On equality of inner and absolute central automorphisms

TL;DR

This paper investigates conditions under which the group of absolute central automorphisms of a finite p-group coincides with its inner automorphism group, providing a precise characterization.

Contribution

It establishes necessary and sufficient conditions for the equality of Aut_l(G) and Inn(G) in finite p-groups, advancing understanding of automorphism structures.

Findings

Characterization of when Aut_l(G) equals Inn(G)

Necessary and sufficient conditions identified

Enhanced understanding of automorphism groups in p-groups

Abstract

Let G be a finite p-group and let Aut_l(G) be the group of absolute central automorphisms of G. We give necessary and sufficient conditions on G such that Aut_l(G) = Inn(G).

Equations

$$L(G)=\{g\in G|[g,\alpha ]=1,\forall \alpha \in \mathrm{Aut}(G)\}$$

$${\mathrm{Aut}}_l(G)\cong \mathrm{Hom}(G/L(G),L(G)).$$

$${\mathrm{Aut}}_{Z(G)}^{L(G)}(G)\cong \mathrm{Hom}(G/Z(G),L(G)).$$

$$|\mathrm{Hom}(G/Z(G),L(G))|\ge |G/Z(G)|p^{r(s-1)}$$

$$L(G)=C_{p^n}\times C_{p^{{\gamma }_2}}\times \cdots \times C_{p^{{\gamma }_s}}$$

$$G/Z(G)=C_{p^c}\times C_{p^{{\beta }_2}}\times \cdots \times C_{p^{{\beta }_r}}$$

$$|{\mathrm{Aut}}_{Z(G)}^{L(G)}(G)|=|\mathrm{Hom}(G/Z(G),L(G))|\le |\mathrm{Hom}(G/L(G),L(G))|=|{\mathrm{Aut}}_l(G)|.$$

$$|G/Z(G)|=|\mathrm{Inn}(G)|$$

$$\mathrm{Hom}(G/Z(G),L(G))\hookrightarrow \mathrm{Hom}(G/L(G)G^{p^n},L(G))\hookrightarrow \mathrm{Hom}(G/L(G),L(G)).$$

$$\mathrm{Inn}(G)\cong G/Z(G)\cong \mathrm{Hom}(G/Z(G),L(G))$$

$$\mathrm{Hom}(G/L(G),L(G))\cong {\mathrm{Aut}}_l(G)=\mathrm{Inn}(G),$$

$$\mathrm{Hom}(G/Z(G),L(G))\cong \mathrm{Hom}(G/L(G)G^{p^n},L(G)),$$

Full text

2010 Mathematics Subject Classification: 20D45; 20D15

1. **Introduction **

An automorphism $\alpha$ of $G$ is called central if $x^{-1}\alpha(x)\in Z(G)$ for each $x\in G$. The central automorphisms of $G$, denoted by $\mathrm{Aut}_{c}(G)$, fix $G^{\prime}$ elementwise and form a normal subgroup of the full automorphism group of $G$. The properties of $\mathrm{Aut}_{c}(G)$ have been well studied. See [1, 2, 4] for example.

Hegarty in [3] generalized the concept of centre into absolute centre. The absolute centre of a group $G$, denoted by $L(G)$, is the subgroup consisting of all those elements that are fixed under all automorphisms of $G$. Also he introduced the absolute central automorphisms. An automorphism $\beta$ of $G$ is called an absolute central automorphism if $x^{-1}\beta(x)\in L(G)$ for each $x\in G$. We denote the set of all absolute central automorphisms of $G$ by $\mathrm{Aut}_{l}(G)$. Notice that $\mathrm{Aut}_{l}(G)$ is a normal subgroup of $\mathrm{Aut}(G)$ contained in $\mathrm{Aut}_{c}(G)$.

In [2] authors gave necessary and sufficient conditions on a $p$-group $G$ such that $\mathrm{Aut}_{c}(G)=\mathrm{Inn}(G)$. In [6] we gave conditions on a finite autonilpotent $p$-group $G$ of class $2$ such that $\mathrm{Aut}_{l}(G)=\mathrm{Inn}(G)$. In this paper we intend to give necessary and sufficient conditions on a non-abelian $p$-group $G$ in which $\mathrm{Aut}_{l}(G)$ and $\mathrm{Inn}(G)$ coincide.

