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 \(b \circ m = m\)
\(\Rightarrow a \circ (b \circ m) = m\) \(\stackrel{associativity}{\Rightarrow} (a \cdot b) \circ m = m \Leftrightarrow (a \cdot b) \in G_m \blacksquare\)