Conjugate L-subgroups of An L-group and Their Applications to Normality and Normalizer

Abstract

In this paper, the notion of the conjugate of an L-subgroup by an L-point has been introduced. Then, several properties of conjugate L-subgroups have been studied analogous to their group-theoretic counterparts. Also, the notion of conjugacy has been investigated in the context of normality of L-subgroups. Furthermore, some important relationships between conjugate L-subgroups and normalizer have also been established. Finally, the normalizer of an L-subgroup has been defined by using the notion of conjugate L-subgroups.

Keywords:

• L-group
• L-subgroup
• Conjugate L-subgroup
• Normal L-subgroup
• Normalizer of An L-subgroup

1. Introduction

The studies of fuzzy algebraic structures began in 1971 when Rosenfeld [1] applied the notion of fuzzy sets to subgroupoids and subgroups. In 1981, Liu [2], replaced the closed unit interval with ‘lattices’ in the definition of fuzzy sets and introduced the notion of lattice valued fuzzy subgroups. Subsequently, a number of researchers investigated fuzzy algebraic structures and generalized various concepts to the fuzzy setting. We mention here that in majority of these studies, the parent structure was considered to be a classical group rather than an L(fuzzy)-group. This setting has a significant limitation that it does not allow the formulation of various concepts from classical group theory to the fuzzy (lattice valued) group theory. This drawback can be easily removed by taking the parent structure to be a L(fuzzy)-group rather than an ordinary group. Indeed, in [3–8], Ajmal and Jahan have introduced and studied various algebraic structures in L-setting specifically keeping in view their compatibility. Additionally, the authors of this paper have continued this research in [9,10] by introducing the maximal and Frattini L-subgroups of an L-group and exploring their relationship with various structures in L-group theory. This paper is a continuation of similar studies.

In classical group theory, the notion of the conjugate has important relationships with the notions of normality and normalizer of subgroups. Moreover, several important notions such as pronormal and abnormal subgroups, etc are defined through the notion of the conjugate. Hence a detailed study of the conjugate was necessitated in L(fuzzy)-group theory that extends the notion while keeping various relationships intact. The notion of the conjugate of a fuzzy group was introduced by Mukherjee and Bhattacharya [11] in 1986. A similar definition has also appeared in [12] in 2013. We note here that in their study, the notion of conjugate fuzzy subgroup of an ordinary group has been discussed rather than conjugate fuzzy subgroups of a fuzzy group. Moreover, the conjugate fuzzy subgroup has been defined with respect to a crisp point of the classical group instead of an element of the fuzzy group itself. This definition has two major limitations. Firstly, in [7], Ajmal and Jahan have provided an improved definition of the normalizer of an L(fuzzy)-subgroup of an L-subgroup that is consistent with the corresponding notions in classical group theory. The conjugate developed by Mukherjee and Bhattacharya in [11] is completely incompatible with this definition of the normalizer. Secondly, the notion of the conjugate in [11] cannot be used to extend several significant concepts to the L(fuzzy) setting such as the notions of pronormal and abnormal subgroups, etc. Hence a new study of conjugate L-subgroups of an L-group was required. In this paper, we have removed these limitations by firstly taking the parent structure to be an L-group μ and secondly by defining the conjugate of an L-subgroup by an L-point in μ rather than an element of the ordinary group G. The new definition of the conjugate thus introduced has been found to be perfectly compatible with the related notions, while at the same time revealing various peculiarities of L-group theory.

We begin our work in Section 3 by introducing the notion of the conjugate of an L-subgroup by an L-point in the parent L-group μ. Moreover, we show that the conjugate L-subset so formed defines an L-subgroup of μ. This notion of conjugate L-subgroup has been illustrated through an example. Next, It has been shown that the image as well as the pre-image of a conjugate L-subgroup of an L-group under a group homomorphism are themselves conjugate L-subgroups. Then, a level subset characterization for conjugate L-subgroups has been developed. We end the section by exploring the properties of maximality under conjugacy.

In Section 4, we investigate the relationship between the notions of normality and conjugacy of L-subgroups. Firstly, we show that η is a normal L-subgroup of μ if and only if every conjugate L-subgroup of η is contained in η. Moreover, the equality holds if tip of η is equal to the tip of the conjugate L-subgroup. Next, we exhibit a significant relationship between the normalizer of a conjugate L-subgroup, $N({\eta }^{a_z})$, and the conjugate of the normalizer of the L-subgroup, $N(\eta {)}^{a_z}$. This has also been demonstrated with the help of an example. Next, we provide a new definition of the normalizer of an L-subgroup using the notion of conjugacy developed in this paper. We conclude the section by exhibiting this definition by an example.

2. Preliminaries

Throughout this paper, L denotes a completely distributive lattice, ‘≤’ denotes the partial ordering on L and ‘$\vee$'and ‘$\wedge$’ denote, respectively, the join and meet of the elements of L. Moreover, the maximal and minimal elements of L are denoted by 1 and 0, respectively. The reader may refer to [13] for the concept of completely distributive lattices.

The notion of a fuzzy subset of a set was introduced by Zadeh [14] in 1965. In 1967, Goguen [15] extended this concept to L-fuzzy sets. In this section, we recall the basic definitions and results associated with L-subsets that shall be used throughout this work. These definitions can be found in chapter 1 of [16].

Definition 2.1

[15]

Let X be a non-empty set. An L-subset of X is a function from X into L.

The set of L-subsets of X is called the L-power set of X and is denoted by $L^X$. For $\mu \in L^X,$ the set $\{\mu (x)\mid x\in X\}$ is called the image of μ and is denoted by Im μ. The tip and tail of μ are defined as $\underset{x\in X}{\bigvee }\mu (x)$ and $\underset{x\in X}{\bigwedge }\mu (x)$, respectively.

Definition 2.2

[14]

An L-subset μ of X is said to be contained in an L-subset η of X if $\mu (x)\le \eta (x)$ for all $x\in X$. This is denoted by $\mu \subseteq \eta$.

Definition 2.3

[14]

For a family $\{{\mu }_i\mid i\in I\}$ of L-subsets in X, where I is a non-empty index set, the union $\bigcup _{i\in I}{\mu }_i$ and the intersection $\bigcap _{i\in I}{\mu }_i$ of $\{{\mu }_i\mid i\in I\}$ are, respectively, defined by $\bigcup _{i\in I}{\mu }_i(x)=\underset{i\in I}{\bigvee }\mu (x)\mathrm{and}\bigcap _{i\in I}{\mu }_i(x)=\underset{i\in I}{\bigwedge }\mu (x)$ for each $x\in X$.

Definition 2.4

[14]

If $\mu \in L^X$ and $a\in L$, then the level subset ${\mu }_a$ of μ is defined as ${\mu }_a=\{x\in X\mid \mu (x)\ge a\}.$

For $\mu ,\nu \in L^X$, it can be verified easily that if $\mu \subseteq \nu$, then ${\mu }_a\subseteq {\nu }_a$ for each $a\in L$.

Definition 2.5

[17]

For $a\in L$ and $x\in X$, we define $a_x\in L^X$ as follows: for all $y\in X$, $a_x(y)=\{\begin{array}{ll}a& \mathrm{if}y=x,\\ 0& \mathrm{if}y\ne x.\end{array}$ $a_x$ is referred to as an L-point or L-singleton.

We say that $a_x$ is an L-point of μ if and only if $\mu (x)\ge a$ and we write $a_x\in \mu$.

Definition 2.6

[17]

Let S be a groupoid. The set product $\mu \circ \eta$ of $\mu ,\eta \in L^S$ is an L-subset of S defined by $\mu \circ \eta (x)=\underset{x=yz}{\bigvee }\{\mu (y)\wedge \eta (z)\}.$

Remark 2.1

If x cannot be factored as x = yz in S, then $\mu \circ \eta (x)$, being the least upper bound of the empty set, is zero.

