Autocommuting probability of a finite group

TL;DR

This paper investigates the autocommuting probability in finite groups, providing formulas, bounds, and characterizations, and shows its invariance under autoisoclinism, enriching the understanding of automorphism actions.

Contribution

It introduces a generalized autocommuting probability for finite groups, derives formulas and bounds, and proves invariance under autoisoclinism, advancing group automorphism theory.

Findings

Derived a computing formula for autocommuting probability

Established bounds and characterizations of finite groups based on this probability

Proved invariance of the generalized autocommuting probability under autoisoclinism

Abstract

Let $G$ be a finite group and $\Aut(G)$ the automorphism group of $G$. The autocommuting probability of $G$, denoted by $\Pr(G, \Aut(G))$, is the probability that a randomly chosen automorphism of $G$ fixes a randomly chosen element of $G$. In this paper, we study $\Pr(G, \Aut(G))$ through a generalization. We obtain a computing formula, several bounds and characterizations of $G$ through $\Pr(G, \Aut(G))$. We conclude the paper by showing that the generalized autocommuting probability of $G$ remains unchanged under autoisoclinism.

Equations

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

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

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

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

$${\mathrm{Pr}}_g(G,\mathrm{Aut}(G))=\frac{1}{|G||\mathrm{Aut}(G)|}\underset{xg\in \mathrm{orb}(x)}{\sum _{x\in G}}|C_{\mathrm{Aut}(G)}(x)|=\frac{1}{|G|}\underset{xg\in \mathrm{orb}(x)}{\sum _{x\in G}}\frac{1}{|\mathrm{orb}(x)|}.$$

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

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

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

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

$$\mathrm{Pr}(G,\mathrm{Aut}(G))=\frac{1}{|G||\mathrm{Aut}(G)|}\sum _{\alpha \in \mathrm{Aut}(G)}|C_G(\alpha )|.$$

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

$$A$$

$$B$$

$${\mathrm{Pr}}_{(g,h)}(G\times H,\mathrm{Aut}(G\times H))={\mathrm{Pr}}_g(G,\mathrm{Aut}(G)){\mathrm{Pr}}_h(H,\mathrm{Aut}(H)).$$

$$\mathcal{X}$$

$$\mathcal{Y}$$

$$\mathcal{Z}$$

$$\frac{|\mathcal{X}|}{|G\times H||\mathrm{Aut}(G\times H)|}=\frac{|\mathcal{Y}|}{|G||\mathrm{Aut}(G)|}\cdot \frac{|\mathcal{Z}|}{|H||\mathrm{Aut}(H)|}.$$

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

$${\mathrm{Pr}}_g(G,\mathrm{Aut}(G))$$

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

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

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

$$\mathrm{Pr}(G,\mathrm{Aut}(G))\ge \frac{|L(G)|}{|G|}+\frac{p(|G|-|X_G|-|L(G)|)+|X_G|}{|G||\mathrm{Aut}(G)|}$$

$$\mathrm{Pr}(G,\mathrm{Aut}(G))\le \frac{(p-1)|L(G)|+|G|}{p|G|}-\frac{|X_G|(|\mathrm{Aut}(G)|-p)}{p|G||\mathrm{Aut}(G)|},$$

$$\sum _{x\in G}|C_{\mathrm{Aut}(G)}(x)|=|X_G|+|\mathrm{Aut}(G)||L(G)|+\sum _{x\in G\setminus (X_G\cup L(G))}|C_{\mathrm{Aut}(G)}(x)|.$$

$$\sum _{x\in G}|C_{\mathrm{Aut}(G)}(x)|\ge |X_G|+|\mathrm{Aut}(G)||L(G)|+p(|G|-|X_G|-|L(G)|)$$

$$\sum _{x\in G}|C_{\mathrm{Aut}(G)}(x)|\le |X_G|+|\mathrm{Aut}(G)||L(G)|+\frac{|\mathrm{Aut}(G)|(|G|-|X_G|-|L(G)|)}{p}.$$

$$\mathrm{Pr}(G,\mathrm{Aut}(G))\le \frac{p+q-1}{pq}.$$

$$\mathrm{Pr}(G,\mathrm{Aut}(G))\le \frac{1}{p}\left(\frac{p-1}{|G:L(G)|}+1\right)\le \frac{p+q-1}{pq}.$$

$$\mathrm{Pr}(G,\mathrm{Aut}(G))\le \frac{q^2+p-1}{p{q}^2}.$$

$$\mathrm{Pr}(G,\mathrm{Aut}(G))\le \frac{1}{p}\left(\frac{p-1}{|G:L(G)|}+1\right)\le \frac{q^2+p-1}{p{q}^2}.$$

