Autocommuting probability of a finite group relative to its subgroups

TL;DR

This paper introduces a generalized concept of autocommuting probability in finite groups, analyzing the likelihood that a randomly chosen pair from a subgroup and an automorphism group commutes, extending previous notions.

Contribution

It generalizes the autocommuting probability to include subgroups and automorphism groups, providing new insights into their interaction within finite groups.

Findings

Derived formulas for autocommuting probability in various group classes

Established bounds for the probability based on subgroup properties

Connected autocommuting probability with group automorphism structures

Abstract

Let $H \subseteq K$ be two subgroups of a finite group $G$ and Aut$(K)$ the automorphism group of $K$. The autocommuting probability of $G$ relative to its subgroups $H$ and $K$, denoted by ${\rm Pr}(H, {\rm Aut}(K))$, is the probability that the autocommutator of a randomly chosen pair of elements, one from $H$ and the other from Aut$(K)$, is equal to the identity element of $G$. In this paper, we study ${\rm Pr}(H, {\rm Aut}(K))$ through a generalization.

Equations

$$\mathrm{Pr}(G,\mathrm{Aut}(G))=\frac{\left|\{(x,\alpha )\in G\times \mathrm{Aut}(G):[x,\alpha ]=1\}\right|}{|G||\mathrm{Aut}(G)|}$$

$${\mathrm{Pr}}_g(H,\mathrm{Aut}(K))=\frac{\left|\{(x,\alpha )\in H\times \mathrm{Aut}(K):[x,\alpha ]=g\}\right|}{|H||\mathrm{Aut}(K)|}$$

$$[x,\beta ]=[x,\sigma \alpha ]=x^{-1}\sigma (\alpha (x))=[x,\sigma ]=g.$$

$$|{\mathrm{orb}}_K(x)|=\frac{|\mathrm{Aut}(K)|}{|C_{\mathrm{Aut}(K)}(x)|}.$$

$${\mathrm{Pr}}_g(H,\mathrm{Aut}(K))=$$

$$=$$

$$|H||\mathrm{Aut}(K)|{\mathrm{Pr}}_g(H,\mathrm{Aut}(K))=|\bigsqcup _{x\in H}(\{x\}\times T_{x,g}(H,K))|=\sum _{x\in H}|T_{x,g}(H,K)|.$$

$$\mathrm{Pr}(H,\mathrm{Aut}(K))=\frac{1}{|H||\mathrm{Aut}(K)|}\sum _{x\in H}|C_{\mathrm{Aut}(K)}(x)|=\frac{|{\mathrm{orb}}_K(H)|}{|H|}$$

$$\mathrm{Pr}(H,\mathrm{Aut}(K))=\frac{1}{|H|}+\frac{1}{|\mathrm{Aut}(K)|}-\frac{1}{|H||\mathrm{Aut}(K)|}.$$

$$|H||\mathrm{Aut}(K)|\mathrm{Pr}(H,\mathrm{Aut}(K))=\sum _{x\in H}|C_{\mathrm{Aut}(K)}(x)|=|\mathrm{Aut}(K)|+|H|-1.$$

$$\mathrm{Pr}(H,\mathrm{Aut}(K))=\frac{1}{|H||\mathrm{Aut}(K)|}\sum _{\alpha \in \mathrm{Aut}(K)}|C_H(\alpha )|.$$

$${\mathrm{Pr}}_{g^{-1}}(H,\mathrm{Aut}(K))={\mathrm{Pr}}_g(H,\mathrm{Aut}(K)).$$

$$A$$

$$B$$

$${\mathrm{Pr}}_{(g_1,g_2)}(H_1\times H_2,\mathrm{Aut}(K_1\times K_2))={\mathrm{Pr}}_{g_1}(H_1,\mathrm{Aut}(K_1)){\mathrm{Pr}}_{g_2}(H_2,\mathrm{Aut}(K_2)).$$

$$\mathcal{X}$$

$$[(x,y),{\alpha }_{K_1\times K_2}]=(g_1,g_2)\},$$

$$\mathcal{Y}$$

$$\mathcal{Z}$$

$$\frac{|\mathcal{X}|}{|H_1\times H_2||\mathrm{Aut}(K_1\times K_2)|}=\frac{|\mathcal{Y}|}{|H_1||\mathrm{Aut}(K_1)|}\cdot \frac{|\mathcal{Z}|}{|H_2||\mathrm{Aut}(K_2)|}.$$

$${\mathrm{Pr}}_g(H,\mathrm{Aut}(K))\le \mathrm{Pr}(H,\mathrm{Aut}(K)).$$

$${\mathrm{Pr}}_g(H,\mathrm{Aut}(K))$$

$$\le \frac{1}{|H||\mathrm{Aut}(K)|}\sum _{x\in H}|C_{\mathrm{Aut}(K)}(x)|=\mathrm{Pr}(H,\mathrm{Aut}(K)).$$