It can be verified that the set product is associative in $L^S$ if S is a semigroup.

Definition 2.7

[14]

Let f be a mapping from a set X to a set Y. If $\mu \in L^X$ and $\nu \in L^Y$, then the image $f(\mu )$ of μ under f and the preimage $f^{-1}(\nu )$ of ν under f are L-subsets of Y and X respectively, defined by $f(\mu )(y)=\underset{x\in f^{-1}(y)}{\bigvee }\{\mu (x)\}$ and $f^{-1}(\nu )(x)=\nu (f(x)).$

Remark 2.2

If $f^{-1}(y)=\varphi$, then $f(\mu )(y)$, being the least upper bound of the empty set, is zero.

Let G be a group and ‘e’ be the identity element of G. For any non-empty set A, let $1_A$ be the characteristic function of A.

Definition 2.8

[2]

Let $\mu \in L^G$. Then, μ is called an L-subgroup of G if for each $x,y\in G$,

1. $\mu (xy)\ge \mu (x)\wedge \mu (y)$,

2. $\mu (x^{-1})=\mu (x)$.

The set of L-subgroups of G is denoted by $L(G)$. Clearly, the tip of an L-subgroup is attained at the identity element of G.

Theorem 2.9

[16, Lemma 1.2.5]

Let $\mu \in L^G$. Then, μ is an L-subgroup of G if and only if each non-empty level subset ${\mu }_a$ is a subgroup of G.

Theorem 2.10

[16, Theorems 1.2.10, 1.2.11]

Let $f:G\to H$ be a group homomorphism. Let $\mu \in L(G)$ and $\nu \in L(H)$. Then, $f(\mu )\in L(H)$ and $f^{-1}(\nu )\in L(G)$.

Theorem 2.11

[1]

The intersection of an arbitrary family of L-subgroups of a group is an L-subgroup of the given group.

The concept of normal fuzzy subgroup of a group was introduced by Liu [17] in 1982. We define the normal L-subgroup of a group G below:

Definition 2.12

[17]

Let $\mu \in L(G)$. Then, μ is called a normal L-subgroup of G if for all $x,y\in G$, $\mu (xy)=\mu (yx)$.

The set of normal L-subgroups of G is denoted by $NL(G)$.

Theorem 2.13

[16, Theorem 1.3.3]

Let $\mu \in L(G)$. Then, $\mu \in NL(G)$ if and only if each non-empty level subset ${\mu }_a$ is a normal subgroup of G.

Let $\eta ,\mu \in L^G$ such that $\eta \subseteq \mu$. Then, η is said to be an L-subset of μ. The set of all L-subsets of μ is denoted by $L^{\mu }$. Moreover, if $\eta ,\mu \in L(G)$ such that $\eta \subseteq \mu$, then η is said to be an L-subgroup of μ. The set of all L-subgroups of μ is denoted by $L(\mu )$. It is well known that the intersection of an arbitray family of L-subgroup of an L-group μ is again an L-subgroup of μ.

Definition 2.14

[4]

Let $\eta \in L^{\mu }$. Then, the L-subgroup of μ generated by η is defined as the smallest L-subgroup of μ which contains η. It is denoted by $⟨\eta ⟩$, that is, $⟨\eta ⟩=\cap \{{\eta }_i\in L(\mu )\mid \eta \subseteq {\eta }_i\}.$

From now onwards, μ denotes an L-subgroup of G which shall be considered as the parent L-group.

Definition 2.15

[8]

Let $\eta \in L(\mu )$ such that η is non-constant and $\eta \ne \mu$. Then, η is said to be a proper L-subgroup of μ.

Clearly, η is a proper L-subgroup of μ if and only if η has distinct tip and tail and $\eta \ne \mu$.

Definition 2.16

[6]

Let $\eta \in L(\mu )$. Let $a_0$ and $t_0$ denote the tip and tail of η, respectively. We define the trivial L-subgroup of η as follows: ${\eta }_{t_0}^{a_0}(x)=\{\begin{array}{ll}{a}_0& \mathrm{if}x=e,\\ t_0& \mathrm{if}x\ne e.\end{array}$

Theorem 2.17

[6, Theorem 2.1]

Let $\eta \in L^{\mu }$. Then, $\eta \in L(\mu )$ if and only if each non-empty level subset ${\eta }_a$ is a subgroup of ${\mu }_a$.

The normal fuzzy subgroup of a fuzzy group was introduced by Wu [18] in 1981. We note that for the development of this concept, Wu[18] preferred L-setting.

Definition 2.18

[18]

Let $\eta \in L(\mu )$. Then, we say that η is a normal L-subgroup of μ if $\eta (yx{y}^{-1})\ge \eta (x)\wedge \mu (y)\mathrm{for}\mathrm{all}x,y\in G.$

The set of normal L-subgroups of μ is denoted by $NL(\mu )$. If $\eta \in NL(\mu )$, then we write $\eta ◃\mu$.

Here, we mention that the arbitrary intersection of a family of normal L-subgroups of an L-group μ is again a normal L-subgroup of μ.

Theorem 2.19

[16, Theorem 1.4.3]

Let $\eta \in L(\mu )$. Then, $\eta \in NL(\mu )$ if and only if each non-empty level subset ${\eta }_a$ is a normal subgroup of ${\mu }_a$.

Lastly, recall the following form [3]:

Theorem 2.20

[3, Theorem 3.1]

Let $\eta \in L^{{}^{\mu }}.$ Let $a_0={\vee }_{x\in G}\{\eta (x)\}$ and define an L-subset $\hat{\eta }$ of G by $\hat{\eta }(x)=\underset{a\le a_0}{\vee }\{a\mid x\in ⟨{\eta }_a⟩\}.$ Then, $\hat{\eta }\in L(\mu )$ and $\hat{\eta }=⟨\eta ⟩$. Moreover, tip $⟨\eta ⟩=\mathrm{tip}\eta$.

3. Conjugate L-subgroups

The notion of conjugate subgroups has played an important role in the evolution of classical group theory. We are motivated to contemplate such progress in L-group theory. For this, we firstly recall the definition of conjugate of an L-subset by an L-subset from [19]:

Definition 3.1

[19]

Let $\eta ,\theta \in L^{\mu }$. Define an L-subset $\theta \eta {\theta }^{-1}$ of G as follows: For all $x\in G$, $\theta \eta {\theta }^{-1}(x)=\underset{x=zy{z}^{-1}}{\bigvee }\{\eta (y)\wedge \theta (z)\}.$ We call $\theta \eta {\theta }^{-1}$ the conjugate of η by θ. Clearly, $\theta \eta {\theta }^{-1}\subseteq \mu .$ Hence the L-subgroup $⟨\theta \eta {\theta }^{-1}⟩\in L(\mu )$ and is denoted by ${\eta }^{\theta }.$ If η and θ are L-subgroups of μ, then clearly, in view of Theorem 2.20, the tip of the L-subgroup ${\eta }^{\theta }$ is ${\eta }^{\theta }(e)=\eta (e)\wedge \theta (e)$.

The notion of conjugate fuzzy subgroups has appeared in [11]. However, in this definition the notion of conjugate fuzzy subgroup of an ordinary group has been discussed rather than conjugate fuzzy subgroups of a fuzzy group. Below, we introduce the notion of conjugate L-subgroups of an L-group. Firstly, observe that in view of the Definition 3.1, the conjugate of an L-subgroup η by an L-point $a_z\in \mu$ is given by: ${\eta }^{a_z}(x)=a\wedge \eta (zx{z}^{-1})\text{for all}x\in G.$ Here, we show that the conjugate of an L-subgroup by an L-point in μ is an L-subgroup of μ.

Theorem 3.2

Let $\eta \in L(\mu )$ and $a_z$ be an L-point of μ. Then, ${\eta }^{a_z}$ is an L-subgroup of μ.

