From a bank of previous masters exams:

Let $G$ be a finite group such that its automorphism group $\operatorname{Aut}(G)$ is cyclic. Prove that $G$ is abelian.

Here's what I was thinking. Let $\phi:G \to G$ be the generator of $\operatorname{Aut}(G)$, with order $n$. Assume that $G$ is not abelian. Then there is a nontrivial inner automorphism $\psi_g(x) = g^{-1}xg$. Since $\phi$ generates all automorphisms, then $\psi_g = \phi^k$ for some $k$. This also implies that $\psi_g^n(x) = g^{-n}xg^n$ is the identity map.

After that, no luck. Any ideas?


If the automorphism group is cyclic, then the Inner automorphism group is cyclic. But the inner automorphism group is isomorphic to $G/Z(G)$, and if $G/Z(G)$ is cyclic, then it is trivial. Therefore, $G=Z(G)$ so $G$ is abelian. (The argument does not require $G$ to be finite, by the by.)

For the latter:

Prop. If $H\subseteq Z(G)$ and $G/H$ is cyclic, then $G$ is abelian.

Suppose $G/H$ is cyclic, with $H\subseteq Z(G)$. Let $g\in G$ be such that $gH$ generates $G/H$. Then every $x\in G$ can be written as $x=g^kh$ for some integer $k$ and some $h\in H$. Given $x,y\in G$, we have $x=g^kh$ and $y=g^{\ell}h'$, so $$\begin{align*} xy &= (g^kh)(g^{\ell}h')\\ &= g^kg^{\ell}hh' &\quad&\text{since }h\in Z(G)\\ &= g^{\ell}g^kh'h\\ &= g^{\ell}h'g^kh &&\text{since }h'\in Z(G)\\ &= (g^{\ell}h')(g^kh)\\ &= yx, \end{align*}$$ hence $G$ is abelian. QED

For more on what groups can occur as central quotients, see this previous question.

Added. Since you mention you did not know that $\mathrm{Inn}(G)\cong G/Z(G)$, let's do that too:

Define a map $G\to \mathrm{Aut}(G)$ by mapping $g\mapsto \varphi_g$, where $\varphi_g$ is "conjugation by $g$". That is, for all $x\in G$, $\varphi_g(x) = gxg^{-1}$.

This map is a group homomorphism: if $g,h\in G$, then we want to show that $\varphi_{gh} = \varphi_g\circ\varphi_h$. To that end, let $x\in G$ be any element, and we show that $\varphi_{gh}(x) = \varphi_g(\varphi_h(x))$. $$\varphi_{gh}(x) = (gh)x(gh)^{-1} = ghxh^{-1}g^{-1}= g(hxh^{-1})g^{-1} = \varphi_g(hxh^{-1}) = \varphi_g(\varphi_h(x)).$$

Therefore, the map $g\mapsto\varphi_g$ is a homomorphism from $G$ onto $\mathrm{Inn}(G)$. By the Isomorphism Theorem, $\mathrm{Inn}(G)$ is isomorphic to $G/N$, where $N$ is the kernel of this homomorphism.

What is $N$? $g\in N$ if and only if $\varphi_g$ is the identity element of $\mathrm{Aut}(G)$, which is the identity; that is, if and only if $\varphi_g(x)=x$ for all $x\in G$. But $\varphi_g(x)=x$ if and only if $gxg^{-1}=x$, if and only if $gx = xg$. So $\varphi_g(x) = x$ if and only if $g$ commutes with $x$. Thus, $\varphi_g(x)=x$ for all $x$ if and only if $g$ commutes with all $x$, if and only if $g\in Z(G)$. Thus, $N=Z(G)$, so $\mathrm{Inn}(G)\cong G/Z(G)$, as claimed.

  • $\begingroup$ Thank you. I did not know that the inner automorphism group is isomorphic to $G/Z(G)$, and I'll have to think about why that is true (although it is intuitively plausible). And your proposition proves that a cyclic $G/Z(G)$ must be trivial. $\endgroup$ Apr 18 '11 at 0:58
  • $\begingroup$ @Michael: It's a standard fact, but if you've never seen it, you've never seen it. I've added the standard proof for this to the post. $\endgroup$ Apr 18 '11 at 1:41
  • $\begingroup$ Ah – I had planned on sketching this out, but you seem to have beat me to it. And I forgot to accept your answer. Thank you; this is above and beyond the call of duty. $\endgroup$ Apr 18 '11 at 2:46

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