$${\mathrm{Pr}}_g(H,\mathrm{Aut}(K))\le \frac{|H|-|L(H,\mathrm{Aut}(K))|}{p|H|}<\frac{1}{p}.$$

$${\mathrm{Pr}}_g(H,\mathrm{Aut}(K))=\frac{1}{|H|}\underset{xg\in {\mathrm{orb}}_K(x)}{\sum _{x\in H\setminus L(H,\mathrm{Aut}(K))}}\frac{1}{|{\mathrm{orb}}_K(x)|}$$

$${\mathrm{Pr}}_g(H_1,\mathrm{Aut}(K))\le |H_2:H_1|{\mathrm{Pr}}_g(H_2,\mathrm{Aut}(K)).$$

$$|H_1||\mathrm{Aut}(K)|{\mathrm{Pr}}_g(H_1,\mathrm{Aut}(K))$$

$$\le \underset{xg\in {\mathrm{orb}}_K(x)}{\sum _{x\in H_2}}|C_{\mathrm{Aut}(K)}(x)|$$

$$=|H_2||\mathrm{Aut}(K)|{\mathrm{Pr}}_g(H_2,\mathrm{Aut}(K)).$$

$${\mathrm{Pr}}_g(H,\mathrm{Aut}(K))\le |K:H|\mathrm{Pr}(K,\mathrm{Aut}(K))$$

$${\mathrm{Pr}}_g(H,\mathrm{Aut}(K))$$

$$=\frac{1}{|H||\mathrm{Aut}(K)|}\sum _{x\in H}|C_{\mathrm{Aut}(K)}(x)|$$

$$\le \frac{1}{|H||\mathrm{Aut}(K)|}\sum _{x\in K}|C_{\mathrm{Aut}(K)}(x)|.$$

$${\mathrm{Pr}}_g(H,K)=\frac{|\{(x,y)\in H\times K:x^{-1}{y}^{-1}xy=g\}|}{|H||K|}.$$

$${\mathrm{Pr}}_g(H,\mathrm{Aut}(K))\le {\mathrm{Pr}}_g(H,K).$$

Full text

*Department of Mathematical Sciences, Tezpur University,

Napaam-784028, Sonitpur, Assam, India.

Emails:  parama@gonitsora.com and rajatkantinath@yahoo.com

Key words: Automorphism group, Autocommuting probability, Autoisoclinism.

2010 Mathematics Subject Classification: 20D60, 20P05, 20F28.

1 Introduction

Let $G$ be a finite group acting on a set $\Omega$. Let $\Pr(G,\Omega)$ denote the probability that a randomly chosen element of $\Omega$ fixes a randomly chosen element of $G$. In 1975, Sherman [11] initiated the study of $\Pr(G,\Omega)$ considering $G$ to be an abelian group and $\Omega=\operatorname{Aut}(G)$, the automorphism group of $G$. Note that

[TABLE]

where $[x,\alpha]$ is the autocommutator of $x$ and $\alpha$ defined as   $x^{-1}\alpha(x)$. The ratio ${\Pr}(G,\operatorname{Aut}(G))$ is called autocommuting probability of $G$.

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. Motivated by the works in [2, 6], we define

[TABLE]

where $g\in K$. That is, ${\Pr}_{g}(H,\operatorname{Aut}(K))$ is the probability that the autocommutator of a randomly chosen pair of elements, one from $H$ and the other from $\operatorname{Aut}(K)$, is equal to a given element $g\in K$. The ratio ${\Pr}_{g}(H,\operatorname{Aut}(K))$ is called generalized autocommuting probability of $G$ relative to its subgroups $H$ and $K$. Clearly, if $H=G$ and $g=1$ then ${\Pr}_{g}(H,\operatorname{Aut}(K))={\Pr}(G,\operatorname{Aut}(G))$. We would like to mention here that the case when $H=G$ is considered in [3]. In this paper, we study ${\Pr}_{g}(H,\operatorname{Aut}(K))$ extensively. In particular, we obtain some computing formulae, various bounds and a few characterizations of $G$ through a subgroup. We conclude the paper describing an invariance property of ${\Pr}_{g}(H,\operatorname{Aut}(K))$.

