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I wonder something about an action of a group $A$ on a group $G$ by a automorphism;

There are many nice result related with some restrictions such as when $(|A|,|G|)=1$ , $G$ is abelian or $[G,A,A]=e$...

I wonder can we say anything when we know $A$ act on $G$ by inner automorphism i.e.for any $a\in A$ and $g\in G$ there exist an $h\in G$ s.t. $g^a=g^h$.

mesel
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1 Answers1

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There is nothing you can say, because any group $A$ has the trivial group action on $G$, and in this case $A$ acts by inner automorphisms.

The assumption that $A$ acts on $G$ by inner automorphisms just gives you some homomorphism $f: A \rightarrow \operatorname{Inn}(G)$. Without any restriction on this action or on the groups $A$ and $G$, you cannot deduce anything from this.

spin
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  • That is a big claim :) (you can ignore the trivial action) – mesel Nov 09 '14 at 19:38
  • My trivial observation is that we can say $(|A|,|G|) \neq 1$ in that case. – mesel Nov 09 '14 at 19:44
  • Well all you get then is that there is some nontrivial homomorphism $A \rightarrow \operatorname{Inn}(G)$. I think this is a very weak condition.. I would say there is nothing nontrivial that you can find out from this – spin Nov 09 '14 at 19:56
  • I have found a lots of article related this problem. The problem first asked by Burnside at 1911. – mesel Nov 15 '14 at 07:38
  • Not sure this is what really Burnside first asked in 1911, if by this you mean what himself first anwered in 1913: https://math.stackexchange.com/a/4564436/1092170 , @mesel. – Kan't Nov 08 '24 at 06:17