2 Results

Let G be a group, and consider a group $\Gamma $ of automorphisms of G. The interaction between $\Gamma $ and the subgroups

$$ \begin{align*}C_G(\Gamma)=\{g\in G\;|\;g^{\gamma}=g\mbox{ for all } \gamma \in \Gamma\}\end{align*} $$

and

$$ \begin{align*}[G, \Gamma]=\langle [g, \gamma]\;|\;g\in G, \gamma \in \Gamma\rangle\end{align*} $$

of G was first investigated in 1952 by Baer [Reference Baer2], who proved in particular the following result.

Our next result gives information of this type concerning the group of central automorphisms of a group.

Proof. Put $|G:A|=n$ . As the subgroup A lies in the centre of G, the transfer homomorphism of G into A is the map

$$ \begin{align*}\tau :g\in G\mapsto g^n\in A.\end{align*} $$

Let g and $\gamma $ be elements of G and $Aut_c(G)$ , respectively. Then $(g^{\gamma })^n=(g^n)^{\gamma }=g^n$ , so

$$ \begin{align*}[g, \gamma]^{\tau}=g^{-\tau}g^{\gamma \tau}=g^{-n}g^n=1.\end{align*} $$

It follows that $[G, Aut_c(G)]$ is contained in the kernel of $\tau $ and $[G, Aut_c(G)]^n=\{1\}$ .

As in many investigations concerning automorphisms, we will use in our arguments some cohomological methods. Let

$$ \begin{align*}\Sigma:A \overset{\mu} {\rightarrowtail}G\overset{\varepsilon} {\twoheadrightarrow}Q\end{align*} $$

be a central extension of a group A by a group Q. Without loss of generality, it can be assumed that A is a central subgroup of G, $\mu $ is the embedding of A into G, Q is the factor group $G/A$ and $\varepsilon $ is the natural projection. Recall that a transversal map

$$ \begin{align*}\tau :Q\longrightarrow G\end{align*} $$

of $\Sigma $ is a function such that $\tau \varepsilon =\iota _Q$ , so that the set

$$ \begin{align*}\{x^{\tau}\;|\;x\in Q\}\end{align*} $$

is a transversal to A in G. If $x,y\in Q$ , we have

$$ \begin{align*}(x^{\tau}y^{\tau})^{\varepsilon}=xy=((xy)^{\tau})^{\varepsilon}.\end{align*} $$

It follows that there exists a unique element $\varphi (x,y)\in A(=\ker \varepsilon $ ) such that

$$ \begin{align*}x^{\tau}y^{\tau}=(xy)^{\tau}\varphi (x,y).\end{align*} $$

If $x,y,z$ are elements of Q, from the equality $x^{\tau }(y^{\tau }z^{\tau })=(x^{\tau }y^{\tau })z^{\tau }$ , it follows that

$$ \begin{align*}\varphi(x,yz)+\varphi(y,z)=\varphi(xy,z)+\varphi(x,y).\end{align*} $$

Therefore, the map

$$ \begin{align*}\varphi :Q\times Q\longrightarrow A\end{align*} $$

is a 2-cocycle of Q in A. The coset

$$ \begin{align*}\Delta=\varphi +B^2(Q,A)\end{align*} $$

is an element of the second cohomology group $H^2(Q,A)$ of Q with coefficients in A, which depends only on the extension $\Sigma $ and not on the choice of the transversal function. The element $\Delta $ is called the cohomology class of $\Sigma $ .

Let $\alpha $ be an automorphism of A. An easy application of [Reference Stammbach13, Proposition II 4.3] shows that $\alpha $ can be extended to an automorphism $\beta $ of G inducing the identity on  $G/A$  if

$$ \begin{align*}\varphi (x,y)^{\alpha}=\varphi(x,y)\end{align*} $$

for all $x,y\in Q$ . Clearly, $\beta $ is a central automorphism G. As an immediate consequence of the previous considerations, we have the following result which is useful for constructing central automorphisms.

Lemma 2.3. Let G be a group and consider the central extension

$$ \begin{align*}\Sigma:A \rightarrowtail G\twoheadrightarrow G/A,\end{align*} $$

where $A=K(G)\cap Z(G)$ . If $\Delta =\varphi +B^2(Q,A)$ is the cohomology class of $\Sigma $ and $\alpha $ is an automorphism of A such that

$$ \begin{align*}\varphi (xA,yA)^{\alpha}=\varphi(xA,yA)\end{align*} $$

for all $xA,yA\in G/A$ , then $\alpha $ is the identity of A.