We write $S(H,\operatorname{Aut}(K))$ to denote the set $\{[x,\alpha]:x\in H\text{ and }\alpha\in\operatorname{Aut}(K)\}$ and $[H,\operatorname{Aut}(K)]:=\langle S(H,\operatorname{Aut}(K))\rangle$. We also write $L(H,\operatorname{Aut}(K)):=\{x\in H:[x,\alpha]=1\text{ for all }\alpha\in\operatorname{Aut}(K)\}$ and $L(G):=L(G,\operatorname{Aut}(G))$, the absolute center of $G$ (see [5]). Note that $L(H,\operatorname{Aut}(K))$ is a normal subgroup of $H$ contained in $H\cap Z(K)$. Further, $L(H,\operatorname{Aut}(K))=\underset{\alpha\in\operatorname{Aut}(K)}{\cap}C_{H}(\alpha)$, where $C_{H}(\alpha)=\{x\in H:[x,\alpha]=1\}$ is a subgroup of $H$. Let $C_{\operatorname{Aut}(K)}(x):=\{\alpha\in\operatorname{Aut}(K):\alpha(x)=x\}$ for $x\in H$ and $C_{\operatorname{Aut}(K)}(H)=\{\alpha\in\operatorname{Aut}(K):\alpha(x)=x\text{ for all }x\in H\}$. Then $C_{\operatorname{Aut}(K)}(x)$ is a subgroup of $\operatorname{Aut}(K)$ and $C_{\operatorname{Aut}(K)}(H)=\underset{x\in H}{\cap}C_{\operatorname{Aut}(K)}(x)$.

Clearly, ${\Pr}_{g}(H,\operatorname{Aut}(K))=1$ if and only if $[H,\operatorname{Aut}(K)]=\{1\}$ and $g=1$ if and only if $H=L(H,\operatorname{Aut}(K))$ and $g=1$. Also, ${\Pr}_{g}(H,\operatorname{Aut}(K))=0$ if and only if $g\notin S(H,\operatorname{Aut}(K))$. Therefore, we consider $H\neq L(H,\operatorname{Aut}(K))$ and $g\in S(H,\operatorname{Aut}(K))$ throughout the paper.

2 Some computing formulae

For any $x\in H$, let us define the set $T_{x,g}(H,K)=\{\alpha\in\operatorname{Aut}(K):[x,\alpha]=g\}$, where $g$ is a fixed element of $K$. Note that $T_{x,1}(H,K)=C_{\operatorname{Aut}(K)}(x)$. The following two lemmas play a crucial role in obtaining computing formula for ${\Pr}_{g}(H,\operatorname{Aut}(K))$.

Lemma 2.1**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. If $T_{x,g}(H,K)\neq\phi$ then $T_{x,g}(H,K)=\sigma C_{\operatorname{Aut}(K)}(x)$ for some $\sigma\in T_{x,g}(H,K)$ and hence $|T_{x,g}(H,K)|=|C_{\operatorname{Aut}(G)}(x)|$.

Proof.

Let $\sigma\in T_{x,g}(H,K)$ and $\beta\in\sigma C_{\operatorname{Aut}(K)}(x)$. Then $\beta=\sigma\alpha$ for some $\alpha\in C_{\operatorname{Aut}(K)}(x)$. We have

[TABLE]

Therefore, $\beta\in T_{x,g}(H,K)$ and so $\sigma C_{\operatorname{Aut}(K)}(x)\subseteq T_{x,g}(H,K)$. Again, let $\gamma\in T_{x,g}(H,K)$ then $\gamma(x)=xg$. We have $\sigma^{-1}\gamma(x)=\sigma^{-1}(xg)=x$ and so $\sigma^{-1}\gamma\in C_{\operatorname{Aut}(K)}(x)$. Therefore, $\gamma\in\sigma C_{\operatorname{Aut}(K)}(x)$ which gives $T_{x,g}(H,K)\subseteq\sigma C_{\operatorname{Aut}(K)}(x)$. Hence, the result follows. ∎

Consider the action of $\operatorname{Aut}(K)$ on $K$ given by $(\alpha,x)\mapsto\alpha(x)$ where $\alpha\in\operatorname{Aut}(K)$ and $x\in K$. Let $\operatorname{orb}_{K}(x):=\{\alpha(x):\alpha\in\operatorname{Aut}(K)\}$ be the orbit of $x\in K$. Then by orbit-stabilizer theorem, we have

[TABLE]

Lemma 2.2**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. Then $T_{x,g}(H,K)\neq\phi$ if and only if $xg\in\operatorname{orb}_{K}(x)$.

Proof.

The result follows from the fact that $\alpha\in T_{x,g}(H,K)$ if and only if $xg\in\operatorname{orb}_{K}(x)$. ∎

The following theorem gives two computing formulae for ${\Pr}_{g}(H,\operatorname{Aut}(K))$.

Theorem 2.3**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. If $g\in K$ then

[TABLE]

Proof.