Proof.

Let $x,y\in G$. Then, $\begin{array}{rl}{\eta }^{a_z}(xy)& =a\wedge \eta (z(xy)z^{-1})\\ & =a\wedge \eta ((zx{z}^{-1})(zy{z}^{-1}))\\ & \ge a\wedge \eta (zx{z}^{-1})\wedge \eta (zy{z}^{-1})(\text{since }\eta \text{ is an }L\text{-subgroup of }\mu )\\ & =(a\wedge \eta (zx{z}^{-1}))\wedge (a\wedge \eta (zy{z}^{-1}))\\ & ={\eta }^{a_z}(x)\wedge {\eta }^{a_z}(y).\end{array}$ Similarly, we can verify that ${\eta }^{a_z}(x^{-1})={\eta }^{a_z}(x)$. Thus ${\eta }^{a_z}\in L(G).$ Now, to show that ${\eta }^{a_z}\subseteq \mu$, let $x\in G$. Then, $\begin{array}{rl}\mu (x)& =\mu (z^{-1}(zx{z}^{-1})z)\\ & \ge \mu (z)\wedge \mu (zx{z}^{-1})\\ & \ge a\wedge \eta (zx{z}^{-1})(\text{since }{a}_z\in \mu \text{ and }\eta \subseteq \mu )\\ & ={\eta }^{a_z}(x).\end{array}$ This proves that ${\eta }^{a_z}\in L(\mu )$.

Remark 3.1

Clearly, $\mathrm{tip}({\eta }^{a_z})=a\wedge \mathrm{tip}(\eta )$, since ${\eta }^{a_z}(e)=a\wedge \eta (ze{z}^{-1})=a\wedge \eta (e)$.

We demonstrate the notion of conjugate L-subgroups with the following example:

Example 3.1

Let $M=\{l,f,a,b,c,d,u\}$ be the lattice given by Figure 1. Let $G=S_4$, the group of all permutations of the set $\{1,2,3,4\}$ with the identity element e. Let $D_4^1=⟨(24),(1234)⟩$, $D_4^2=⟨(12),(1324)⟩$, and $D_4^3=⟨(23),(1342)⟩$ denote the dihedral subgroups of G and $V_4=\{e,(12)(34),(13)(24),(14)(23)\}$ denote the Klein-4 subgroup of G. Define the L-subgroup μ of G as follows: $\mu (x)=\{\begin{array}{ll}u& \mathrm{if}x\in V_4,\\ d& \mathrm{if}x\in S_4\setminus V_4.\end{array}$ Next, let η be the L-subset of μ be defined by $\eta (x)=\{\begin{array}{ll}u& \mathrm{if}x=e,\\ d& \mathrm{if}x\in V_4\setminus \{e\},\\ a& \mathrm{if}x\in D_4^1\setminus V_4,\\ b& \mathrm{if}x\in D_4^2\setminus V_4,\\ c& \mathrm{if}x\in D_4^3\setminus V_4,\\ l& \mathrm{if}x\in S_4\setminus \underset{i=1}{\overset{3}{\cup }}{D}_4^i.\end{array}$ By Theorem 2.17, η is an L-subgroup of μ. Note that $d_{(123)}\in \mu$. We determine ${\eta }^{d_{(123)}}$.

By definition, ${\eta }^{d_{(123)}}(x)=d\wedge \eta ((123)x(132))$ for all $x\in S_4$. Hence ${\eta }^{d_{(123)}}(x)=\{\begin{array}{ll}d\wedge u& \mathrm{if}x=e,\\ d\wedge d& \mathrm{if}x\in V_4\setminus \{e\},\\ d\wedge b& \mathrm{if}x\in D_4^1\setminus V_4,\\ d\wedge c& \mathrm{if}x\in D_4^2\setminus V_4,\\ d\wedge a& \mathrm{if}x\in D_4^3\setminus V_4,\\ d\wedge l& \mathrm{if}x\in S_4\setminus \underset{i=1}{\overset{3}{\cup }}{D}_4^i.\end{array}$ Thus ${\eta }^{d_{(123)}}(x)=\{\begin{array}{ll}d& \mathrm{if}x\in V_4,\\ b& \mathrm{if}x\in D_4^1\setminus V_4,\\ c& \mathrm{if}x\in D_4^2\setminus V_4,\\ a& \mathrm{if}x\in D_4^3\setminus V_4,\\ l& \mathrm{if}x\in S_4\setminus \underset{i=1}{\overset{3}{\cup }}{D}_4^i.\end{array}$

short-legendFigure 1.

Below, we discuss the set product of conjugate L-subgroups:

Theorem 3.3

Let $\eta ,\nu \in L(\mu )$ and $a_z$ be an L-point of μ. Then, $(\eta \circ \nu {)}^{a_z}={\eta }^{a_z}\circ {\nu }^{a_z}.$

Proof.

Let $g\in G$. Then, $\begin{array}{rl}({\eta }^{a_z}\circ {\nu }^{a_z})(g)& =\vee \{{\eta }^{a_z}(x)\wedge {\nu }^{a_z}(y)\mid xy=g\}\\ & =\vee \{\{a\wedge \eta (zx{z}^{-1})\}\wedge \{a\wedge \nu (zy{z}^{-1})\}\mid xy=g\}\\ & =a\wedge \{\vee \{\eta (zx{z}^{-1})\wedge \nu (zy{z}^{-1})\}\mid (zx{z}^{-1})(zy{z}^{-1})=zg{z}^{-1}\}\\ & =a\wedge \{\vee \{\eta (x_1)\wedge \nu (y_1)\}\mid x_1{y}_1=zg{z}^{-1}\}\\ & =a\wedge (\eta \circ \nu )(zg{z}^{-1})\\ & =(\eta \circ \nu {)}^{a_z}(g).\end{array}$ Hence the result.

In Theorems 3.4 and 3.5, we study the properties of conjugate L-subgroups under group homomorphisms.

Theorem 3.4

Let $f:G\to H$ be a group homomorphism and $\mu \in L(G)$. Then, for $\eta \in L(\mu )$ and $a_z\in \mu$, the L-subgroup $f({\eta }^{a_z})$ is a conjugate L-subgroup of $f(\eta )$ in $f(\mu )$. In fact, $f({\eta }^{a_z})=f(\eta {)}^{a_{f(z)}}.$

Proof.

Firstly, note that $a_{f(z)}\in f(\mu )$, since $(f(\mu ))(f(z))=\vee \{\mu (x)\mid f(x)=f(z)\}\ge \mu (z)\ge a.$ Next, let $y\in H$. Then, $\begin{array}{rl}f({\eta }^{a_z})(y)& =\vee \{{\eta }^{a_z}(x)\mid f(x)=y\}\\ & =\vee \{a\wedge \eta (zx{z}^{-1})\mid f(x)=y\}\\ & =\vee \{a\wedge \eta (zx{z}^{-1})\mid f(zx{z}^{-1})=f(z)yf(z{)}^{-1}\}\\ & =a\wedge \{\vee \{\eta (u)\mid f(u)=f(z)yf(z{)}^{-1}\}\}\\ & (\text{since }L\text{ is completely distributive})\\ & =a\wedge f(\eta )(f(z)yf(z{)}^{-1})\\ & =f(\eta {)}^{a_{f(z)}}(y).\end{array}$ Hence the result.

Theorem 3.5

Let $f:G\to H$ be a surjective group homomorphism and let $\mu \in L(H)$. Then, for $\eta \in L(\mu )$ and $a_z\in \mu$, the L-subgroup $f^{-1}({\eta }^{a_z})$ is a conjugate L-subgroup of $f^{-1}(\eta )$ in $f^{-1}(\mu )$. In fact, $f^{-1}({\eta }^{a_z})=f^{-1}(\eta {)}^{a_s},$ where $s\in f^{-1}(z)$.