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 and $\operatorname{Aut}(G)$ be its automorphism group. For any $x\in G$ and $\alpha\in\operatorname{Aut}(G)$ the element $x^{-1}\alpha(x)$, denoted by $[x,\alpha]$, is called an autocommutator in $G$. We write $S(G,\operatorname{Aut}(G))$ to denote the set $\{[x,\alpha]:x\in G\text{ and }\alpha\in\operatorname{Aut}(G)\}$ and $K(G):=\langle S(G,\operatorname{Aut}(G))\rangle$. We also write $L(G):=\{x:[x,\alpha]=1\text{ for all }\alpha\in\operatorname{Aut}(G)\}$. Note that $K(G)$ and $L(G)$ are characteristic subgroups of $G$ known as the autocommutator subgroup and absolute center of $G$. It is easy to see that $K(G)$ contains the commutator subgroup $G^{\prime}$ and $L(G)$ is contained in the center $Z(G)$ of $G$. Further, $L(G)=\underset{\alpha\in\operatorname{Aut}(G)}{\cap}C_{G}(\alpha)$, where $C_{G}(\alpha)=\{x\in G:[x,\alpha]=1\}$ is a subgroup of $G$ known as acentralizer of $\alpha\in\operatorname{Aut}(G)$. The subgroups $K(G)$ and $L(G)$ were defined and studied by Hegarty in [6]. Let $C_{\operatorname{Aut}(G)}(x):=\{\alpha\in\operatorname{Aut}(G):\alpha(x)=x\}$ for $x\in G$ then $C_{\operatorname{Aut}(G)}(x)$ is a subgroup of $\operatorname{Aut}(G)$ and $C_{\operatorname{Aut}(G)}(G)=\{\alpha\in\operatorname{Aut}(G):\alpha(x)=x\text{ for all }x\in G\}$. It follows that $C_{\operatorname{Aut}(G)}(G)=\underset{x\in G}{\cap}C_{\operatorname{Aut}(G)}(x)$.

Let $G$ be a finite group acting on a set $\Omega$. In the year 1975, Sherman [13] introduced the probability that a randomly chosen element of $\Omega$ fixes a randomly chosen element of $G$. We denote this probability by $\Pr(G,\Omega)$. If $\Omega=\operatorname{Aut}(G)$ then $\Pr(G,\operatorname{Aut}(G))$ is nothing but the probability that the autocommutator of a randomly chosen pair of elements, one from $G$ and the other from $\operatorname{Aut}(G)$, is equal to $1$ (the identity element of $G$). Thus

[TABLE]

${\Pr}(G,\operatorname{Aut}(G))$ is called autocommuting probability of $G$. Sherman [13] considered the case when $G$ is abelian and $\Omega=\operatorname{Aut}(G)$. Note that if we take $\Omega=\operatorname{Inn}(G)$, the inner automorphim group of $G$, then ${\Pr}(G,\operatorname{Inn}(G))$ gives the probability that a randomly chosen pair of elements of $G$ commute. ${\Pr}(G,\operatorname{Inn}(G))$ is known as commuting probability of $G$. It is also denoted by $\Pr(G)$. The study of $\Pr(G)$ was initiated by Erd$\ddot{\rm o}s$ and Tur$\acute{\rm a}$n [4]. After Erd$\ddot{\rm o}s$ and Tur$\acute{\rm a}$n many authors have worked on $\Pr(G)$ and its generalizations (conf. [3] and the references therein). Somehow the study of ${\Pr}(G,\operatorname{Aut}(G))$ was neglected for long time. At this moment we have only a handful of papers on ${\Pr}(G,\operatorname{Aut}(G))$ (see [2], [8], [12]). In this paper, we study ${\Pr}(G,\operatorname{Aut}(G))$ through a generalization.

In the year 2008, Pournaki and Sobhani [10] have generalized $\Pr(G)$ and considered the probability that the commutator of a randomly chosen pair of elements of $G$ equals a given element $g\in G$. We write ${\Pr}_{g}(G)$ to denote this probability. Motivated by their work, in this paper, we consider the probability that the autocommutator of a randomly chosen pair of elements, one from $G$ and the other from $\operatorname{Aut}(G)$, is equal to a given element $g\in G$. We write ${\Pr}_{g}(G,\operatorname{Aut}(G))$ to denote this probability. Thus,

[TABLE]

Notice that ${\Pr}_{1}(G,\operatorname{Aut}(G))={\Pr}(G,\operatorname{Aut}(G))$. Hence ${\Pr}_{g}(G,\operatorname{Aut}(G))$ is a generalization of ${\Pr}(G,\operatorname{Aut}(G))$. Clearly, ${\Pr}_{g}(G,\operatorname{Aut}(G))=1$ if and only if $K(G)=\{1\}$ and $g=1$ if and only if $G=L(G)$ and $g=1$. Also, ${\Pr}_{g}(G,\operatorname{Aut}(G))=0$ if and only if $g\notin S(G,\operatorname{Aut}(G))$. Therefore, we consider $G\neq L(G)$ and $g\in S(G,\operatorname{Aut}(G))$ throughout the paper.

