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I think I managed to prove that, if a finite group $G$ has an automorphism $\varphi$ of order $2$ which doesn't fix any nontrivial elements, then:

  1. $G$ has odd order;
  2. $G$ has odd conjugacy classes, and each of them is of odd size;
  3. $\varphi$ acts on the set of the conjugacy classes of $G$ likewise it acts on $G$ itself, namely it doesn't fix any nontrivial conjugacy class.

This is not surprising, since such a group turns out to be Abelian (see e.g. here and here). I'd expect 1-2-3 to lead to an alternative proof of $G$'s Abelianiness, by obtaining from them that all the conjugacy classes must be singletons. This is where I'm stuck.

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    It's hard to answer questions of the form "Can $A$ be used to prove $B$?" when it's unclear what else is fair game in such a proof. To make your question precise, you could ask whether any group $G$ with properties 1, 2, 3 must be abelian, but you need to say what you are assuming about $\varphi$ in 3: is it an automorphism, or just a bijection of $G$? If you assume it's an automorphism of order 2 then it has no nontrivial fixed points and you can just use the old proof. – Sean Eberhard Feb 18 '22 at 13:42
  • @SeanEberhard, assumptions are like in the old proof. What I feel uncomfortable with it, it's the very starting point, i.e. the map $x\mapsto x^{-1}\varphi(x)$, which I'd personally have never thought about (but of course this is a problem of mine). Items 1-2-3 are the facts that I could derive myself from those assumptions, and I have the feeling that they could be used to get that the conjugacy classes must be all singletons, thus proving differently the claim. –  Feb 18 '22 at 13:52
  • Well, "Young man, in mathematics you don't understand things. You just get used to them." (https://math.stackexchange.com/questions/11267/what-are-some-interpretations-of-von-neumanns-quote) – Sean Eberhard Feb 18 '22 at 14:09

1 Answers1

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Let $G$ be the Heisenberg group mod $p$ for odd prime $p>3$. Then $|G|=p^3$ and $G$ has $p^2-1$ conjugacy classes of size $p$ and $p$ conjugacy classes of size $1$, so $G$ satisfies 1 and 2. Also:

  1. There is a bijection $\varphi : G \to G$ (but not an automorphism) such that $\varphi^2 = \mathrm{id}$ fixing the identity, permuting the conjugacy classes, and not fixing any nontrivial conjugacy class: there are an even number of conjugacy classes of size $p$ and an even number of nontrivial conjugacy classes of size $1$, so there are many ways of doing this.

  2. There is an automorphism $\varphi : G \to G$ (but not satisfying $\varphi^2 = \mathrm{id}$) fixing no nontrivial conjugacy class: take $\varphi$ to be conjugation by a diagonal matrix with entries $a^{-1},1,b$ where $a, b, ab \neq 1$. Then $$ \begin{pmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{pmatrix}^\varphi = \begin{pmatrix} 1 & ax & abz \\ 0 & 1 & by \\ 0 & 0 & 1 \end{pmatrix},$$ and it follows from this that no nontrivial conjugacy class is fixed.

And as mentioned in the comment section, if you assume $\varphi$ is both an automorphism and order-2 then $\varphi$ has no nontrivial fixed points so $G$ is abelian.

Sean Eberhard
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  • Incidentally, in 2, $\varphi$ can have prime order for any odd prime, giving examples of Frobenius groups with nonabelian kernel. – Sean Eberhard Feb 18 '22 at 15:53