# A finite group containing an element with some property is a $p$-group

Let $G$ be a finite group. Suppose there exists a non-trivial element $g \in G$ such that $gxg^{-1}=x^{p+1}$ for all $x\in G$. Prove that $G$ is a $p$-group.

-
In other words, $x\mapsto x^{p+1}$ is an inner automorphism. –  Hagen von Eitzen May 14 '14 at 6:28
@HagenvonEitzen: is that it? Does that unlock a proof? If so, please write it up as an answer! Curiosity is eating me alive as I type these words! Whoops! There goes my right arm! Oh well, can still type; glad I'm left handed! ;-) –  Robert Lewis May 14 '14 at 6:35

By the condition, $gx=xg\iff x^p=1$. Hence the centralizer $C(g)$ is a $p$-group and is given by $$C(g)=\{\,x\in G\mid x^p=1\,\}.$$ Especially, we have $g^{-1}\in C(g)$ and for arbitrary $x$ we have $x^{-1}gx\in C(g)$, hence also $$x^p=x^{-1}gxg^{-1}\in C(g).$$ We conclude that $x^{p^2}=1$ for all $x\in G$, hence $G$ is a $p$-group.