We have $\{(x,\alpha)\in H\times\operatorname{Aut}(K):[x,\alpha]=g\}=\underset{x\in H}{\sqcup}(\{x\}\times T_{x,g}(H,K))$, where $\sqcup$ represents the union of disjoint sets. Therefore, by (1.1), we have

[TABLE]

Hence, the result follows from Lemma 2.1, Lemma 2.2 and (2.1). ∎

Considering $g=1$ in Theorem 2.3, we get the following computing formulae for ${\Pr}(H,\operatorname{Aut}(K))$.

Corollary 2.4**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. Then

[TABLE]

where $\operatorname{orb}_{K}(H)=\{\operatorname{orb}_{K}(x):x\in H\}$.

Corollary 2.5**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. If $C_{\operatorname{Aut}(K)}(x)=\{I\}$ for all $x\in H\setminus\{1\}$, where $I$ is the identity element of $\operatorname{Aut}(K)$, then

[TABLE]

Proof.

By Corollary 2.4, we have

[TABLE]

Hence, the result follows. ∎

We also have $|\{(x,\alpha)\in H\times\operatorname{Aut}(K):[x,\alpha]=1\}|=\underset{\alpha\in\operatorname{Aut}(K)}{\sum}|C_{H}(\alpha)|$ and hence

[TABLE]

We conclude this section with the following two results.

Proposition 2.6**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. If $g\in K$ then

[TABLE]

Proof.

Let

[TABLE]

Then $(x,\alpha)\mapsto(\alpha(x),\alpha^{-1})$ gives a bijection between $A$ and $B$. Therefore $|A|=|B|$. Hence, the result follows from (1.1).

∎

Proposition 2.7**.**

Let $G_{1}$ and $G_{2}$ be two finite groups. Let $H_{1},K_{1}$ and $H_{2},K_{2}$ be subgroups of $G_{1}$ and $G_{2}$ respectively such that $H_{1}\subseteq K_{1}$, $H_{2}\subseteq K_{2}$ and $\gcd(|K_{1}|,|K_{2}|)=1$. If $(g_{1},g_{2})\in K_{1}\times K_{2}$ then

[TABLE]

Proof.

Let

[TABLE]

Since $\gcd(|K_{1}|,|K_{2}|)=1$, by Lemma 2.1 of [1], we have $\operatorname{Aut}(K_{1}\times K_{2})=\operatorname{Aut}(K_{1})$ $\times\operatorname{Aut}(K_{2})$. Therefore, for every $\alpha_{K_{1}\times K_{2}}\in\operatorname{Aut}(K_{1}\times K_{2})$ there exist unique $\alpha_{K_{1}}\in\operatorname{Aut}(K_{1})$ and $\alpha_{K_{2}}\in\operatorname{Aut}(K_{2})$ such that $\alpha_{K_{1}\times K_{2}}=\alpha_{K_{1}}\times\alpha_{K_{2}}$, where $\alpha_{K_{1}}\times\alpha_{K_{2}}((x,y))=(\alpha_{K_{1}}(x),\alpha_{K_{2}}(y))$ for all $(x,y)\in H_{1}\times H_{2}$. Also, for all $(x,y)\in H_{1}\times H_{2}$, we have $[(x,y),\alpha_{K_{1}\times K_{2}}]=(g_{1},g_{2})$ if and only if $[x,\alpha_{K_{1}}]=g_{1}$ and $[y,\alpha_{K_{2}}]=g_{2}$. These leads to show that $\mathcal{X}=\mathcal{Y}\times\mathcal{Z}$. Therefore

[TABLE]

Hence, the result follows from (1.1). ∎

3 Various bounds

In this section, we obtain various bounds for ${\Pr}_{g}(H,\operatorname{Aut}(K))$. We begin with the following lower bounds.

Proposition 3.1**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. Then, for $g\in K$, we have

1. (a)

${\Pr}_{g}(H,\operatorname{Aut}(K))\geq\frac{|L(H,\operatorname{Aut}(K))|}{|H|}+\frac{|C_{\operatorname{Aut}(K)}(H)|(|H|-|L(H,\operatorname{Aut}(K))|)}{|H||\operatorname{Aut}(K)|}$* if $g=1$.* 2. (b)

${\Pr}_{g}(H,\operatorname{Aut}(K))\geq\frac{|L(H,\operatorname{Aut}(K))||C_{\operatorname{Aut}(K)}(H)|}{|H||\operatorname{Aut}(K)|}$* if $g\neq 1$.*

Proof.

Let $\mathcal{C}$ denote the set $\{(x,\alpha)\in H\times\operatorname{Aut}(K):[x,\alpha]=g\}$.

