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

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.

Thanks!

• 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. – Arturo Magidin Jul 2 '19 at 2:58
• 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 $G/Z(G)$) – Thomas Browning Jul 2 '19 at 3:03
• See also this question. This exercise of Rotman's book has been solved on this site. – Dietrich Burde Jul 2 '19 at 11:26

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 $$(2^n-1)(2^n-2)\cdots(2^n-2^{n-1})=2^{\frac{n(n-1)}{2}}(2^n-1)(2^{n-1}-1)\cdots(2^1-1).$$