Throughout this paper all groups are assumed to be finite and $p$ denotes a prime number. Also if $G$ is a group, then $\mathrm{exp}(G)$, $\mathrm{Hom}(G,H)$ and $G^{p^{n}}$ stand for the exponent, the group of homomorphisms of $G$ into an abelian group $H$ and the subgroup generated by all $p^{n}th$ powers of elements of $G$, respectively. Let $M$ and $N$ be two normal subgroups of $G$. We denote the subgroup of $\mathrm{Aut}(G)$ consisting of all automorpisms centralizing $G/M$ by $\mathrm{Aut}^{M}(G)$ and the subgroup of $\mathrm{Aut}(G)$ consisting of all automorphisms which act trivially on $N$ by $\mathrm{Aut}_{N}(G)$. Also, we consider $\mathrm{Aut}_{N}^{M}(G)=\mathrm{Aut}^{M}(G)\cap\mbox{Aut}_{N}(G)$.

2. Preliminary results

In this section, we give some results that will be used in the proof of the main results.

Lemma 2.1**.**

[6, Lemma 2.3]** Suppose $H$ is an abelian $p$-group of exponent $p^{c}$, and $K$ is a cyclic group of order divisible by $p^{c}$. Then $\mathrm{Hom}(H,K)$ is isomorphic to $H$.

Let $G$ be a group. Then the autocommutator of an element $g\in G$ and automorphism $\alpha\in\mathrm{Aut}(G)$ is defined as $[g,\alpha]=g^{-1}g^{\alpha}$.

Definition 2.2**.**

The absolute centre of a group $G$, denoted by $L(G)$, is defined as

[TABLE]

Clearly, $L(G)$ is a central characteristic subgroup of $G$.

Definition 2.3**.**

An automorphism $\alpha$ of $G$ is called absolute central, if $g^{-1}\alpha(g)\in L(G)$ for each $g\in G$. The set of all absolute central automorphisms of $G$ is denoted by $\mathrm{Aut}_{l}(G)$.

Clearly, $\mathrm{Aut}_{l}(G)$ is an abelian normal subgroup of $\mathrm{Aut}(G)$.

Proposition 2.4**.**

[5, Proposition 1]** Let $G$ be a group. Then

[TABLE]

Lemma 2.5**.**

Let $G$ be a group. Then

[TABLE]

Proof.

Consider the map $\bar{\alpha}:G/Z(G)\longrightarrow L(G)$ defined by $\bar{\alpha}(gZ(G))=g^{-1}\alpha(g)$ for all $g\in G$ and each $\alpha\in\mbox{Aut}_{Z(G)}^{L(G)}(G)$. Clearly, $\bar{\alpha}$ is a well-defined homomorphism of $G/Z(G)$ into $L(G)$. Now, a simple verification shows that the map $\phi:\mathrm{Aut}_{Z(G)}^{L(G)}(G)\longrightarrow\mathrm{Hom}(G/Z(G),L(G))$ defined by $\phi(\alpha)=\bar{\alpha}$, for any $\alpha\in\mbox{Aut}_{Z(G)}^{L(G)}(G)$, is an isomorphism. ∎

Lemma 2.6**.**

Let $G$ be a $p$-group of class $2$ such that $\mathrm{exp}(G/Z(G))\leq\mathrm{exp}(L(G))$. Then

[TABLE]

where $r=\mathrm{rank}(G/Z(G))$ and $s=\mathrm{rank}(L(G))$.

Proof.

Let $\mathrm{exp}(L(G))=p^{n}$ and $\mathrm{exp}(G/Z(G))=p^{c}$. Clearly, $r\geq 2$. Now let

[TABLE]

and

[TABLE]

where $n\geq\gamma_{2}\geq\dots\geq\gamma_{s}>0$ and $c\geq\beta_{2}\geq\dots\geq\beta_{r}>0$. Since $G$ is a $p$-group and $\mathrm{exp}(G/Z(G))\leq\mbox{exp}(L(G))$, $\mathrm{exp}(G/Z(G))$ divides $\mathrm{exp}(L(G))$. The rest of proof is similar to that of [6, Lemma 2.8] so we omit the details. ∎

Now we establish a lower bound for $|\mathrm{Aut}_{l}(G)|$.