Proof.

Let $s\in f^{-1}(z)$. Firstly, note that $a_s\in f^{-1}(\mu )$, since $f^{-1}(\mu )(s)=\mu (f(s))=\mu (z)\ge a.$ Next, let $x\in G$. Then, $\begin{array}{rl}{f}^{-1}({\eta }^{a_z})(x)& ={\eta }^{a_z}(f(x))\\ & =a\wedge \eta (zf(x)z^{-1})\\ & =a\wedge \eta (f(s)f(x)f(s{)}^{-1})\\ & =a\wedge \eta (f(sx{s}^{-1}))\\ & =a\wedge f^{-1}(\eta )(sx{s}^{-1})\\ & =f^{-1}(\eta {)}^{a_s}(x).\end{array}$ This proves that $f^{-1}({\eta }^{a_z})=f^{-1}(\eta {)}^{a_s}$.

In Theorem 3.6, we discuss the level subset characterization for conjugate L-subgroups.

Theorem 3.6

Let $\eta ,\nu \in L(\mu )$ and $a\in L$ such that $tip(\nu )=a\wedge tip(\eta )$. Then, $\nu ={\eta }^{a_z}$ for $a_z\in \mu$ if and only if ${\nu }_t={{\eta }_t}^{z^{-1}}$ for all $t\le tip(\nu )$.

Proof.

(⇒) Let $\nu ={\eta }^{a_z}$. Let $t\le \mathrm{tip}(\nu )=a\wedge \mathrm{tip}(\eta )$ and let $x\in {\nu }_t$. Then, $\nu (x)\ge t$, that is, ${\eta }^{a_z}(x)=a\wedge \eta (zx{z}^{-1})\ge a\wedge \eta (e)=\mathrm{tip}(\nu )\ge t.$ This implies $\eta (zx{z}^{-1})\ge t$, that is, $zx{z}^{-1}\in {\eta }_t$. Thus $x\in {{\eta }_t}^{z^{-1}}$. Hence ${\nu }_t\subseteq {{\eta }_t}^{z^{-1}}.$ To show the reverse inclusion, let $x\in {{\eta }_t}^{z^{-1}}$. Then, $x=z^{-1}gz$ for some $g\in {\eta }_t$. This implies $zx{z}^{-1}=g\in {\eta }_t$, that is, $\eta (zx{z}^{-1})\ge t$. Moreover, by assumption, $t\le a$. Hence $\nu (x)=a\wedge \eta (zx{z}^{-1})\ge t.$ Thus $x\in {\nu }_t$ and we conclude that ${\nu }_t={\eta }_t^{z^{-1}}$.

(⇐) Suppose that ${\nu }_t={{\eta }_t}^{z^{-1}}$ for all $t\le \mathrm{tip}(\nu )$. Let $x\in G$ and let $\nu (x)=b$. Then, $b\le \mathrm{tip}(\nu )$ and by the hypothesis, ${\nu }_b={{\eta }_b}^{z^{-1}}$. Thus $x\in {{\eta }_b}^{z^{-1}}$, that is, $x=z^{-1}gz$ for some $g\in {\eta }_b$. This implies $zx{z}^{-1}=g\in {\eta }_b$, and hence $\eta (zx{z}^{-1})\ge b=\nu (x)$. Thus ${\eta }^{a_z}(x)=a\wedge \eta (zx{z}^{-1})\ge a\wedge \nu (x)=\nu (x)$ and we conclude that $\nu \subseteq {\eta }^{a_z}$. For the reverse inclusion, let $x\in G$ and let $b=a\wedge \eta (zx{z}^{-1})\le a\wedge \eta (e)=\mathrm{tip}\nu$. Then, by hypothesis, ${\nu }_b={{\eta }_b}^{z^{-1}}$, or equivalently, ${\eta }_b={{\nu }_b}^z$. Now, $zx{z}^{-1}\in {\eta }_b={{\nu }_b}^z$ implies $zx{z}^{-1}=zg{z}^{-1}$ for some $g\in {\nu }_b$. Thus $x=g\in {\nu }_b$ and hence $\nu (x)\ge b=a\wedge \eta (zx{z}^{-1})={\eta }^{a_z}(x).$ Thus ${\eta }^{a_z}\subseteq \nu$ and we conclude that $\nu ={\eta }^{a_z}$.

Theorem 3.7

Let H and K be subgroups of G. Then, K is conjugate to H in G if and only if $1_K$ is conjugate to $1_H$ as L-subgroups of $1_G$.

Proof.

$(\Rightarrow )$ Since H and K are conjugate in G, there exists $z\in G$ such that $K=H^z$. Since $z\in G$, $1_{z^{-1}}\in 1_G$. We claim that $1_K=(1_H{)}^{1_{z^{-1}}}$.

Let $x\in G$. If $x\notin K$, then $1_K(x)=0\le (1_H{)}^{1_{z^{-1}}}(x)$. If $x\in K$, then $x=zh{z}^{-1}$ for some $h\in H$. This implies $(1_H{)}^{1_{z^{-1}}}(x)=1\wedge 1_H(z^{-1}xz)=1=1_K(x).$ So we conclude that $1_K\subseteq (1_H{)}^{1_{z^{-1}}}$. Similarly we can prove that $(1_H{)}^{1_{z^{-1}}}\subseteq 1_K$.

(⇐) Since $1_H$ and $1_K$ are conjugate L-subgroups in $1_G$, there exists $a_z\in 1_G$ such that $1_K={1_H}^{a_z}$. Note that a>0, for if a = 0, then $1_K(x)=a\wedge 1_H(zx{z}^{-1})=0$ for all $x\in G$, which contradicts the fact that $1_K(e)=1$. We claim that $K=H^{z^{-1}}$.

Let $x\in K$. Then, $1_K(x)=1$. This implies $1={1_H}^{a_z}(x)=a\wedge 1_H(zx{z}^{-1}).$ Thus a = 1 and $1_H(zx{z}^{-1})=1$. Hence $zx{z}^{-1}\in H$, that is, $x\in H^{z^{-1}}$. Consequently, $K\subseteq H^{z^{-1}}$. For the reverse inclusion, let $x\in H^{z^{-1}}$. This implies, ${1_H}^{a_z}(x)=a\wedge 1_H(zx{z}^{-1})=a.$ Thus $1_K(x)=a>0$. It follows that $1_K(x)=1$, that is, $x\in K$. Therefore, $H^{z^{-1}}\subseteq K$. Hence the result.

The maximal L-subgroup of an L-group has been introduced in [9]. We recall the definition below:

Definition 3.8

[9]

Let $\mu \in L(G)$. A proper L-subgroup η of μ is said to be a maximal L-subgroup of μ if, whenever $\eta \subseteq \theta \subseteq \mu$ for some $\theta \in L(\mu )$, then either $\theta =\eta$ or $\theta =\mu$.

Here, we discuss conjugate of a maximal L-subgroup of an L-group.

Theorem 3.9

Let L be a chain. Let η be a maximal L-subgroup of an L-group μ and $a_z$ be an L-point of μ. Then, either ${\eta }^{a_z}={\mu }^{a_z}$ or ${\eta }^{a_z}$ is a maximal L-subgroup of ${\mu }^{a_z}$.

Proof.