Now we are in a position to prove our main result.

Proof. First we note that the factor group $G/Z(G)$ is finite by [Reference Catino, de Giovanni and Miccoli4, Theorem A]. As $A=K(G)\cap Z(G)$ is a characteristic subgroup of G, then every automorphism $\gamma $ of G induces an automorphism $\bar \gamma $ on the finite group $G/A$ . Therefore, we may consider the homomorphism

$$ \begin{align*}f:\gamma \in Aut(G) \mapsto \bar \gamma \in Aut(G/A).\end{align*} $$

Clearly, the kernel $\Gamma $ of f is a subgroup of $Aut_c(G)$ , and hence the subgroup $[G, \Gamma ]$ has finite exponent by Lemma 2.2.

Assume for a contradiction that $\Gamma $ is infinite, so that $[G, \Gamma ]$ is likewise infinite by Lemma 2.1. Since $[G, \Gamma ]$ is a subgroup of A with finite exponent, it follows that the p-component P of A has infinite rank for some prime p. Let D be the largest divisible subgroup of P. Then A splits over D, so that $A=D\times E$ and hence $P=D\times R$ , where $R=E\cap P$ is a reduced subgroup. First suppose that the subgroup D has finite rank. Then there exists a sequence $(X_n)_{n\in \mathbb {N}}$ of cyclic nontrivial subgroups of R such that

$$ \begin{align*}R=X_1\times \cdots \times X_n\times R_n\end{align*} $$

for all positive integers n and suitable subgroups $R_n$ (see [Reference Robinson12, Propositions 4.3.3 and 4.3.8]). Moreover, since $X_1\times \cdots \times X_n$ is a finite direct factor of P and P is pure in A for each n, there exists a subgroup $A_n$ of A such that

$$ \begin{align*}A=X_1\times \cdots \times X_n\times A_n.\end{align*} $$

Clearly, if D has infinite rank, we have a similar decomposition of A for each n, where all the subgroups $X_1, \ldots , X_n$ are of type $p^{\infty }$ .

Let $\Delta =\varphi +B^2(G/A,A)$ be the cohomology class of the central extension

$$ \begin{align*}A \rightarrowtail G\twoheadrightarrow G/A.\end{align*} $$

By hypothesis, the set $\{\varphi (xA,yA) \mid xA,yA\in G/A\}$ is finite. It follows that for a sufficiently large n, there exist two direct factors U and V of the decomposition $A=X_1\times \cdots \times X_n\times A_n$ of A such that

$$ \begin{align*}\langle \varphi(xA,yA)\rangle\cap (U\times V)=\{1\}\end{align*} $$

for all $xA,yA\in G/A$ . Let $\alpha $ be a nonidentity automorphism of $U\times V$ . Clearly, $\alpha $ can be extended to an automorphism $\beta $ of A acting trivially over all direct factors other than U and V of the decomposition $A=X_1\times \cdots \times X_n\times A_n$ of A. This is a contradiction by Lemma 2.3. Therefore, the full automorphism group $Aut(G)$ of G is finite as required.

Finally, as the commutator subgroup $G'$ of G is finite by Schur’s theorem, it follows that the set T of all elements of finite order of G is a subgroup, and hence T is finite by a result of Nagrebeckiı̆ [Reference Nagrebeckiĭ10].

We note that the above result has been proved with different arguments also in [Reference Catino, de Giovanni and Miccoli4, Proposition 2.1].

Let G be a group. Following [Reference Hegarty9], the absolute centre (or autocentre) of G is the characteristic subgroup $C_G(Aut(G))$ of G consisting of all elements of G fixed by every automorphism of G. Clearly, $C_G(Aut(G))$ is contained in the central kernel $K(G)$ of G. Therefore, if the index $|G:C_G(Aut(G))|$ is finite, then Theorem 2.4 yields that $Aut(G)$ is finite, and hence the autocommutator subgroup $[G,Aut(G)]$ of G is likewise finite by Lemma 2.1. Thus, we have obtained the following result that was first proved by Hegarty [Reference Hegarty9].

