Group Theory/The action by conjugation and p-groups

Definition (global stabilizer):

Let $G$ be a group that acts on $A$, where $A$ belongs to some algebraic variety ${\mathcal {A}}$. Let $S\subseteq A$ be a subset. Then the global stabilizer of $S$ is the set

$G(S):=\{g\in G|gS\subseteq S\}$,

where the notation $gS$ stands for the set $\{gs|s\in S\}$.

Definition (p-group):

Let $p\in \mathbb {N}$ be a prime number. Then a $p$-group is a group of order $p^{k}$ for some $k\in \mathbb {N}$.

Proposition (cardinality of fixed point set of a p-group equals cardinality of set mod p):

Let $G$ be a $p$-group that acts on a set $X$. Then

$|\operatorname {Fix} (G)|\equiv |X|\mod p$.

Proof: By the class equation,

$|X|=\sum _{k=1}^{n}[G:G_{x_{k}}]$,

where for each orbit of the action of $G$ on $X$ we pick one representative $x_{k}$ of that orbit. Since $G$ is a $p$-group, whenever $[G:G_{x_{k}}]$ is not $1$, it is divisible by $p$ by Lagrange's theorem. Hence, by taking the above equation $\mod p$, we get

$|X|\equiv \sum _{j=1}^{l}1\equiv l\mod p$,

where $l$ is the number of those $x_{k}$ for which $[G:G_{x_{k}}]=1$. But $[G:G_{x_{k}}]=1$ means precisely that the orbit of $x_{k}$ is trivial, that is, that $x_{k}$ is fixed by all of $G$. $\Box$

Proposition (p-groups have nontrivial center):

Let $G$ be a $p$-group. Then $Z(G)\neq \{e\}$, where $e\in G$ denotes the identity.

Proof: $G$ acts on itself via conjugation. Furthermore,

$h\in Z(G)\Leftrightarrow \forall g\in G:g^{-1}hg=h$,

so that $Z(G)$ is precisely the fixed point set of $G$ under that action. But since the cardinality of the fixed point set of a p-group equals the cardinality of the whole set mod p, we get that

$0\equiv |G|\equiv |\operatorname {Fix} (G)|\equiv |Z(G)|\mod p$,

which would be impossible if $|Z(G)|=1$. $\Box$