# Group Theory/The action by conjugation and p-groups

Definition (global stabilizer):

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

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

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

Definition (p-group):

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

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

Let ${\displaystyle G}$ be a ${\displaystyle p}$-group that acts on a set ${\displaystyle X}$. Then

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

Proof: By the class equation,

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

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

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

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

Proposition (p-groups have nontrivial center):

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

Proof: ${\displaystyle G}$ acts on itself via conjugation. Furthermore,

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

so that ${\displaystyle Z(G)}$ is precisely the fixed point set of ${\displaystyle G}$ 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

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

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