Lemma 2.7**.**

Let $G$ be a $p$-group of class $2$ such that $\mathrm{exp}(G/Z(G))\leq\mathrm{exp}(L(G))$. Then $|\mathrm{Aut}_{l}(G)|\geq|G/Z(G)|p^{r(s-1)}$, where $r$ and $s$ are as defined before.

Proof.

We have

[TABLE]

So by Lemma $2.6$, $|\mathrm{Aut}_{l}(G)|\geq|G/Z(G)|p^{r(s-1)}$. ∎

Lemma 2.8**.**

[6, Lemma 2.7]** Let $G$ be a group. Then $G/L(G)$ is abelian if and only if $\mathrm{Inn}(G)\leq\mathrm{Aut}_{l}(G)$.

3. Main results

In this section, we obtain some properties of the group $G$ when $\mathrm{Aut}_{l}(G)=\mathrm{Inn}(G)$, and then give necessary and sufficient conditions under which $\mathrm{Aut}_{l}(G)=\mathrm{Inn}(G)$.

Lemma 3.1**.**

Let $G$ be a non-abelian $p$-group. If $\mathrm{Aut}_{l}(G)=\mathrm{Inn}(G)$, then $L(G)$ is cyclic.

Proof.

Let $\mathrm{Aut}_{l}(G)=\mathrm{Inn}(G)$. Thus $\mathrm{Hom}(G/L(G),L(G))\cong G/Z(G)$. Hence $\mathrm{exp}(G/Z(G))\leq\mathrm{exp}(L(G))$. Also, $\mathrm{Inn}(G)$ is abelian so that $G$ is nilpotent of class $2$. Now, by Lemma $2.7$, we have $p^{r(s-1)}\leq\frac{|\mathrm{Aut}_{l}(G)|}{|\mathrm{Inn}(G)|}$, where $r=\mathrm{rank}(G/Z(G))$ and $s=\mathrm{rank}(L(G))$. Thus $p^{r(s-1)}\leq 1$. Since $r\geq 2$ we must have $s=1$, that is, $L(G)$ is cyclic. ∎

Proposition 3.2**.**

Let $G$ be a non-abelian $p$-group ($p$ odd) such that $\mathrm{Aut}_{l}(G)=\mathrm{Inn}(G)$. Then $\exp(G/Z(G))=\exp(L(G))$.

Proof.

Suppose $p$ is an odd prime number and $\mathrm{exp}(L(G))=p^{n}$. Consider the map $\theta:G\longrightarrow G$ defined by $\theta(x)=x^{1+p^{n-1}}$ for all $x\in G$. Clearly, $\theta$ is an automorphism when $\mathrm{exp}(G/Z(G))<\mathrm{exp}(L(G))$. In this case, $g^{p^{n-1}}=1$ for all $g$ in $L(G)$. Hence $\mathrm{exp}(L(G))<p^{n}$, which is a contradiction. Thus $\mathrm{exp}(G/Z(G))=\mathrm{exp}(L(G))$. ∎

Theorem 3.3**.**

Let $G$ be a non-abelian $p$-group. Then $\mathrm{Inn}(G)=\mathrm{Aut}_{Z(G)}^{L(G)}(G)$ if and only if $G^{\prime}\leq L(G)$ and $L(G)$ is cyclic.

Proof.

