1. ## p-group help

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

I saw there is a lemma on a book say:
Let G be finite p-group (ie $\displaystyle |G|=p^n$) which acts on finite set $\displaystyle X$.
Let $\displaystyle F=${$\displaystyle x \in X | x^g =x \forall g \in G$} $\displaystyle =$ set of fixed points of $\displaystyle G$.
Then $\displaystyle |F| \equiv |X|$(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 $\displaystyle Z(G) \neq <1>$.

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

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