TITLE: A finite group containing an element with some property is a $p$-group
QUESTION [9 upvotes]: 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.

REPLY [8 votes]: 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.