In Section 2, we obtain a computing formula for ${\Pr}_{g}(G,\operatorname{Aut}(G))$ and deduce some of its consequences. In Section 3, we obtain several bounds for ${\Pr}_{g}(G,\operatorname{Aut}(G))$. In Section 4, we obtain some characterizations of $G$ through ${\Pr}(G,\operatorname{Aut}(G))$. Finally, in the last section, we show that ${\Pr}_{g}(G,\operatorname{Aut}(G))$ is an invariant under autoisoclinism of groups.

2 A computing formula

For any $x,g\in G$ let $T_{x,g}$ denotes the set $\{\alpha\in\operatorname{Aut}(G):[x,\alpha]=g\}$. Note that $T_{x,1}=C_{\operatorname{Aut}(G)}(x)$. The following two lemmas play a crucial role in obtaining the computing formula for ${\Pr}_{g}(G,\operatorname{Aut}(G))$.

Lemma 2.1**.**

Let $G$ be a finite group. If $T_{x,g}\neq\phi$ then $T_{x,g}=\sigma C_{\operatorname{Aut}(G)}(x)$ for some $\sigma\in T_{x,g}$. Hence, $|T_{x,g}|=|C_{\operatorname{Aut}(G)}(x)|$.

Proof.

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

[TABLE]

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

We know that $\operatorname{Aut}(G)$ acts on $G$ by the action $(\alpha,x)\mapsto\alpha(x)$ where $\alpha\in\operatorname{Aut}(G)$ and $x\in G$. Let $\operatorname{orb}(x):=\{\alpha(x):\alpha\in\operatorname{Aut}(G)\}$ be the orbit of $x\in G$. Then by orbit-stabilizer theorem, we have

[TABLE]

Lemma 2.2**.**

Let $G$ be a finite group. Then $T_{x,g}\neq\phi$ if and only if $xg\in\operatorname{orb}(x)$.

Proof.

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

Now we state and prove the main theorem of this section.

Theorem 2.3**.**

Let $G$ be a finite group. If $g\in G$ then

[TABLE]

Proof.

We have $\{(x,\alpha)\in G\times\operatorname{Aut}(G):[x,\alpha]=g\}=\underset{x\in G}{\sqcup}(\{x\}\times T_{x,g})$, where $\sqcup$ represents the union of disjoint sets. Therefore, by (1.2), we have

[TABLE]

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

Taking $g=1$, in Theorem 2.3, we get the following corollary.

Corollary 2.4**.**

Let $G$ be a finite group. Then

[TABLE]

where $\operatorname{orb}(G)=\{\operatorname{orb}(x):x\in G\}$.

Corollary 2.5**.**

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

[TABLE]

Proof.

By Corollary 2.4, we have

[TABLE]

Hence, the result follows. ∎

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

[TABLE]

We conclude this section with the following two results.

Proposition 2.6**.**

Let $G$ be a finite group. If $g\in G$ 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 (1.2).

∎

Proposition 2.7**.**

Let $G$ and $H$ be two finite groups such that $\gcd(|G|,|H|)=1$. If $(g,h)\in G\times H$ then

[TABLE]

Proof.

Let

[TABLE]

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

[TABLE]

Hence, the result follows from (1.1). ∎

3 Some bounds

We begin with the following lower bounds.

Proposition 3.1**.**

Let $G$ be a finite group. Then

1. (a)

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

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

Proof.

Let $\mathcal{C}:=\{(x,\alpha)\in G\times\operatorname{Aut}(G):[x,\alpha]=g\}$.

(a) We have $(L(G)\times\operatorname{Aut}(G))\cup(G\times C_{\operatorname{Aut}(G)}(G))\subseteq\mathcal{C}$ and $|(L(G)\times\operatorname{Aut}(G))\cup(G\times C_{\operatorname{Aut}(G)}(G))|$ is equal to $|L(G)||\operatorname{Aut}(G)|+|C_{\operatorname{Aut}(G)}(G)||G|-|L(G)||C_{\operatorname{Aut}(G)}(G)|$. Hence, the result follows from (1.2).

