Let $(G, \cdot)$ be a group and $M$ a set. A group action is a function:
$G \circ M \rightarrow M$
that satisfies the following two conditions
- Identity: $\forall m \in M: e_G \circ m = m$
- Associativity: $\forall g, h \in G, m \in M: (g \cdot h) \circ m= g \circ (h \circ m)$
Let $m \in M$. Then $G_m := \{g \in G | g \circ m = m\}$ is called the stabilizer of $m$.
Theorem: $G_m$ is a group
Proof:
Let $m \in M$.
$\stackrel{identity}{\Rightarrow}e_G \in G_m$
Let $a \in G_m$. This means, that $a \circ m = m$. And $\exists a^{-1} \in G$, as $G$ is a group. $a^{-1} \cdot a = e_G$, this means $(a^{-1} \cdot a) \circ m = m$. $\stackrel{associativity}{\Rightarrow}a^{-1} \circ (a \circ m) = m \Leftrightarrow a^{-1} \circ m = m \Leftrightarrow a^{-1} \in G_m$
Let $a, b \in G_m$. Then: $a \circ m = m$ and $b \circ m = m$
$\Rightarrow a \circ (b \circ m) = m$ $\stackrel{associativity}{\Rightarrow} (a \cdot b) \circ m = m \Leftrightarrow (a \cdot b) \in G_m \blacksquare$