Suppose $G^{\prime}\leq L(G)$ and $L(G)$ is cyclic. Since $G^{\prime}\leq L(G)$, $G$ is nilpotent of class $2$ and $\mathrm{exp}(G^{\prime})=\mathrm{exp}(G/Z(G))$. Hence $\mathrm{exp}(G/Z(G))$ divides $\mathrm{exp}(L(G))$ and by Lemma $2.1$, $\mathrm{Hom}(G/Z(G),L(G))\cong G/Z(G)$. Therefore by Lemma $2.5$, $\mathrm{Aut}_{Z(G)}^{L(G)}(G)\cong\mbox{Inn}(G)$. On the other hand, by Lemma $2.8$, we have $\mathrm{Inn}(G)\leq\mathrm{Aut}_{l}(G)$. Now since $\mathrm{Inn}(G)$ fixes the centre element-wise, we conclude that $\mathrm{Inn}(G)\leq\mathrm{Aut}_{Z(G)}^{L(G)}(G)$. Hence $\mathrm{Inn}(G)=\mathrm{Aut}_{Z(G)}^{L(G)}(G)$.

To prove the converse, assume that $\mathrm{Inn}(G)=\mathrm{Aut}_{Z(G)}^{L(G)}(G)$. Let $x\in G$ and $\theta_{x}$ be the inner automorphism of $G$ induced by $x$. Then, $g^{-1}\theta_{x}(g)=[g,x]\in L(G)$ for all $g$ in $G$, and consequently $G^{\prime}\leq L(G)$. Hence,

[TABLE]

where $r=\mathrm{rank}(G/Z(G))$ and $s=\mathrm{rank}(L(G))$. Thus $p^{r(s-1)}=1$. Since $r\geq 2$, we must have $s=1$ so that $L(G)$ is cyclic. ∎

Theorem 3.4**.**

Let $G$ be a non-abelian $p$-group. Then $\mathrm{Aut}_{l}(G)=\mathrm{Inn}(G)$ if and only if $G^{\prime}\leq L(G)$, $L(G)$ is cyclic and $Z(G)=L(G)G^{p^{n}}$, where $p^{n}=\mathrm{exp}(L(G))$.

Proof.

Suppose first that $G^{\prime}\leq L(G)$, $L(G)$ is cyclic and $Z(G)=L(G)G^{p^{n}}$. By Theorem $3.3$, $\mathrm{Inn}(G)=\mathrm{Aut}_{Z(G)}^{L(G)}(G)$. Now suppose $\alpha\in\mathrm{Aut}_{l}(G)$. Let $g\in G$ and $l\in L(G)$ be such that $\alpha(g)=gl$. Then $\alpha(g^{p^{n}})=g^{p^{n}}l^{p^{n}}$. Since $\mathrm{exp}(L(G))=p^{n}$, we have $l^{p^{n}}=1$ and so $\alpha(g^{p^{n}})=g^{p^{n}}$. Hence $\alpha$ acts trivially on $Z(G)$, that is, $\alpha\in\mathrm{Aut}_{Z(G)}^{L(G)}(G)$. Thus $\mathrm{Aut}_{l}(G)\leq\mathrm{Aut}_{Z(G)}^{L(G)}(G)$ so that $\mathrm{Aut}_{l}(G)=\mathrm{Aut}_{Z(G)}^{L(G)}(G)=\mathrm{Inn}(G)$, as required.

Conversely, suppose $\mathrm{Aut}_{l}(G)=\mathrm{Inn}(G)$. By Lemma $3.1$, $L(G)$ is cyclic and by Lemma $2.8$, $G^{\prime}\leq L(G)$. Now we have $\mathrm{exp}(G^{\prime})\leq\mathrm{exp}(L(G))=p^{n}$. Hence for all $a,b\in G$, $[a^{p^{n}},b]=1$. This means $a^{p^{n}}\in Z(G)$, for all $a\in G$. Therefore $G^{p^{n}}\leq Z(G)$ whence $L(G)G^{p^{n}}\leq Z(G)$. Since $L(G)\leq Z(G)$, it follows that

[TABLE]

As

[TABLE]

and

[TABLE]

we observe that

[TABLE]

hence $|G/Z(G)|=|G/L(G)G^{p^{n}}|$ from which it follows that $Z(G)=L(G)G^{p^{n}}$.The proof is complete. ∎

Acknowledgments

The authors are grateful to Dr. M. Farrokhi Derakhshandeh Ghouchan for his valuable suggestions and help in carrying out this work.