I came across this question in a past exam in an Algebraic structures course. It's a first course in group theory, and I had some difficulty solving this problem.
Let $ p \in \mathbb{N}$ be a prime number, and $G$ is a finite group.
Assume for all $g \in G \setminus \{e_G\}$ there exists $r \in \mathbb{N}$ such that $\left| C_G(g)\right| = p^r$. where $C_G(g) = \{x \in G \mid xgx^{-1}=g\}$.
Show that G is a p-group.
I have spent a lot of time trying to solve it, but without success...
Any suggestions would be greatly appreciated!
Assume by contradiction that $G$ is not a p-group. Thus, there must be some prime $q\not=p$ which divides the order of $\left|G\right|$. By Cauchy's theorem, there exists some $x\in G$ of order $o(x)=q$. But $x\in C_{G}(x)$ because $$xxx^{-1}=xe_{G}=x,$$and by Lagrange's theorem we have that $o(x)\mid \left|C_{G}(x)\right|=p^{r}$ reaching a contradiction since $q\not\mid p^{r}$. Thus $G$ must be a p-group.