A few questions on $\mathrm{Aut}(G)$ when $G$ is finite.

1) $\mathrm{Aut}(G)$ is not cyclic when $G$ is not abelian

I proved this by showing the contra-positive, that if $\mathrm{Aut}(G)$ is cyclic then $G$ is abelian. I'm trying to show it directly now.

Since $G$ is not abelian, there exists a nontrivial inner automorphism of finite order. This inner autmorphism will generate a cyclic subgroup of $\mathrm{Inn}(G) \leq \mathrm{Aut}(G)$ If I can show that $\mathrm{Inn}(G)$ is a proper subgroup of $\mathrm{Aut}(G)$ I will be done.

2) $\mathrm{Aut}(G)$ is never cyclic of odd order $> 1$

I need help with this one.


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    $\begingroup$ You say that if you can show that $\mathrm{Inn}(G)$ is a proper subgroup of $\mathrm{Aut}(G)$ you will be done, after showing that that $\mathrm{Inn}(G)$ contains a cyclic subgroup, I simply do not see how you think that proving that $\mathrm{Inn}(G)$ is a proper subgroup, you will be able to conclude that $\mathrm{Aut}(G)$ is not cyclic. $\endgroup$ Jul 2, 2019 at 2:58
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    $\begingroup$ to prove (1) "directly", you pretty much just reverse the proof. If $G$ is not abelian then $Inn(G)$ is not cyclic so $Aut(G)$ is not cyclic. I don't know if there is any other way to go (besides slight modifications involving $G/Z(G)$) $\endgroup$ Jul 2, 2019 at 3:03
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    $\begingroup$ See also this question. This exercise of Rotman's book has been solved on this site. $\endgroup$ Jul 2, 2019 at 11:26

1 Answer 1


Suppose that $Aut(G)$ is cyclic of odd order. You have already shown that $G$ is abelian. We will use additive notation for $G$. Consider the inversion map $\varphi\colon G\to G$ defined by $\varphi(g)=-g$. Since $G$ is abelian, $\varphi$ is an automorphism of $G$. Then $\varphi\in Aut(G)$ with $\varphi^2=\varphi\circ\varphi=id_G$. Since $Aut(G)$ has odd order, $\varphi=id_G$. Thus, $g=-g$ for all $g\in G$ so $2g=0$ for all $g\in G$. By the classification of finite abelian groups (or by treating $G$ as a vector space over the finite field $\mathbb{Z}/2\mathbb{Z}$), $G\cong\mathbb{Z}/2\mathbb{Z}\times\cdots\times\mathbb{Z}/2\mathbb{Z}$.

If there is only one copy of $\mathbb{Z}/2\mathbb{Z}$ then $G\cong\mathbb{Z}/2\mathbb{Z}$ and $Aut(G)$ is the trivial group. Otherwise, there are at least two copies of $\mathbb{Z}/2\mathbb{Z}$. Swapping two of those copies of $\mathbb{Z}/2\mathbb{Z}$ gives an automorphism of $G$ order 2. This is impossible since $Aut(G)$ has odd order.

In general, linear algebra shows that $Aut(\mathbb{Z}/2\mathbb{Z}\times\cdots\times\mathbb{Z}/2\mathbb{Z})\cong GL(n,\mathbb{Z}/2\mathbb{Z})$ is the group of $n\times n$ invertible matrices with entries in the finite field $\mathbb{Z}/2\mathbb{Z}$. This group has order $$(2^n-1)(2^n-2)\cdots(2^n-2^{n-1})=2^{\frac{n(n-1)}{2}}(2^n-1)(2^{n-1}-1)\cdots(2^1-1).$$


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