(b) Since $g\in S(G,\operatorname{Aut}(G))$ we have $\mathcal{C}$ is non-empty. Let $(y,\beta)\in\mathcal{C}$ then $(y,\beta)\notin L(G)\times C_{\operatorname{Aut}(G)}(G)$ otherwise $[y,\beta]=1$. It is easy to see that the coset $(y,\beta)(L(G)\times C_{\operatorname{Aut}(G)}(G))$ having order $|L(G)||C_{\operatorname{Aut}(G)}(G)|$ is a subset of $\mathcal{C}$. Hence, the result follows from (1.2). ∎

Proposition 3.2**.**

Let $G$ be a finite group. 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 $xg\in\operatorname{orb}(x)$ for all $x\in G$ if and only if $g=1$. ∎

Proposition 3.3**.**

Let $G$ be a finite group and $p$ the smallest prime dividing $|\operatorname{Aut}(G)|$. If $g\neq 1$ then

[TABLE]

Proof.

By Theorem 2.3, we have

[TABLE]

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

The remaining part of this section is devoted in obtaining some lower and upper bounds for $\Pr(G,\operatorname{Aut}(G))$. The following theorem is an improvement of [8, Theorem 2.3 (ii)].

Theorem 3.4**.**

Let $G$ be a finite group and $p$ the smallest prime dividing $|\operatorname{Aut}(G)|$. Then

[TABLE]

and

[TABLE]

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

Proof.

We have $X_{G}\cap L(G)=\phi$. Therefore

[TABLE]

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

[TABLE]

and

[TABLE]

Hence, the result follows from Corollary 2.4, (3.2) and (3.3). ∎

We have the following two corollaries.

Corollary 3.5**.**

Let $G$ be a finite group. If $p$ and $q$ are the smallest primes dividing $|\operatorname{Aut}(G)|$ and $|G|$ respectively then

[TABLE]

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

Proof.

Since $G\neq L(G)$ we have $|G:L(G)|\geq q$. Therefore, by Theorem 3.4, we have

[TABLE]

∎

Corollary 3.6**.**

Let $G$ be a finite group and $p$, $q$ be the smallest primes dividing $|\operatorname{Aut}(G)|$ and $|G|$ respectively. If $G$ is non-abelian then

[TABLE]

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

Proof.

Since $G$ is non-abelian we have $|G:L(G)|\geq q^{2}$. Therefore, by Theorem 3.4, we have

[TABLE]

∎

Theorem 3.7**.**

Let $G$ be a finite group. Then

[TABLE]

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

Proof.

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

[TABLE]

Hence, the result follows. ∎

The following lemma is useful in obtaining the next corollary.

Lemma 3.8**.**

Let $G$ be a finite group. Then, for any two integers $m\geq n$, we have

[TABLE]

If $L(G)\neq G$ then equality holds if and only if $m=n$.

Proof.

The proof is an easy exercise. ∎

Corollary 3.9**.**

Let $G$ be a finite group. Then

[TABLE]

If $G\neq L(G)$ then the equality holds if and only if $K(G)=S(G,\operatorname{Aut}(G))$ and $\operatorname{orb}(x)$ is equal to $S(G,\operatorname{Aut}(G))$ for all $x\in G\setminus L(G)$.

Proof.

Since $|K(G)|\geq|S(G,\,\operatorname{Aut}(G))|$, the result follows from Theorem 3.7 and Lemma 3.8.

Note that the equality holds if and only if equality holds in Theorem 3.7 and Lemma 3.8. ∎

This lower bound is also obtained in [2]. Note that the lower bound in Theorem 3.7 is better than that in Corollary 3.9. Also

[TABLE]

Hence, the lower bound in Theorem 3.7 is also better than that in [8, Theorem 2.3(ii)].

4 Some Characterizations

In [2], Arora and Karan obtain a characterization of a finite group $G$ if equality holds in Corollary 3.9. In this section, we obtain some characterizations of $G$ if equality holds in Corollary 3.5 and Corollary 3.6. We begin with the following result.

Proposition 4.1**.**

Let $G$ be a finite group with $\Pr(G,\operatorname{Aut}(G))=\frac{p+q-1}{pq}$ for some primes $p$ and $q$. Then $pq$ divides $|G||\operatorname{Aut}(G)|$. If $p$ and $q$ are the smallest primes dividing $|\operatorname{Aut}(G)|$ and $|G|$ respectively, then

[TABLE]

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

Proof.

