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Let $G$ be a group and $a \in Aut(G)$ with $o(a)=2.$ If $C_G(a)=1$, then $x^a=x^{-1}$ for all $x \in G$. In particular, $G$ is abelian.

Hello, does anyone have an example where this theorem can be applied? At first, it seems like a very powerful statement, but after further thinking, I couldn't come up with a meaningful example. I'm having trouble understanding when it might be useful, and a quick, simple example would really help.

Thank you in advance for your help :D

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  • $\begingroup$ Well, any automorphism fixes the identity, so do you mean "no fixed points other than the identity"? $\endgroup$
    – lulu
    Commented Jun 22 at 16:26
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    $\begingroup$ Well, we have $(xx^a)^a=x^ax$, so if $x$ commutes with $x^a$, then $x^a=x^{-1}$. But it is not clear what would happen if $x$ does not commute with $x^a$. $\endgroup$ Commented Jun 22 at 16:52
  • $\begingroup$ @lulu yes exactly an automorphim of $G$, where $\forall g\in G \setminus {1}$ it ist $a(g) \neq g$ (I think this is implied by the Centraliser being equal to 1 since this means that the identity is the only element of $G$ where $a(g)=g$) $\endgroup$ Commented Jun 22 at 16:54
  • $\begingroup$ @GeoffreyTrang I do not understand yor comment, the theorem is already proofed i just wondered if it was somehow possible to create an example of a group which has an automorphism $a$ with $o(a)$ s.t. $C_G(a)=1$. So we can conclude $G$ is abelian. $\endgroup$ Commented Jun 22 at 17:01
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    $\begingroup$ @Kan't: I was also confused. From the way in which the OP denotes the action of $a \in \text{Aut}(G)$ using the symbols $x^a$ --- which is also how one would write conjugation of $x$ by another group element $a$ --- I suspect that $C_G(a)$ is the set of all $x \in G$ such that $x^a=x$. In other words, $C_G(a)$ the subgroup of all group elements that are fixed by $a$. $\endgroup$
    – Lee Mosher
    Commented Jun 22 at 18:40

2 Answers 2

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Maybe you want a non-trivial application of this theorem?

A finite group $G$ is a Frobenius group if it has a proper subgroup $H$ such that $H\cap H^g=\{1\}$ whenever $g\in G\setminus H$. Any book on character theory will prove that $G$ is actually a semi direct product $G=K\rtimes H$, with $K$ called the kernel, and $H$ the complement. It's not hard to show that no non-trivial element of $H$ centralizes any non-trivial element of $K$.

Thus, we can use your quoted result to conclude that if $|H|$ is even, then $K$ is Abelian!

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  • $\begingroup$ Thank you, I really like this one! I am still thinking about why $C_K(H)=1$, you said it was not hard to show - is there a little trick to get there? $\endgroup$ Commented Jun 22 at 19:49
  • $\begingroup$ @Stippinator If $h$ centralizes $k$, what can you say about $H^k\cap H$? $\endgroup$
    – Steve D
    Commented Jun 22 at 21:23
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    $\begingroup$ Ah I think I see it. When H acts on K by conjugation you can say that from $H$ centralizes $k \in K$, you can conclude that $h =h^k$ so $H \cap H^k \neq \{1\}$. But this contradicts $H \cap H^g= \{1\}$ for all $g \in G \setminus H$. $\endgroup$ Commented Jun 22 at 22:48
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With the mere excuse to use the theorem within a basic/elementary context, you can take $G:=(\mathbb Z/p\mathbb Z,+)$, with $p$ a prime, and note that: \begin{alignat}{1} &a(0) &&= 0 \\ &a_{|G\setminus\{0\}} &&= (1,p-1)(2,p-2)\dots\bigl(\frac{p-1}{2},\frac{p+1}{2}\bigr) \end{alignat} (cycle notation) is an automorphism of $G$ of order $2$, which solely fixes $0$. So, by the theorem, $G$ is abelian :)

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    $\begingroup$ Thank you, this is also a very nice example and easy! $\endgroup$ Commented Jun 23 at 22:11

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