(a) We have $(L(H,\operatorname{Aut}(K))\times\operatorname{Aut}(K))\cup(H\times C_{\operatorname{Aut}(K)}(H))$ is a subset of $\mathcal{C}$ and $|(L(H,\operatorname{Aut}(K))\times\operatorname{Aut}(K))\cup(H\times C_{\operatorname{Aut}(K)}(H))|=|L(H,\operatorname{Aut}(K))||\operatorname{Aut}(K)|+|C_{\operatorname{Aut}(K)}(H)||H|-|L(H,\operatorname{Aut}(K))||C_{\operatorname{Aut}(K)}(H)|$. Hence, the result follows from (1.1).

(b) Since $g\in S(H,\operatorname{Aut}(K))$ we have $\mathcal{C}$ is non-empty. Let $(y,\beta)\in\mathcal{C}$ then $(y,\beta)\notin L(H,\operatorname{Aut}(K))\times C_{\operatorname{Aut}(K)}(H)$ otherwise $[y,\beta]=1$. It is easy to see that the coset $(y,\beta)(L(H,\operatorname{Aut}(K))\times C_{\operatorname{Aut}(K)}(H))$ is a subset of $\mathcal{C}$ having order $|L(H,\operatorname{Aut}(K))||C_{\operatorname{Aut}(K)}(H)|$. Hence, the result follows from (1.1). ∎

Proposition 3.2**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. If $g\in K$ then

[TABLE]

The equality holds if and only if $g=1$.

Proof.

By Theorem 2.3, we have

[TABLE]

The equality holds if and only if $xg\in\operatorname{orb}_{K}(x)$ for all $x\in H$ if and only if $g=1$. ∎

Proposition 3.3**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. Let $g\in K$ and $p$ the smallest prime dividing $|\operatorname{Aut}(K)|$. If $g\neq 1$ then

[TABLE]

Proof.

By Theorem 2.3, we have

[TABLE]

noting that for $x\in L(H,\operatorname{Aut}(K))$ we have $xg\notin\operatorname{orb}_{K}(x)$. Also, for $x\in H\setminus L(H,\operatorname{Aut}(K))$ and $xg\in\operatorname{orb}_{K}(x)$ we have $|\operatorname{orb}_{K}(x)|>1$. Since $|\operatorname{orb}_{K}(x)|$ is a divisor of $|\operatorname{Aut}(K)|$ we have $|\operatorname{orb}_{K}(x)|\geq p$. Hence, the result follows from (3.1). ∎

Proposition 3.4**.**

Let $H_{1}$, $H_{2}$ and $K$ be subgroups of a finite group $G$ such that $H_{1}\subseteq H_{2}\subseteq K$. Then

[TABLE]

The equality holds if and only if $xg\notin\operatorname{orb}_{K}(x)$ for all $x\in H_{2}\setminus H_{1}$.

Proof.

By Theorem 2.3, we have

[TABLE]

Hence, the result follows. ∎

Proposition 3.5**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. If $g\in K$ then

[TABLE]

with equality if and only if $g=1$ and $H=K$.

Proof.

By Proposition 3.2, we have

[TABLE]

Hence, the result follows from Corollary 2.4. ∎

Note that if we replace $\operatorname{Aut}(K)$ by $\operatorname{Inn}(K)$, the inner automorphism group of $K$, in (1.1) then ${\Pr}_{g}(H,\operatorname{Inn}(K))={\Pr}_{g}(H,K)$ where

[TABLE]

A detailed study on ${\Pr}_{g}(H,K)$ can be found in [2]. The following proposition gives a relation between ${\Pr}_{g}(H,\operatorname{Aut}(K))$ and ${\Pr}_{g}(H,K)$ for $g=1$.

Proposition 3.6**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. If $g=1$ then

[TABLE]

Proof.

If $g=1$ then by [2, Theorem 2.3], we have

[TABLE]

where $\operatorname{cl}_{K}(x)=\{\alpha(x):\alpha\in\operatorname{Inn}(K)\}$. Since $\operatorname{cl}_{K}(x)\subseteq\operatorname{orb}_{K}(x)$ for all $x\in H$, the result follows from (3.2) and Theorem 2.3. ∎

Theorem 3.7**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$ and $p$ the smallest prime dividing $|\operatorname{Aut}(K)|$. Then

[TABLE]

and

[TABLE]

where $X_{H}=\{x\in H:C_{\operatorname{Aut}(K)}(x)=\{I\}\}$.

Proof.

We have $X_{H}\cap L(H,\operatorname{Aut}(K))=\phi$. Therefore

[TABLE]

For $x\in H\setminus(X_{H}\cup L(H,\operatorname{Aut}(K)))$ we have $\{I\}\neq C_{\operatorname{Aut}(K)}(x)\neq\operatorname{Aut}(K)$ which implies $p\leq|C_{\operatorname{Aut}(K)}(x)|\leq\frac{|\operatorname{Aut}(K)|}{p}$. Therefore

[TABLE]

and

[TABLE]

Hence, the result follows from Corollary 2.4, (3) and (3). ∎

We have the following two corollaries.