If ${\eta }^{a_z}={\mu }^{a_z}$, then there is nothing to prove. So, let ${\eta }^{a_z}⊊{\mu }^{a_z}$. Suppose that ${\eta }^{a_z}$ is not a maximal L-subgroup of ${\mu }^{a_z}$. Then, there exists an L-subgroup θ of ${\mu }^{a_z}$ such that (1) ${\eta }^{a_z}⊊\theta ⊊{\mu }^{a_z}.$(1) We shall construct an L-subgroup γ of μ such that $\eta ⊊\gamma ⊊\mu$ which will contradict maximality of η. Define $\gamma :G\to L$ as follows: $\gamma (x)=\eta (x)\vee \theta (z^{-1}xz)\text{for all}x\in G.$ Firstly, we claim that $\eta \subseteq \gamma \subseteq \mu$. Clearly, $\eta \subseteq \gamma$. Next, as $\theta ⊊{\mu }^{a_z}$, it follows that for all $x\in G$, $\theta (z^{-1}xz)\le {\mu }^{a_z}(z^{-1}xz)=a\wedge \mu (x)\le \mu (x)$. Also, $\eta (x)\le \mu (x)$. Consequently, $\gamma (x)=\eta (x)\vee \theta (z^{-1}xz)\le \mu (x)\text{for all}x\in G.$ This proves the claim. Now, we shall show that $\gamma \in L(\mu )$. So let $x,y\in G$ and consider $\begin{array}{rl}\gamma (xy)& =\eta (xy)\vee \theta (z^{-1}xyz)\\ & =\eta (xy)\vee \theta ((z^{-1}xz)(z^{1}yz))\\ & \ge \{\eta (x)\wedge \eta (y)\}\vee \{\theta (z^{-1}xz)\wedge \theta ((z^{-1}yz))\}\\ & \ge \{\eta (x)\vee \theta (z^{-1}xz)\}\wedge \{\eta (y)\vee \theta (z^{-1}yz)\}\\ & =\gamma (x)\wedge \gamma (y).\end{array}$ Similarly, we can verify that $\gamma (x^{-1})=\gamma (x)$. Thus $\gamma \in L(\mu )$. Next, we claim that $\eta ⊊\gamma ⊊\mu .$ In view of (1(1) ${\eta }^{a_z}⊊\theta ⊊{\mu }^{a_z}.$(1) ), there exist $x_1,x_2\in G$ such that ${\eta }^{a_z}(x_1)<\theta (x_1)\mathrm{and}\theta (x_2)<{\mu }^{a_z}(x_2).$ This implies (2) $a\wedge \eta (z{x}_1{z}^{-1})<\theta (x_1)\mathrm{and}\theta (x_2)<a\wedge \mu (z{x}_2{z}^{-1})\le a.$(2) Note that by (1(1) ${\eta }^{a_z}⊊\theta ⊊{\mu }^{a_z}.$(1) ), $\mathrm{tip}{\eta }^{a_z}=a\wedge \eta (e)\le \mathrm{tip}\theta =\theta (e)\le \mathrm{tip}{\mu }^{a_z}=a\wedge \mu (e)\le a$. Hence ${\eta }^{a_z}(x_1)<\theta (x_1)\le \theta (e)\le a$ implies $a\wedge \eta (z{x}_1{z}^{-1})<a$. As L is a chain, we must have $a\wedge \eta (z{x}_1{z}^{-1})=\eta (z{x}_1{z}^{-1})$. Hence by (2(2) $a\wedge \eta (z{x}_1{z}^{-1})<\theta (x_1)\mathrm{and}\theta (x_2)<a\wedge \mu (z{x}_2{z}^{-1})\le a.$(2) ), (3) $\eta (z{x}_1{z}^{-1})<\theta (x_1).$(3) Now, consider $\begin{array}{rl}\gamma (z{x}_1{z}^{-1})& =\eta (z{x}_1{z}^{-1})\vee \theta (z^{-1}(z{x}_1{z}^{-1})z)\\ & =\eta (z{x}_1{z}^{-1})\vee \theta (x_1)\\ & =\theta (x_1)(\text{by (3)})\\ & >\eta (z{x}_1{z}^{-1}).\end{array}$ This implies $\eta ⊊\gamma$. Further, by (2(2) $a\wedge \eta (z{x}_1{z}^{-1})<\theta (x_1)\mathrm{and}\theta (x_2)<a\wedge \mu (z{x}_2{z}^{-1})\le a.$(2) ), we have (4) $\theta (x_2)<{\mu }^{a_z}(x_2)=a\wedge \mu (z{x}_2{z}^{-1})\le \mu (z{x}_2{z}^{-1}).$(4) In view of (1(1) ${\eta }^{a_z}⊊\theta ⊊{\mu }^{a_z}.$(1) ) (5) ${\eta }^{a_z}(x_2)=a\wedge \eta (z{x}_2{z}^{-1})\le \theta (x_2).$(5) Since L is a chain, either $a\wedge \eta (z{x}_2{z}^{-1})=a$ or $a\wedge \eta (z{x}_2{z}^{-1})=\eta (z{x}_2{z}^{-1})$. In the first case, we have $a\le \theta (x_2)\le \theta (e)\le a.$ Thus $\theta (x_2)=a$. However, by (4(4) $\theta (x_2)<{\mu }^{a_z}(x_2)=a\wedge \mu (z{x}_2{z}^{-1})\le \mu (z{x}_2{z}^{-1}).$(4) ), $\theta (x_2)<{\mu }^{a_z}(x_2)=a\wedge \mu (z{x}_2{z}^{-1})\le a$. That is $\theta (x_2)<a$. So there is a contradiction. Hence we must have, $a\wedge \eta (z{x}_2{z}^{-1})=\eta (z{x}_2{z}^{-1})$. Now by (5(5) ${\eta }^{a_z}(x_2)=a\wedge \eta (z{x}_2{z}^{-1})\le \theta (x_2).$(5) ), $\eta (z{x}_2{z}^{-1})\le \theta (x_2).$ Therefore, by (2(2) $a\wedge \eta (z{x}_1{z}^{-1})<\theta (x_1)\mathrm{and}\theta (x_2)<a\wedge \mu (z{x}_2{z}^{-1})\le a.$(2) ), $\gamma (z{x}_2{z}^{-1})=\eta (z{x}_2{z}^{-1})\vee \theta (x_2)=\theta (x_2)<\mu (z{x}_2{z}^{-1}).$ Consequently, $\gamma ⊊\mu$. This completes the proof of the claim. However, this a contradiction to the maximality of η in μ. Hence the result.

4. Conjugacy and Normality of L-subgroups

In this section, we explore the inter-connections between the concepts of conjugacy and normality of L-subgroups. We prove significant results pertaining to the notions of conjugate L-subgroups and normalizer [7] and explore the various similarities as well as peculiarities of these concepts compared to their group theoretic counterparts. We end this section by providing a new definition of the normalizer using the concept of conjugacy of L-subgroups. Thus this section demonstrates the compatibility of the conjugacy of L-subgroups with the several concepts developed so far in the study of L-subgroups of L-groups.

Proposition 4.1

Let $\eta \in L(\mu )$. Then, η is a normal L-subgroup of μ if and only if ${\eta }^{a_z}\subseteq \eta$ for every L-point $a_z\in \mu$. Moreover, if $\eta \in NL(\mu )$ and $\mathrm{tip}({\eta }^{a_z})=\mathrm{tip}(\eta )$, then ${\eta }^{a_z}=\eta$.

Proof.

Let η be a normal L-subgroup of μ and $a_z$ be an L-point of μ. Then, for all $x\in G$, $\begin{array}{rl}\eta (x)& =\eta (z^{-1}(zx{z}^{-1})z)\\ & \ge \eta (zx{z}^{-1})\wedge \mu (z)(\text{since }\eta \text{ is normal in }\mu )\\ & \ge \eta (zx{z}^{-1})\wedge a(\text{since }{a}_z\in \mu )\\ & ={\eta }^{a_z}(x).\end{array}$ Hence ${\eta }^{a_z}\subseteq \eta$. Conversely, suppose that ${\eta }^{a_z}\subseteq \eta$ for all L-points $a_z\in \mu$. Let $x,g\in G$ and let $a=\mu (g)$. Then $a_{g^{-1}}\in \mu$ and by the hypothesis, ${\eta }^{a_{g^{-1}}}\subseteq \eta$. Thus $\begin{array}{rl}\eta (gx{g}^{-1})& \ge {\eta }^{a_{g^{-1}}}(gx{g}^{-1})\\ & =a\wedge \eta (g^{-1}(gx{g}^{-1})g)\\ & =\mu (g)\wedge \eta (x).\end{array}$ Therefore η is a normal L-subgroup of μ.