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

If $p$ and $q$ are the smallest primes dividing $|\operatorname{Aut}(G)|$ and $|G|$ respectively then, by Theorem 3.4, we have

[TABLE]

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

Proposition 4.2**.**

Let $G$ be a finite non-abelian group with $\Pr(G,\operatorname{Aut}(G))=\frac{q^{2}+p-1}{pq^{2}}$ for some primes $p$ and $q$. Then $pq$ divides $|G||\operatorname{Aut}(G)|$. If $p$ and $q$ are the smallest primes dividing $|\operatorname{Aut}(G)|$ and $|G|$ respectively then

[TABLE]

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

Proof.

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

If $p$ and $q$ are the smallest primes dividing $|\operatorname{Aut}(G)|$ and $|G|$ respectively then, by Theorem 3.4, we have

[TABLE]

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

We conclude this section with the following partial converses of Proposition 4.1 and 4.2.

Proposition 4.3**.**

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

1. (a)

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

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

Proof.

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

[TABLE]

Thus

[TABLE]

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

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

5 Autoisoclinism of groups

In the year 1940, Hall defined isoclinism between two groups (see [5]). In the year 1995, Lescot [7] showed that the commuting probability of two isoclinic finite groups are same. Later on Pournaki and Sobhani [10] showed that ${\Pr}_{g}(G)={\Pr}_{\beta(g)}(H)$ if $(\alpha,\beta)$ is an isoclinism between the finite groups $G$ and $H$. Following Hall, Moghaddam et al. [9] have defined autoisoclinism between two groups. Recall that two groups $G$ and $H$ are said to be autoisoclinic if there exist isomorphisms $\psi:\frac{G}{L(G)}\to\frac{H}{L(H)}$, $\beta:K(G)\to K(H)$ and $\gamma:\operatorname{Aut}(G)\to\operatorname{Aut}(H)$ such that the following diagram commutes

\begin{CD}\frac{G}{L(G)}\times\operatorname{Aut}(G)@>{\psi\times\gamma}>{}>\frac{H}{L(H)}\times\operatorname{Aut}(H)\\ @V{}V{a_{(G,\operatorname{Aut}(G))}}V@V{}V{a_{(H,\operatorname{Aut}(H))}}V\\ K(G)@>{\beta}>{}>K(H)\end{CD}

where the maps $a_{(G,\operatorname{Aut}(G))}:\frac{G}{L(G)}\times\operatorname{Aut}(G)\to K(G)$ and $a_{(H,\operatorname{Aut}(H))}:\frac{H}{L(H)}\times\operatorname{Aut}(H)\to K(H)$ are given by

[TABLE]

respectively. In this case, the pair $(\psi\times\gamma,\beta)$ is called an autoisoclinism between the groups $G$ and $H$.

Recently, in [12], Rismanchain and Sepehrizadeh have shown that ${\Pr}(G,\operatorname{Aut}(G))={\Pr}(H,\operatorname{Aut}(H))$ if $G$ and $H$ are autoisoclinic finite groups. We conclude this paper with the following generalization of [12, Lemma 2.5].

Theorem 5.1**.**

Let $G$ and $H$ be two finite groups and $(\psi\times\gamma,\beta)$ an autoisoclinism between them. Then

[TABLE]

Proof.

Let ${\mathcal{S}}_{g}=\{(xL(G),\alpha_{1})\in\frac{G}{L(G)}\times\operatorname{Aut}(G):[x,\alpha_{1}]=g\}$ and ${\mathcal{T}}_{\beta(g)}=\{(yL(H),\alpha_{2})$ $\in\frac{H}{L(H)}\times\operatorname{Aut}(H):[y,\alpha_{2}]=\beta(g)\}$. Since $(\psi\times\gamma,\beta)$ an autoisoclinism between $G$ and $H$, the mapping $\lambda:{\mathcal{S}}_{g}\to{\mathcal{T}}_{\beta(g)}$ given by $(xL(G),\alpha_{1})\mapsto(\psi(xL(G)),\gamma(\alpha_{1}))$ gives an one to one correspondence between ${\mathcal{S}}_{g}$ and ${\mathcal{T}}_{\beta(g)}$. Hence, $|\mathcal{S}_{g}|=|{\mathcal{T}}_{\beta(g)}|$. Also

[TABLE]

and

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Therefore, by (1.1) and (5.1), we have

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Since $|G/L(G)|=|H/L(H)|$ and $|\operatorname{Aut}(G)|=|\operatorname{Aut}(H)|$ we have

[TABLE]

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