Automorphisms of groups; basic remarks

Posted: August 15, 2021 in Basic Algebra, Groups
Tags: automorphism group, center of groups, inner automorphism, p-groups

Throughout this post, $G$ is a group with the center $Z(G).$

Definition 1. An automorphism of $G$ is any group isomorphism $G \to G.$ The set of all automorphisms of $G$ is denoted by $\rm{Aut}(G).$

Notation. Given $x \in G,$ the map $\tau_x : G \to G$ is defined by $\tau_x(g)=xgx^{-1}$ for all $g \in G.$

Definition 2. It is easy to see that the map $\tau_x,$ defined above, is an automorphism of $G.$ We say that $\tau_x$ is an inner automorphism of $G$ and we denote by $\rm{Inn}(G)$ the set of all inner automorphisms of $G,$ i.e

$\rm{Inn}(G):=\{\tau_x: \ \ x \in G\}.$

Remarks. i) $\rm{Aut}(G)$ is a group under composition of functions, and $\rm{Inn}(G)$ is a normal subgroup of it.

ii) If $g \in G, \ \alpha \in \rm{Aut}(G),$ then $g$ and $\alpha(g)$ have the same order.

iii) If $G$ is generated by a set $X$ and $\alpha \in \text{Aut}(G),$ then $\alpha$ is completely determined by $\{\alpha(x): \ x \in X\}.$

iv) $\rm{Inn}(G) \cong G/Z(G).$

v) $G/Z(G)$ is cyclic if and only if $|G/Z(G)|=1$ if and only if $G$ is abelian.

vi) If $G$ is finite, then $\rm{Aut}(G)$ is finite.

vii) If $|G| \ge 3,$ then $|\rm{Aut}(G)| \ge 2.$

viii) If $G$ is finite, abelian, and $|G| \ge 3,$ then $|\text{Aut}(G)|$ is even.

ix) If $G$ is not abelian, then $\text{Aut}(G)$ is not cyclic.

x) There is no finite group $G$ such that $\rm{Aut}(G)$ is a non-trivial cyclic group of odd order.

xi) If $p$ is a prime number and $G$ is a finite non-abelian $p$ -group, then $p^2 \mid |\rm{Aut}(G)|.$

xii) If $H$ is a subgroup of $G$ and $G=H \times K$ for some subgroup $K$ of $G,$ then $\rm{Aut}(H)$ is isomorphic to a subgroup of $\rm{Aut}(G).$

xiii) If $|Z(G)|=1,$ then $|Z(\rm{Aut}(G))|=1.$

Proof. i) Let $\alpha \in \rm{Aut}(G)$ and $\tau_x \in \text{Inn}(G).$ Then for any $g \in G$ we have

$\alpha \tau_x \alpha^{-1}(g)=\alpha(x\alpha^{-1}(g)x^{-1})=\alpha(x)g\alpha(x^{-1})=\alpha(x)g(\alpha(x))^{-1}=\tau_{\alpha(x)}(g).$

So $\alpha \tau_x \alpha^{-1}=\tau_{\alpha(x)} \in \rm{Inn}(G)$ for all $x \in G$ and so $\rm{Inn}(G)$ is normal in $\rm{Aut}(G).$

ii) If $n$ is a positive integer, then $(\alpha(g))^n=\alpha(g^n)$ and so since $\alpha(g^n)=1$ if and only if $g^n=1,$ because $\alpha$ is injective, we get that $(\alpha(g))^n=1$ if and only if $g^n=1,$ i.e. $\alpha(g)$ and $g$ have the same order.

iii) Let $g \in G.$ Since $X$ generates $G, \ g=x_1 \cdots x_n,$ for some $x_i \in X,$ and so $\alpha(x)=\alpha(x_1) \cdots \alpha(x_n).$ So if we know what $\alpha(x), \ x \in X$ are, then we know what $\alpha(g), \ g \in G,$ are, and so we know what $\alpha$ is.

iv) Define the map $f: G \to \rm{Inn}(G)$ by $f(x)=\tau_x$ for all $x \in G.$ Then for any $x,y,g \in G,$ we have

$\begin{aligned}f(xy)(g)=\tau_{xy}(g)=xyg(xy)^{-1}=xygy^{-1}x^{-1}=x\tau_y(g)x^{-1}=\tau_x\tau_y(g)=f(x)f(y)(g).\end{aligned}$

So $f(xy)=f(x)f(y)$ for all $x,y \in G,$ i.e. $f$ is a group homomorphism. Obviously $f$ is onto, and so the only thing left to prove is that $\ker f=Z(G).$ Well, $x \in \ker f$ if and only if $\tau_x$ is the identity map if and only if $xgx^{-1}=\tau_x(g)=g$ for all $g \in G$ if and only if $xg=gx$ for all $g \in G,$ i.e. $x \in Z(G).$

v) Let $gZ(G), \ g \in G,$ be a generator of $G/Z(G)$ and let $x,y \in G.$ So there exist integers $m,n$ such that $xZ(G)=g^nZ(G), \ yZ(G)=g^mZ(G),$ and so $x=g^nz_1, \ y=g^mz_2$ for some $z_1,z_2 \in Z(G).$ Hence

$xy=g^nz_1g^mz_2=g^ng^mz_1z_2=g^mg^nz_2z_1=g^mz_2g^nz_1=yx.$

vi) An element of $\rm{Aut}(G)$ is a bijection $G \to G$ and there are only $|G|!$ such bijections. In fact, clearly, if $|G|=n,$ then $\rm{Aut}(G)$ is (isomorphic to) a subgroup of the symmetric group $S_n.$

vii) See this post.

viii) As shown in the post linked in v), $\rm{Aut}(G)$ always has an element of order $2$ in this case, and so $2$ divides $|\rm{Aut}(G)|.$

ix) If $\rm{Aut}(G)$ is cyclic, then $\rm{Inn}(G),$ which is a subgroup of $\rm{Aut}(G),$ is cyclic too. But then, by v), $G$ would be abelian, contradiction.

x) Suppose, to the contrary, that $G$ exists. Then, by ix), $G$ is abelian and so, by viii), $|G| \le 2.$ Thus either $G$ is trivial or $G \cong \mathbb{Z}_2,$ and so either way $|\rm{Aut}(G)|=1,$ contradicting our assumption that $\rm{Aut}(G)$ is non-trivial.

xi) Since $|\rm{Inn}(G)|$ divides $|\rm{Aut}(G)|,$ because $\rm{Inn}(G)$ is a subgroup of $\rm{Aut}(G),$ we only need to show that $p^2$ divides $|\rm{Inn}(G)|.$ Thus, by iv), we only need to show that $p^2$ divides $|G/Z(G)|.$ Well, since $G$ is a $p$ -group, $|G/Z(G)|=p^m$ for some integer $m \ge 0.$ If $m=0,1,$ then $G/Z(G)$ is cyclic and hence, by v), $G$ is abelian, which is a contradiction. So $m \ge 2$ and hence $p^2$ divides $|G/Z(G)|.$

xii) For any $\alpha \in \rm{Aut}(H),$ define $\overline{\alpha}: G \to G$ by $\overline{\alpha}(h,k)=(\alpha(h),k)$ for all $h \in H, k \in K.$ See that $\overline{\alpha} \in \rm{Aut}(G).$ Now define the map $f: \rm{Aut}(H) \to \rm{Aut}(G)$ by $f(\alpha)=\overline{\alpha}$ and see that $f$ is an injective group homomorphism.