Next, let η be a normal L-subgroup of μ and $a_z$ be an L-point of μ such that $\mathrm{tip}({\eta }^{a_z})=\mathrm{tip}(\eta )$. Then, by Remark 3.1, $a\ge \mathrm{tip}(\eta )$. Thus for all $x\in G$, $\begin{array}{rl}{\eta }^{a_z}(x)& =a\wedge \eta (zx{z}^{-1})\\ & \ge a\wedge \eta (x)\wedge \mu (z)(\text{since }\eta \text{ is normal in }\mu )\\ & =a\wedge \eta (x)(\text{since }{a}_z\in \mu )\\ & =\eta (x).(\text{since }a\ge \eta (e))\end{array}$ Hence $\eta \subseteq {\eta }^{a_z}$ and we conclude that $\eta ={\eta }^{a_z}$.

In [7], Ajmal and Jahan have introduced the notion of the normalizer of an L-subgroup by introducing the coset of an L-subgroup with respect to an L-point. We recall these concepts below:

Definition 4.2

[7]

Let $\eta \in L(\mu )$ and let $a_x$ be an L-point of μ. The left (respectively, right) coset of η in μ with respect to $a_x$ is defined as the set product $a_x\circ \eta$ ($\eta \circ a_x$).

From the definition of set product of two L-subsets, it can be easily seen that for all $z\in G$, $(a_x\circ \eta )(z)=a\wedge \eta (x^{-1}z)\mathrm{and}(\eta \circ a_x)(z)=a\wedge \eta (z{x}^{-1}).$

Definition 4.3

[7]

Let $\eta \in L(\mu )$. The normalizer of η in μ, denoted by $N(\eta$), is the L-subgroup defined as follows: $N(\eta )=\bigcup \{a_x\in \mu \mid a_x\circ \eta =\eta \circ a_x\}.$ $N(\eta )$ is the largest L-subgroup of μ such that η is a normal L-subgroup of $N(\eta )$. Also, it has been established in [7] that η is a normal L-subgroup of μ if and only if $N(\eta )=\mu$.

Below, we demonstrate the conjugate of the normalizer $N(\eta )$ of the L-subgroup η:

Theorem 4.4

Let $\eta \in L(\mu )$ and $a_z$ be an L-point of μ. Then, for all $g\in G$, $N(\eta {)}^{a_z}(g)=a\wedge N({\eta }^{a_z})(g).$

Proof.

Let $g\in G$. Then, $\begin{array}{rl}N(\eta {)}^{a_z}(g)& =a\wedge N(\eta )(zg{z}^{-1})\\ & =a\wedge \{\vee \{b\mid b_{zg{z}^{-1}}\in \mu \mathrm{and}{b}_{zg{z}^{-1}}\circ \eta =\eta \circ b_{zg{z}^{-1}}\}\}\\ & =\vee \{a\wedge b\mid b_{zg{z}^{-1}}\in \mu \mathrm{and}{b}_{zg{z}^{-1}}\circ \eta =\eta \circ b_{zg{z}^{-1}}\},\end{array}$ since L is completely distributive. Similarly, $N({\eta }^{a_z})(g)=\vee \{b\mid b_g\in \mu \mathrm{and}{b}_g\circ {\eta }^{a_z}={\eta }^{a_z}\circ b_g\}.$ Let $L_1=\{b\in L\mid b_g\in \mu \text{ and }{b}_g\circ {\eta }^{a_z}={\eta }^{a_z}\circ b_g\}$ and $L_2=\{b\in L\mid b_{zg{z}^{-1}}\in \mu \text{ and }{b}_{zg{z}^{-1}}\circ \eta =\eta \circ b_{zg{z}^{-1}}\}.$ We claim that if $b\in L_2$, then $a\wedge b\in L_1$. For this, let $b\in L_2$. Then, $b_{zg{z}^{-1}}\in \mu \mathrm{and}{b}_{zg{z}^{-1}}\circ \eta =\eta \circ b_{zg{z}^{-1}}.$ Firstly, since $b_{zg{z}^{-1}}\in \mu$, $\mu (zg{z}^{-1})\ge b$. This, and the fact that $a_z\in \mu$, implies $\mu (g)=\mu (z^{-1}(zg{z}^{-1})z)\ge \mu (z)\wedge \mu (zg{z}^{-1})\ge a\wedge b.$ Hence $(a\wedge b{)}_g\in \mu$. Next, let $x\in G$. Since $b_{zg{z}^{-1}}\circ \eta =\eta \circ b_{zg{z}^{-1}}$, we have $(b_{zg{z}^{-1}}\circ \eta )(x)=(\eta \circ b_{zg{z}^{-1}})(x),$ that is, $b\wedge \eta ((zg{z}^{-1}{)}^{-1}x)=b\wedge \eta (x(zg{z}^{-1}{)}^{-1}).$ Hence $b\wedge \eta (z{g}^{-1}{z}^{-1}x)=b\wedge \eta (xz{g}^{-1}{z}^{-1}).$ This implies $a\wedge (b\wedge \eta (z{g}^{-1}{z}^{-1}x))=a\wedge (b\wedge \eta (xz{g}^{-1}{z}^{-1})),$ or equivalently (6) $(a\wedge b)\wedge (a\wedge \eta (z{g}^{-1}{z}^{-1}x))=(a\wedge b)\wedge (a\wedge \eta (xz{g}^{-1}{z}^{-1}))$(6) for all $x\in G$. Now, let y be any arbitrary element of G. Since Equation (6(6) $(a\wedge b)\wedge (a\wedge \eta (z{g}^{-1}{z}^{-1}x))=(a\wedge b)\wedge (a\wedge \eta (xz{g}^{-1}{z}^{-1}))$(6) ) holds for all $x\in G$, taking $x=zy{z}^{-1}$, we get $(a\wedge b)\wedge (a\wedge \eta (z{g}^{-1}y{z}^{-1}))=(a\wedge b)\wedge (a\wedge \eta (zy{g}^{-1}{z}^{-1})).$ This implies $(a\wedge b)\wedge {\eta }^{a_z}(g^{-1}y)=(a\wedge b)\wedge {\eta }^{a_z}(y{g}^{-1}).$ Therefore $((a\wedge b{)}_g\circ {\eta }^{a_z})(y)=({\eta }^{a_z}\circ (a\wedge b{)}_g)(y).$ Since, by assumption, y is an arbitrary element of G, we conclude that $(a\wedge b{)}_g\circ {\eta }^{a_z}={\eta }^{a_z}\circ (a\wedge b{)}_g.$ Therefore we have shown that $(a\wedge b{)}_g\in \mu \mathrm{and}(a\wedge b{)}_g\circ {\eta }^{z_x}={\eta }^{a_z}\circ (a\wedge b{)}_g.$ Hence $(a\wedge b)\in L_1$. Finally, $\begin{array}{rl}N(\eta {)}^{a_z}(g)& =a\wedge N(\eta )(zg{z}^{-1})\\ & =a\wedge (\vee \{b\mid b\in L_2\})\\ & =a\wedge \{\vee \{a\wedge b\mid b\in L_2\}\}\\ & \le a\wedge \{\vee \{c\mid c\in L_1\}\}\\ & =a\wedge N({\eta }^{a_z})(g).\end{array}$ For the reverse inequality, let $b\in L_1$. We show that $a\wedge b\in L_2$. Firstly, since $b\in L_1$, $b_g\in \mu$. Thus $\mu (zg{z}^{-1})\ge \mu (z)\wedge \mu (g)\ge a\wedge b.$ Therefore $(a\wedge b{)}_{zg{z}^{-1}}\in \mu$. Next, let $x\in G$. Since $b_g\circ {\eta }^{a_z}={\eta }^{a_z}\circ b_g$, we have $b\wedge {\eta }^{a_z}(g^{-1}x)=b\wedge {\eta }^{a_z}(x{g}^{-1}),$ that is, $b\wedge (a\wedge \eta (z{g}^{-1}x{z}^{-1}))=b\wedge (a\wedge \eta (zx{g}^{-1}{z}^{-1})).$ Thus for all $x\in G$, (7) $(a\wedge b)\wedge \eta (z{g}^{-1}x{z}^{-1})=(a\wedge b)\wedge \eta (zx{g}^{-1}{z}^{-1}).$(7) Let y be any arbitrary element of G. Since Equation (7(7) $(a\wedge b)\wedge \eta (z{g}^{-1}x{z}^{-1})=(a\wedge b)\wedge \eta (zx{g}^{-1}{z}^{-1}).$(7) ) holds for all $x\in G$, taking $x=z^{-1}yz$, we get $(a\wedge b)\wedge \eta (z{g}^{-1}{z}^{-1}y)=(a\wedge b)\wedge \eta (yz{g}^{-1}{z}^{-1}),$ or equivalently, $((a\wedge b{)}_{zg{z}^{-1}}\circ \eta )(y)=(\eta \circ (a\wedge b{)}_{zg{z}^{-1}})(y).$ Since, by assumption, y is an arbitrary element of G, we conclude that $(a\wedge b{)}_{zg{z}^{-1}}\circ \eta =\eta \circ (a\wedge b{)}_{zg{z}^{-1}}.$ Hence we have shown that $(a\wedge b{)}_{zg{z}^{-1}}\in \mu$ and $(a\wedge b{)}_{zg{z}^{-1}}\circ \eta =\eta \circ (a\wedge b{)}_{zg{z}^{-1}}$. Thus $(a\wedge b)\in L_2$. Therefore $\begin{array}{rl}N(\eta {)}^{a_z}(g)& =a\wedge N(\eta )(zg{z}^{-1})\\ & =a\wedge (\vee \{b\mid b\in L_2\})\\ & \ge a\wedge \{\vee \{a\wedge b\mid b\in L_1\}\}\\ & =a\wedge \{\vee \{c\mid c\in L_1\}\}\\ & =a\wedge N({\eta }^{a_z})(g).\end{array}$ We conclude that $N(\eta {)}^{a_z}(g)=a\wedge N({\eta }^{a_z})(g).\text{■}$

