), K] ≅ Q_{8}$ and therefore $2 \in ρ$ . We let $r \in \tilde{π}$ be such that a Sylow r-subgroup R of K acts faithfully on $O_{2} (J)$ . Then we must have that $| R | = 3$ because $O_{2} (J) ≅ Q_{8}$ . Moreover Lemma 7.8(a) yields that R centralizes $O_{2'} (N) \geq H$ . This contradicts (*).

Hence J is abelian. For every $q \in π$ and every subgroup $Q \leq L_{1}$ of order q, the definition of σ and Lemma 7.7 provide some $p \in σ$ such that $[N, Q] \leq O_{p} (N) = P$ . Then we see, using (6) and Corollary 5.6, that $C_{P} (Q) = 1$ . Furthermore (10) yields that Q centralizes $J = O_{ρ} (N)$ , and then it follows that $1 = C_{P} (Q) \geq C_{P} (C_{L} (J))$ . Since $H = O_{σ} (N)$ is abelian, we conclude that $C_{H} (C_{L} (J)) = 1$ .

The previous argument also yields that ${q \in π (L) | C_{O_{q} (L)} (H) < C_{O_{q} (L)} (J)} \supseteq π$ . Let $r \in \tilde{π}$ and suppose that $R \leq L$ has order r. We recall the definition of ρ and apply Lemma 7.7: Then we see that $1 \neq [N, R] \leq O_{ρ} (N) = J$ . Thus $r \notin {q \in π (L) | C_{O_{q} (L)} (H) < C_{O_{q} (L)} (J)}$ and it follows that ${q \in π (L) | C_{O_{q} (L)} (H) < C_{O_{q} (L)} (J)} = π$ .

Since $G \in L$ , we can use Property $(L 4)$ , which gives some $g \in {(HJ)}^{#}$ that centralizes $O_{π} (L) = : L_{1}$ or $O_{π'} (L) = : L_{2}$ . Let $i \in {1, 2}$ be such that $[L_{i}, g] = 1$ . We may suppose that g has prime order p. Then $p \in σ \cup ρ$ and hence there is a subgroup $Q \leq L$ of prime order such that $[N, Q] \leq O_{p} (N) = : P$ . Lemma 5.5 gives that $g \in C_{P} (L_{i}) = C_{P} (K)$ or that L_i centralizes P.

In the first case Corollary 5.6 yields that p = 2, and then K does not induce power automorphisms on P. Using (6) we deduce that $p \in ρ$ and $g \in O_{p} (J) \cap C_{P} (K) = [P, K] \cap C_{P} (K) \cap J$ . Since J is abelian, we obtain a contradiction in this case.

It follows that the second case above holds, i.e. $[P, L_{i}] = 1$ . This means that i = 1 if $p \in ρ$ and i = 2 if $p \in σ$ . In addition we see, from (9), that $i \neq 2$ . We conclude that $L_{i} = L_{1}, p \in ρ$ and $q \notin π$ . In particular $q \notin π (A)$ . Since $G \overset{(1)}{=} F = 〈 A, U 〉 = 〈 A, B 〉$ , Lemma 7.10 implies that B and U contain a conjugate of Q. In addition (5) shows that $P \leq D \leq B$ and therefore Lemma 7.8(c) gives that $Q^{G} \subseteq B$ . Finally, we obtain a contradiction, because $B \cap U = E = 1$ by (3). This concludes the proof. □

Main Theorem. A finite group is in $L$ if and only if it is L₉-free.

Proof. Let G be a finite group. If G is L₉-free, then Theorem 6.5 shows that $G \in L$ .

Conversely, if $G \in L$ , then G is L₉-free by Theorem 8.1. □

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L₉-free groups

Abstract

Introduction

Notation and preliminary results

2 Battens and batten groups

3 L₁₀ and its sublattices

4 Group orders with few prime divisors

5 Avoiding L₉

6 The first implication

7 The class $L$

8 The main result

References

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L9-free groups

Abstract

Introduction

Notation and preliminary results

2 Battens and batten groups

3 L10 and its sublattices

4 Group orders with few prime divisors

5 Avoiding L9

6 The first implication

7 The class L

8 The main result

References

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L₉-free groups

3 L₁₀ and its sublattices

5 Avoiding L₉

7 The class $L$