Corollary 3.8**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. If $p$ and $q$ are the smallest primes dividing $|\operatorname{Aut}(K)|$ and $|H|$ respectively then

[TABLE]

In particular, if $p=q$ then $\Pr(H,\operatorname{Aut}(K))\leq\frac{2p-1}{p^{2}}\leq\frac{3}{4}$.

Proof.

Since $H\neq L(H,\operatorname{Aut}(K))$ we have $|H:L(H,\operatorname{Aut}(K))|\geq q$. Therefore, by Theorem 3.7, we have

[TABLE]

∎

Corollary 3.9**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$ and $p$, $q$ be the smallest primes dividing $|\operatorname{Aut}(K)|$ and $|H|$ respectively. If $H$ is non-abelian then

[TABLE]

In particular, if $p=q$ then $\Pr(H,\operatorname{Aut}(K))\leq\frac{p^{2}+p-1}{p^{3}}\leq\frac{5}{8}$.

Proof.

Since $H$ is non-abelian we have $|H:L(H,\operatorname{Aut}(K))|\geq q^{2}$. Therefore, by Theorem 3.7, we have

[TABLE]

∎

Now we obtain two lower bounds analogous to the lower bounds obtained in [9, Theorem A] and [8, Theorem 1].

Theorem 3.10**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. Then

[TABLE]

The equality holds if and only if $\operatorname{orb}_{K}(x)=xS(H,\operatorname{Aut}(K))$ for all $x\in H\setminus L(H,\operatorname{Aut}(K))$.

Proof.

For all $x\in H\setminus L(H,\operatorname{Aut}(K))$ we have $\alpha(x)=x[x,\alpha]\in xS(H,\operatorname{Aut}(K))$. Therefore $\operatorname{orb}_{K}(x)\subseteq xS(H,\operatorname{Aut}(K))$ and so $|\operatorname{orb}_{K}(x)|\leq|S(H,\operatorname{Aut}(K))|$ for all $x\in H\setminus L(H,\operatorname{Aut}(K))$. Now, by Corollary 2.4, we have

[TABLE]

Hence, the result follows. ∎

Lemma 3.11**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. Then, for any two integers $m\geq n$, we have

[TABLE]

If $L(H,\operatorname{Aut}(K))\neq H$ then equality holds if and only if $m=n$.

Proof.

The proof is an easy exercise. ∎

Corollary 3.12**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. Then

[TABLE]

If $H\neq L(H,\operatorname{Aut}(K))$ then the equality holds if and only if $[H,\operatorname{Aut}(K)]=S(H,\operatorname{Aut}(K))$ and $\operatorname{orb}_{K}(x)=x[H,\operatorname{Aut}(K)]$ for all $x\in H\setminus L(H,\operatorname{Aut}(K))$.

Proof.

Since $|[H,\operatorname{Aut}(K)]|\geq|S(H,\,\operatorname{Aut}(K))|$, the result follows from Theorem 3.10 and Lemma 3.11.

Note that the equality holds if and only if equality holds in Theorem 3.10 and Lemma 3.11. ∎

It is worth mentioning that Theorem 3.10 gives better lower bound than the lower bound given by Corollary 3.12. Also

[TABLE]

Hence, Theorem 3.10 gives better lower bound than the lower bound given by Theorem 3.7.

4 A few Characterizations

In this section, we obtain some characterizations of a subgroup $H$ of $G$ if equality holds in Corollary 3.8 and Corollary 3.9. We begin with the following result.

Theorem 4.1**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. If $\Pr(H,\operatorname{Aut}(K))=\frac{p+q-1}{pq}$ for some primes $p$ and $q$. Then $pq$ divides $|H||\operatorname{Aut}(K)|$. Further, if $p$ and $q$ are the smallest primes dividing $|\operatorname{Aut}(K)|$ and $|H|$ respectively, then

[TABLE]

In particular, if $H$ and $\operatorname{Aut}(K)$ are of even order and $\Pr(H,\operatorname{Aut}(K))=\frac{3}{4}$ then $\frac{H}{L(H,\operatorname{Aut}(K))}\cong\mathbb{Z}_{2}$.

Proof.

By (1.1), we have $(p+q-1)|H||\operatorname{Aut}(K)|=pq|\{(x,\alpha)\in H\times\operatorname{Aut}(K):[x,\alpha]=1\}|$. Therefore, $pq$ divides $|H||\operatorname{Aut}(K)|$.