Corollary 4.5

Let H be a subgroup of a group G and $x\in G$. Then, $N({(1_H)}^{1_x})=(N(1_H){)}^{1_x}.$

Example 4.1

Let G be the dihedral group $D_{16}$ of order 16, that is, $D_{16}=⟨r,s|r^8=s^2=e,rs=s{r}^{-1}⟩.$ The dihedral group $D_8$ of order 8 be the dihedral subgroup of $D_{16}$, where $D_8=⟨r^2,s|(r^2{)}^4=s^2=e⟩.$ Define $\mu :D_{16}\to [0,1]$ as follows: for all $z\in D_{16}$, $\mu (z)=\{\begin{array}{ll}\frac{1}{2}& \mathrm{if}z\in D_8,\\ \frac{1}{8}& \mathrm{if}z\in D_{16}\setminus D_8.\end{array}$ By Theorem 2.9, $\mu \in L(D_{16})$. Now, we define $\eta \in L^{\mu }$ as follows: $\eta (z)=\{\begin{array}{ll}\frac{1}{4}& \mathrm{if}z\in ⟨s⟩,\\ \frac{1}{16}& \mathrm{if}z\in D_8\setminus ⟨s⟩,\\ \frac{1}{32}& \mathrm{if}z\in D_{16}\setminus D_8.\end{array}$ Clearly, by Theorem 2.17, $\eta \in L(\mu )$. Now, let $a=1/12$. Then, $a_r\in \mu$, since $\mu (r)=1/8\ge 1/12=a$. We evaluate $N({\eta }^{a_r})$ and $N(\eta {)}^{a_r}$ and show that $N(\eta {)}^{a_r}(z)=a\wedge N({\eta }^{a_r})(z)$ for all $z\in D_{16}$.

Firstly, if H denotes the subgroup $\{e,r^4,s,s{r}^4\}$ of $D_{16}$, then it is easy to see that $N(\eta )(z)=\{\begin{array}{ll}\frac{1}{2}& \mathrm{if}z\in H,\\ \frac{1}{16}& \mathrm{if}z\in D_{16}\setminus H.\end{array}$ Thus the conjugate L-subgroup $N(\eta {)}^{a_r}$ of $N(\eta )$, defined as $N(\eta {)}^{a_r}(z)=a\wedge N(\eta )(rz{r}^{-1}),$ is given by $N(\eta {)}^{a_r}(z)=\{\begin{array}{ll}\frac{1}{12}& \mathrm{if}z\in H,\\ \frac{1}{16}& \mathrm{if}z\in D_{16}\setminus H.\end{array}$ Now, the conjugate L-subgroup ${\eta }^{a_r}$ of η is given by ${\eta }^{a_r}(z)=\{\begin{array}{ll}\frac{1}{12}& \mathrm{if}z\in ⟨s{r}^6⟩,\\ \frac{1}{16}& \mathrm{if}z\in D_8\setminus ⟨s{r}^2⟩,\\ \frac{1}{32}& \mathrm{if}z\in D_{16}\setminus D_8.\end{array}$ Thus the normalizer of $N({\eta }^{a_r})$ is $N({\eta }^{a_r})(z)=\{\begin{array}{ll}\frac{1}{2}& \mathrm{if}z\in H,\\ \frac{1}{16}& \mathrm{if}z\in D_{16}\setminus H.\end{array}$ From this, we can easily see that $N(\eta {)}^{a_r}(z)=a\wedge N({\eta }^{a_r})(z)$ for all $z\in D_{16}$.

Proposition 4.6

Let $\eta ,\nu \in L(\mu )$. Then, $\eta \subseteq \nu$ if and only if ${\eta }_t\subseteq {\nu }_t$ for all $t\le tip(\eta )$.

Lemma 4.7

Let η be an L-subgroup of μ and $a_z\in \mu$. Then, ${\eta }^{a_z}\subseteq \eta$ if and only if ${\eta }^{a_{z^{-1}}}\subseteq \eta$.

Proof.

Suppose that ${\eta }^{a_z}\subseteq \eta$. Then, by Proposition 4.6, $({\eta }^{a_z}{)}_t\subseteq {\eta }_t$ for all $t\le \mathrm{tip}({\eta }^{a_z})=a\wedge \eta (e)$. By Theorem 3.6, $({\eta }^{a_z}{)}_t={{\eta }_t}^{z^{-1}}$ for all $t\le a\wedge \eta (e)$. Thus ${{\eta }_t}^{z^{-1}}\subseteq {\eta }_t\text{for all}t\le a\wedge \eta (e).$ This implies $z^{-1}\in N({\eta }_t)$ for all $t\le a\wedge \eta (e)$. Since the normalizer of ${\eta }_t$ is a subgroup of G, $z\in N({\eta }_t)$ for all $t\le a\wedge \eta (e)$. Thus ${{\eta }_t}^z\subseteq {\eta }_t\text{for all}t\le a\wedge \eta (e).$ Again, by Theorem 3.6, $({\eta }^{a_{z^{-1}}}{)}_t={{\eta }_t}^z$ for all $t\le a\wedge \eta (e)$. Thus $({\eta }^{a_{z^{-1}}}{)}_t\subseteq {\eta }_t$ for all $t\le a\wedge \eta (e)=\mathrm{tip}({\eta }^{a_z})$. Hence by Proposition 4.6, ${\eta }^{a_{z^{-1}}}\subseteq \eta$.