We point out that another generalisation of Hegarty’s theorem was obtained by de Giovanni, Newell and the author [Reference de Giovanni, Newell and Russo6] in 2014.

Let G be a nilpotent group with class at most $2$ . Then ${\mathit {Inn}}(G)\leq Aut_c(G)$ and hence $K(G)\leq Z(G)$ . Conversely, as a central automorphism acts trivially over the commutator subgroup $G'$ of G, we see that G is nilpotent with class at most $2$ if the central kernel of G is contained in $Z(G)$ . Now we construct an infinite nonabelian group G such that $Z(G)=K(G)\neq C_G(Aut(G))$ .

Let $A=\langle x\rangle $ and $B=\langle y\rangle $ be a cyclic group of order 9 and an infinite cyclic group, respectively. Consider the semidirect product

$$ \begin{align*}G=B\ltimes A,\end{align*} $$

where $x^y=x^4$ and $[x,B^3]=\{1\}$ . Clearly, $Z(G)=B^3\times \langle x^3\rangle $ and $G'=\langle x^3\rangle $ . Let $\gamma $ be a central automorphism of G. Then $y^{\gamma }=yz$ for some central element z. Moreover, $\gamma $ induces an automorphism over the characteristic subgroup $B^9$ . First suppose that $(y^9)^{\gamma }=y^{-9}$ . If $z=y^{3i}x^{3j}$ for some integers i and j, then

$$ \begin{align*}y^{-9}=(y^9)^{\gamma}=y^9z^9=y^{27i+9}x^{27j}=y^{27i+9},\end{align*} $$

and so $y^{27i+18}=1$ , which is a contradiction. Therefore, $y^9=(y^9)^{\gamma }=y^9z^9$ and hence $z=x^{3t}$ for some nonnegative integer t. It follows that $(y^3)^{\gamma }=y^3$ so that $\gamma $ acts trivially on $Z(G)$ . Thus, $K(G)=Z(G)$ . Clearly, every inner automorphism of G is central. However, it is easy to show that $Aut_c(G)\simeq \mbox {Hom}(G/Z(G),Z(G))\simeq \mathbb {Z}_3\times \mathbb {Z}_3 $ , so that ${\mathit {Inn}}(G)=Aut_c(G)$ .

Finally, consider the automorphism $\alpha :x\mapsto x^2$ of A and $\beta =\iota _B$ of B. Since

$$ \begin{align*}(x^y)^{\alpha}=(x^4)^{\alpha}=x^8=(x^2)^y=(x^{\alpha})^{y^{\kern1.3pt\beta}},\end{align*} $$

there exists an automorphism $\gamma $ of G inducing $\alpha $ and $\beta $ . We note that $\gamma $ cannot be central since $(x^3)^{\alpha }=x^{-3}$ . It follows that $C_G(Aut(G))\neq K(G)$ .

Let G be a group, and denote by $\overline K(G)$ the set of all elements x of G such that $x^{\alpha }=x$ for every power automorphism $\alpha $ of G. Clearly, $\overline K(G)$ is a characteristic subgroup of G containing the central kernel $K(G)$ of G and the subgroup $G[2]=\langle g\in G\;|\;g^2=1\rangle $ . Note that the consideration of the direct product $G=A\times B$ , where A is a cyclic group of order $3$ and B is a countably infinite abelian group of exponent $2$ , shows that we cannot replace in Theorem 2.4 the central kernel by the subgroup $\overline K(G)$ . Nevertheless, it is easy to prove the following result.

Proof. First suppose that G is a nonperiodic group. By hypothesis, the term $\gamma _n(G)$ of the lower central series of G is finite for some positive integer n. It follows that the set of periodic elements of G is a subgroup, so that G is a weak group. Thus, $PAut(G)$ is finite of order at most $2$ (see [Reference Cooper5, Corollary 4.2.3]).

If G is periodic, then there exists a finite subgroup F of G such that $G=F\overline K(G)$ . Let $\alpha $ be a power automorphism of G. Then $F^{\alpha }=F$ and $x^{\alpha }=x$ for every $x\in \overline K(G)$ . It follows that the map

$$ \begin{align*}f:\alpha \in PAut(G)\mapsto \alpha _F\in Aut(F)\end{align*} $$

is injective and hence $PAut(G)$ is again finite.