If $p$ and $q$ are the smallest primes dividing $|\operatorname{Aut}(K)|$ and $|H|$ respectively then, by Theorem 3.7, we have

[TABLE]

which gives $|H:L(H,\operatorname{Aut}(K))|\leq q$. Hence, $\frac{H}{L(H,\operatorname{Aut}(K))}\cong\mathbb{Z}_{q}$. ∎

Theorem 4.2**.**

Let $H\subseteq K$ be two subgroups of a finite group $G$ such that $H$ is non-abelian and $\Pr(H,\operatorname{Aut}(K))=\frac{q^{2}+p-1}{pq^{2}}$ for some primes $p$ and $q$. Then $pq$ divides $|H||\operatorname{Aut}(K)|$. Further, if $p$ and $q$ are the smallest primes dividing $|\operatorname{Aut}(K)|$ and $|H|$ respectively then

[TABLE]

In particular, if $H$ and $\operatorname{Aut}(K)$ are of even order and $\Pr(H,\operatorname{Aut}(K))=\frac{5}{8}$ then $\frac{H}{L(H,\operatorname{Aut}(K))}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.

Proof.

By (1.1), we have $(q^{2}+p-1)|H||\operatorname{Aut}(K)|=pq^{2}|\{(x,\alpha)\in H\times\operatorname{Aut}(K):[x,\alpha]=1\}|$. Therefore, $pq$ divides $|H||\operatorname{Aut}(K)|$.

If $p$ and $q$ are the smallest primes dividing $|\operatorname{Aut}(K)|$ and $|H|$ respectively then, by Theorem 3.7, we have

[TABLE]

which gives $|H:L(H,\operatorname{Aut}(K))|\leq q^{2}$. Since $H$ is non-abelian we have $|H:L(H,\operatorname{Aut}(K))|\neq 1,q$. Hence, $\frac{H}{L(H,\operatorname{Aut}(K))}\cong\mathbb{Z}_{q}\times\mathbb{Z}_{q}$. ∎

The following two results give partial converses of Theorem 4.1 and 4.2 respectively.

Proposition 4.3**.**

Let $H$ and $K$ be two subgroups of a finite group $G$ such that $H\subseteq K$. Let $p,q$ be the smallest prime divisors of $|\operatorname{Aut}(K)|$, $|H|$ respectively and $|\operatorname{Aut}(K):C_{\operatorname{Aut}(K)}(x)|=p$ for all $x\in H\setminus L(H,\operatorname{Aut}(K))$.

1. (a)

If $\frac{H}{L(H,\operatorname{Aut}(K))}\cong\mathbb{Z}_{q}$ then $\Pr(H,\operatorname{Aut}(K))=\frac{p+q-1}{pq}$. 2. (b)

If $\frac{H}{L(H,\operatorname{Aut}(K))}\cong\mathbb{Z}_{q}\times\mathbb{Z}_{q}$ then $\Pr(H,\operatorname{Aut}(K))=\frac{q^{2}+p-1}{pq^{2}}$.

Proof.

Since $|\operatorname{Aut}(K):C_{\operatorname{Aut}(K)}(x)|=p$ for all $x\in H\setminus L(H,\operatorname{Aut}(K))$ we have $|C_{\operatorname{Aut}(K)}(x)|=\frac{|\operatorname{Aut}(K)|}{p}$ for all $x\in H\setminus L(H,\operatorname{Aut}(K))$. Therefore, by Corollary 2.4, we have

[TABLE]

Thus

[TABLE]

(a) If $\frac{H}{L(H,\operatorname{Aut}(K))}\cong\mathbb{Z}_{q}$ then (4.1) gives $\Pr(H,\operatorname{Aut}(K))=\frac{p+q-1}{pq}$.

(b) If $\frac{H}{L(H,\operatorname{Aut}(K))}\cong\mathbb{Z}_{q}\times\mathbb{Z}_{q}$ then (4.1) gives $\Pr(H,\operatorname{Aut}(K))=\frac{q^{2}+p-1}{pq^{2}}$. ∎

5 Autoisoclinic pairs

In the year 1940, Hall [4] introduced the concept of isoclinism between two groups. Following Hall, Moghaddam et al. [7] have defined autoisoclinism between two groups, in the year 2013. Recall that two groups $G_{1}$ and $G_{2}$ are said to be autoisoclinic if there exist isomorphisms $\psi:\frac{G_{1}}{L(G_{1})}\to\frac{G_{2}}{L(G_{2})}$, $\beta:[G_{1},\operatorname{Aut}(G_{1})]\to[G_{2},\operatorname{Aut}(G_{2})]$ and $\gamma:\operatorname{Aut}(G_{1})\to\operatorname{Aut}(G_{2})$ such that the following diagram commutes

\begin{CD}\frac{G_{1}}{L(G_{1})}\times\operatorname{Aut}(G_{1})@>{\psi\times\gamma}>{}>\frac{G_{2}}{L(G_{2})}\times\operatorname{Aut}(G_{2})\\ @V{}V{a_{(G_{1},\operatorname{Aut}(G_{1}))}}V@V{}V{a_{(G_{2},\operatorname{Aut}(G_{2}))}}V\\ [G_{1},\operatorname{Aut}(G_{1})]@>{\beta}>{}>[G_{2},\operatorname{Aut}(G_{2})]\end{CD}

where the maps $a_{(G_{i},\operatorname{Aut}(G_{i}))}:\frac{G_{i}}{L(G_{i})}\times\operatorname{Aut}(G_{i})\to[G_{i},\operatorname{Aut}(G_{i})]$, for $i=1,2$, are given by

