# Automorphism group cyclic implies abelian group, do we have more?

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.

• $\varphi_g$ 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 $G$. Sep 29 at 18:24

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.