1. ## p-group help

Let G be p-group and G acts on G by conjugation, show $Z(G) \neq <1>$.

I saw there is a lemma on a book say:
Let G be finite p-group (ie $|G|=p^n$) which acts on finite set $X$.
Let $F=${ $x \in X | x^g =x \forall g \in G$} $=$ set of fixed points of $G$.
Then $|F| \equiv |X|$(mod p).

I think this lemma may help to solve this problem, but I need some help for that. Thank you

2. Originally Posted by kleenex
Let G be p-group and G acts on G by conjugation, show $Z(G) \neq <1>$.

I saw there is a lemma on a book say:
Let G be finite p-group (ie $|G|=p^n$) which acts on finite set $X$.
Let $F=${ $x \in X | x^g =x \forall g \in G$} $=$ set of fixed points of $G$.
Then $|F| \equiv |X|$(mod p).

I think this lemma may help to solve this problem, but I need some help for that. Thank you
Let $G$ act upon itself (call this set $X$) by conjugation. Since $G$ and $X$ are finite and $G$ is a p-group it means $|G| \equiv |X^G| (\bmod p)$ where $X^G$ is the invariant subset fixed by $G$. In fact, $X^G$ is precisely $Z(G)$ by how we defined the group action. This tells us that $|G| \equiv |Z(G)| (\bmod p)$ since LHS is divisible by $p$ it means $p$ divides $Z(G)$.