Group Actions
Table of Contents
1 Definition
Let \( G \) be a group and \( X \) a set, then we say a function \( F: G \times X \to X \), \( F(g, x) = g \cdot x \), is an action of \( G \) on \( X \) if it satisfies:
- \( \forall g_1, g_2 \in G, \ x \in X \) we have that \( g_1 \cdot (g_2 \cdot x) = \left( g_2g_1 \right) \cdot x \)
- \( \forall x \in X, \ 1 \cdot x = x \)
We also say \( X \) is a G-set.
2 Examples
3 Orbits and Stablizers
Definition Let \( G \) be a group that acts on the set \( X \). For \( x \in X \) we define the stabilizer of \( x \) in \( G \), \( stab_G(x) \) as:
\[ stab_G(x) = \{ g \in G : g \cdot x = x \} \]
Definition Let \( G \) be a group that acts on the set \( X \). For \( x \in X \) we define the orbit of \( x \) in \( G \), \( orb_G(x) \) as:
\[ orb_G(x) = \{ g \cdot x : g \in G \} \]
The following definitions will also be useful in the next section, they are the flipside of the previous two definitions:
Definition Let \( G \) be a group that acts on the set \( X \). Let \( g \in G \), then we define \( X^g \) as:
\[ X^G = \{ x : g \cdot x = x \} \]
ie the set of all fixed points of \( X \) with respect to \( g \).
Definition Let \( G \) be a group that acts on the set \( X \). We define the set of orbits of \( G \), \( X / G \) as:
\[ X / G = \{ orb_G(x) : g \in G \} \]
3.1 Example
Let \( G = \{ (1), \ (1 \ 2), \ (3 \ 4 \ 6), \ (3 \ 6 \ 4), \ (1 \ 2)(3 \ 4 \ 6), \ (1 \ 2)(3 \ 6 \ 4)\} \), so that \( G \) acts on \( \{ 1,2,3,4,5,6 \} \). Since 5 is fixed by all elements of \( G \) we have that:
\begin{align*} stab_G(5) &= G \\ org_G(5) &= \{ 5 \} \end{align*}Similarly, we have that:
\begin{align*} stab_G(3) &= \{ (1), \ (1 \ 2) \} \\ org_G(3) &= \{ 3, \ 4, \ 6 \} \end{align*}3.2 Orbit-Stabilizer Theorem
Supppose \( G \) is a finite group which acts on the set \( X \). Then for all \( x \in X \) we have:
\[ \mid G \mid = \mid stab_G(x) \mid * \mid orb_G(x) \mid \]
Proof Fix \( x \in X \). We first show that \( stab_G(x) \le G \). First we know that \( e \in stab_G(x) \) since \( e \cdot x = x \). Now let \( g, h \in G \). We have:
\begin{align*} g^{-1} \cdot (g \cdot x) = g^{-1} \cdot x &\implies (gg^{-1}) \cdot x = g^{-1} * x \\ &\implies e \cdot x = g^{-1} \cdot x \\ &\implies g^{-1} \cdot x = x \\ &\implies g^{-1} \in stab_G(x) \end{align*}Secondly:
\[ (gh) \cdot x = g \cdot (h \cdot x) = g \cdot x = x \]
Hence \( gh \in stab_G(x) \) thus \( stab_G(x) \le G \).
It then follows from Lagrange's Theorem that the number of left cosets of \( stab_G(x) \) in \( G \) is given by \( [G:stab_G(x)] = \mid G \mid / \mid stab_G(x) \mid \). Letting \( x \) be fixed as before, we define:
\[ f : G \to X \text{ by } f(g) = g \cdot x \]
Note the image of \( f \) is given by \( orb_G(x) \) by definition. Now suppose two elements map to the same value:
\[ f(g) = f(h) \iff g \cdot x = h \cdot x \iff g^{-1}h \cdot x = x \iff g^{-1}h \in stab_G(x) \]
Which is equivalent to \( g \) and \( h \) having the same left coset with respect to \( stab_G(x) \). From this it is apparent that there is a bijection between the left cosets of \( stab_G(x) \) and \( orb_G(x) \). Hence:
\[ \mid G \mid / \mid stab_G(x) \mid = \mid orb_G(x) \mid \]
which rearranges to the required equality.