The converse part can be obtained by replacing z by $z^{-1}$ in the above exposition.

We wish to characterize the concept of normalizer of an L-subgroup with the help of the notion of conjugacy. In order to achieve this goal, we prove the following lemma:

Lemma 4.8

Let η be an L-subgroup of μ and $a_z\in \mu$. Then, $\eta \circ a_z=a_z\circ \eta \text{if and only if}{\eta }^{a_z}\subseteq \eta .$

Proof.

Let $\eta \circ a_z=a_z\circ \eta$ and $x\in G$. Then, consider $\begin{array}{rl}{\eta }^{a_z}(x)& =a\wedge \eta (zx{z}^{-1})\\ & =(\eta \circ a_z)(zx)\\ & =(a_z\circ \eta )(zx)\text{(by the hypothesis)}\\ & =a\wedge \eta (z^{-1}(zx))\\ & =a\wedge \eta (x)\\ & \le \eta (x).\end{array}$ Thus ${\eta }^{a_z}\subseteq \eta$.

Let ${\eta }^{a_z}\subseteq \eta$. We shall show that $\eta \circ a_z=a_z\circ \eta$. Let $x\in G$ and consider $\begin{array}{rl}(\eta \circ a_z)(x)& =a\wedge \eta (x{z}^{-1})\\ & =a\wedge \eta (z(z^{-1}x)z^{-1})\\ & =a\wedge {\eta }^{a_z}(z^{-1}x)\\ & \le a\wedge \eta (z^{-1}x)(\text{by the hypothesis, }{\eta }^{a_z}\subseteq \eta )\\ & =(a_z\circ \eta )(x).\end{array}$ Thus $\eta \circ a_z\subseteq a_z\circ \eta$. For the reverse inclusion, note that by Lemma 4.7, ${\eta }^{a_{z^{-1}}}\subseteq \eta$. Thus $\begin{array}{rl}(a_z\circ \eta )(x)& =a\wedge \eta (z^{-1}x)\\ & =a\wedge \eta (z^{-1}(x{z}^{-1})z)\\ & =a\wedge {\eta }^{a_{z^{-1}}}(x{z}^{-1})\\ & \le a\wedge \eta (x{z}^{-1})(\text{since }{\eta }^{a_{z^{-1}}}\subseteq \eta )\\ & =(\eta \circ a_z)(x).\end{array}$ Thus $a_z\circ \eta \subseteq \eta \circ a_z$ and we conclude that $a_z\circ \eta =\eta \circ a_z$.

Thus in view of the above lemma, we have the following definition of the normalizer $N(\eta )$ of an L-subgroup η of μ:

Definition 4.9

Let $\eta \in L(\mu )$. The normalizer of η in μ, denoted by $N(\eta$), is the L-subgroup defined as follows: $N(\eta )=\bigcup \{a_z\in \mu |{\eta }^{a_z}\subseteq \eta \}.$

We demonstrate the above definition of the normalizer with the following example:

Example 4.2

Let $G=D_{16}$ and let μ and η be the L-subgroups of G defined in Example 4.1. We use Definition 4.9 to determine the normalizer of η in μ. Note that for $x\in G$, (8) $\bigcup \{a_z\in \mu \mid {\eta }^{a_z}\subseteq \eta \}(x)=\bigvee \{a\in L\mid a\le \mu (x)\mathrm{and}{\eta }^{a_x}\subseteq \eta \}.$(8) Taking x = s in Equation 8(8) $\bigcup \{a_z\in \mu \mid {\eta }^{a_z}\subseteq \eta \}(x)=\bigvee \{a\in L\mid a\le \mu (x)\mathrm{and}{\eta }^{a_x}\subseteq \eta \}.$(8) , we see that for all $g\in G$, ${\eta }^{a_s}(g)=a\wedge \eta (sg{s}^{-1})=a\wedge \eta (g).$ Thus ${\eta }^{a_s}(g)\le \eta (g)$ for all $a\in L$. This implies $\bigvee \{a\in L\mid a\le \mu (s)\text{ and }{\eta }^{a_s}\subseteq \eta \}=\mu (s)=\frac{1}{2}.$ Next, taking x = r in Equation 8(8) $\bigcup \{a_z\in \mu \mid {\eta }^{a_z}\subseteq \eta \}(x)=\bigvee \{a\in L\mid a\le \mu (x)\mathrm{and}{\eta }^{a_x}\subseteq \eta \}.$(8) , $\begin{array}{rl}{\eta }^{a_r}(g)& =a\wedge \eta (rg{r}^{-1})\\ & =\{\begin{array}{ll}a\wedge \frac{1}{4}& \mathrm{if}x\in ⟨s{r}^6⟩,\\ a\wedge \frac{1}{16}& \mathrm{if}x\in D_8\setminus ⟨s{r}^6⟩,\\ a\wedge \frac{1}{32}& \mathrm{if}x\in D_{16}\setminus D_8.\end{array}\end{array}$ Hence it is clear that ${\eta }^{a_r}\subseteq \eta$ if and only if $a\le \frac{1}{16}$. Thus $\bigvee \{a\in L\mid a\le \mu (r)\text{ and }{\eta }^{a_r}\subseteq \eta \}=\frac{1}{16}.$ Using similar computations, we get $\bigcup \{a_z\in \mu \mid {\eta }^{a_z}\subseteq \eta \}(x)=\{\begin{array}{ll}\frac{1}{2}& \text{if }x\in H,\\ \frac{1}{16}& \text{if }x\in D_{16}\setminus H,\end{array}$ which is the normalizer N(η) of η in μ obtained in Example 4.1.

5. Conclusion

In classical group theory, the conjugate subgroups play an indispensable role in the studies of normality. In fact, the normalizer of a subgroup H of a group G can be defined as the collection of all elements x in G such that the conjugate $H^x$ of H with respect to x is contained in H. While the notion of the normalizer of an L-subgroup was efficiently introduced in [7], the concept of the conjugate of an L-subgroup that was compatible with the normality and the normalizer of L-subgroups was absent. We have, in this paper, succeeded in providing such a notion of the conjugate. The relation of the conjugate with the normalizer provides us with a new method to evaluate the nomalizer of L-subgroups and thus can be easily applied in future studies of these and related topics. Moreover, the notion of the conjugate developed in this study can be utilized to develop the concepts of pronormal, abnormal and contranormal L-subgroups. These notions are closely related to the concept of normality. However, a proper research on these topics is lacking due to the absence of a comprehensive recent study of the conjugate. The conjugate developed in this paper has removed this limitation and opens the door to research on these topics.

While the research in areas such as fuzzy topology [20,21] and fuzzy ring theory [22,23] has been thriving, the research in the discipline of fuzzy group theory came to a halt after Tom Head's metatheorem and subdirect product theorems. This is because most of the concepts and results in the studies of fuzzy algebra could be established through simple applications of the metatheorem and the subdirect product theorem. However, the metatheorem and the subdirect product theorems are not applicable in the L-setting. Hence we suggest the researchers pursuing studies in these areas to investigate the properties of L-subalgebras of an L-algebra rather than L-subalgebras of classical algebra.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

The second author of this paper was supported by the Senior Research Fellowship jointly funded by CSIR and UGC, India during the course of development of this paper.

Notes on contributors

Iffat Jahan

Iffat Jahan did her Ph.D. from Department of Mathematics, University of Delhi, India, in 2014. She is working as a Professor in the Department of Mathematics, Ramjas College, University of Delhi, India. She has authored more than 15 research papers in the area of L-Group Theory. Her areas of interest and research are Group Theory, Ring Theory, Lattice Theory, and Fuzzy Sets.

Ananya Manas

Ananya Manas is a doctoral student at Department of Mathematics, University of Delhi. His areas of research are Group Theory, Lattice Theory, and L-Groups.