I am working on this exercise in Lang's Algebra:

Exercise I.7: Let $G$ be a group such that $\text{Aut}(G)$ is cyclic. Prove that $G$ is abelian

I have shown the set of inner-automorphisms form a (normal) subgroup in $\text{Aut}(G)$, so let $H$ be this subgroup. As a subgroup of a cyclic group, it is itself cyclic, so let $\varphi_x$ be its generator. Here $\varphi_x(g) = xgx^{-1}$ for any $g\in G$. But now for arbitrary $g$, the inner-automorphism $\varphi_g$ must be generated by $\varphi_x$, in other words $g = x^k$ for some integer $k$.

My question: Doesn't this show that not only is $G$ abelian, it is actually cyclic? I don't see anything wrong with my proof, but I find it strange the author didn't ask to show $G$ is cyclic instead.

  • 1
    $\begingroup$ $\varphi_g$ generated by $\varphi_x$ only gives you $\varphi_g=\varphi_{x^k}$ not $g=x^k$. In particular, if $G$ is abelian, then all $\varphi$'s are trivial automorphisms of $G$. $\endgroup$ Sep 29 at 18:24

1 Answer 1


If $G$ is abelian, then $\phi_g=\mathrm{id}_G$ for all $g\in G$.

Your error is in concluding that if $\phi_g = (\phi_x)^{\circ k}$ (that is, $\phi_g$ is $\phi_x$ composed with itself $k$ times), then it must be the case that $g=x^k$.

In general, if $\phi_g=\phi_h$, that tells you that for all $x\in G$, $gxg^{-1}=hxh^{-1}$, so $h^{-1}gx = xh^{-1}g$. That is, $g$ and $h$ lie in the same coset of $Z(G)$ in $G$; you can also understand that from the fact that $\mathrm{Inn}(G)\cong G/Z(G)$. So if $g$ and $h$ have the same image in $G/Z(G)$, then they are congruent modulo the center, but not necessarily equal.

Here, because $G$ is abelian, $G=Z(G)$. So the fact that an arbitrary $g$ is congruent to $x^k$ for some $k$ modulo the center is unsurprising: everything is congruent to everything else modulo $Z(G)$. You do not conclude $g=x^k$, you conclude that $g\in x^kZ(G) = x^kG = G$... which is of course true.

That said, it is in fact true that for finite $G$, if $\mathrm{Aut}(G)$ is cyclic then $G$ is cyclic, and of order of the form $2^mp^k$ for some odd prime $p$, $m=0$ or $1$, and $k\geq 0$. You can find a proof here. If $G$ is infinite the conclusion need not hold; see here and here.


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