[TABLE]

Such a pair $(\psi\times\gamma,\beta)$ is called an autoisoclinism between the groups $G_{1}$ and $G_{2}$. We generalize the notion of autoisoclinism in the following way:

Let $H_{1},K_{1}$ and $H_{2},K_{2}$ be subgroups of the groups $G_{1}$ and $G_{2}$ respectively. The pairs of subgroups $(H_{1},K_{1})$ and $(H_{2},K_{2})$ such that $H_{1}\subseteq K_{1}$ and $H_{2}\subseteq K_{2}$ are said to be autoisoclinic if there exist isomorphisms $\psi:\frac{H_{1}}{L(H_{1},\operatorname{Aut}{K_{1}})}\to\frac{H_{2}}{L(H_{2},\operatorname{Aut}(K_{2}))}$, $\beta:[H_{1},\operatorname{Aut}(K_{1})]\to[H_{2},\operatorname{Aut}(K_{2})]$ and $\gamma:\operatorname{Aut}(K_{1})\to\operatorname{Aut}(K_{2})$ such that the following diagram commutes

\begin{CD}\frac{H_{1}}{L(H_{1},\operatorname{Aut}(K_{1}))}\times\operatorname{Aut}(K_{1})@>{\psi\times\gamma}>{}>\frac{H_{2}}{L(H_{2},\operatorname{Aut}(K_{2}))}\times\operatorname{Aut}(K_{2})\\ @V{}V{a_{(H_{1},\operatorname{Aut}(K_{1}))}}V@V{}V{a_{(H_{2},\operatorname{Aut}(K_{2}))}}V\\ [H_{1},\operatorname{Aut}(K_{1})]@>{\beta}>{}>[H_{2},\operatorname{Aut}(K_{2})]\end{CD}

where the maps $a_{(H_{i},\operatorname{Aut}(K_{i}))}:\frac{H_{i}}{L(H_{i},\operatorname{Aut}(K_{i}))}\times\operatorname{Aut}(K_{i})\to(H_{i},\operatorname{Aut}(K_{i}))$, for $i=1,2$, are given by

[TABLE]

Such a pair $(\psi\times\gamma,\beta)$ is said to be an autoisoclinism between the pairs of groups $(H_{1},K_{1})$ and $(H_{2},K_{2})$. We conclude this paper with the following generalization of [3, Theorem 5.1] and [10, Lemma 2.5].

Theorem 5.1**.**

Let $G_{1}$ and $G_{2}$ be two finite groups with subgroups $H_{1},K_{1}$ and $H_{2},K_{2}$ respectively such that $H_{1}\subseteq K_{1}$ and $H_{2}\subseteq K_{2}$. If $(\psi\times\gamma,\beta)$ is an autoisoclinism between the pairs $(H_{1},K_{1})$ and $(H_{2},K_{2})$ then, for $g\in K_{1}$,

[TABLE]

Proof.

Let us consider the sets $\mathcal{S}_{g}=\{(x_{1}L(H_{1},\operatorname{Aut}(K_{1})),\alpha_{1})\in\frac{H_{1}}{L(H_{1},\operatorname{Aut}(K_{1}))}\times\operatorname{Aut}(K_{1}):[x_{1}L(H_{1},\operatorname{Aut}(K_{1})),\alpha_{1}]=g\}$ and $\mathcal{T}_{\beta(g)}=\{(x_{2}L(H_{2},\operatorname{Aut}(K_{2})),\alpha_{2})\in\frac{H_{2}}{L(H_{2},\operatorname{Aut}(K_{2}))}\times\operatorname{Aut}(K_{2}):[x_{2}L(H_{2},\operatorname{Aut}(K_{2})),\alpha_{2}]=\beta(g)\}$. Since $(H_{1},K_{1})$ is autoisoclinic to $(H_{2},K_{2})$ we have $|\mathcal{S}_{g}|=|\mathcal{T}_{\beta(g)}|$. Again, it is clear that

[TABLE]

and

[TABLE]

Hence, the result follows from (1.1), (5.1) and (5.2). ∎