Orbit Stabilizer Theorem

Theorem

When $G$ acts on a set $X$ , for each $x \in X$ there's a bijection between the Orbit of $x$ and the cosets of the Stabilizers of $x$ , denoted $G_{x}$ , in $G$ :

O_{x} ≅ \frac{G}{G_{x}}

Thus $| O_{x} | = | G : G_{x} |$ (see Index (Groups)). This map is done by $g \cdot x \mapsto g G_{x}$ .

The idea here is that the 25 The Lattice of Subgroups of a Group are themselves vertices of, say, a Dihedral Groups, which themselves can be analyzed! The orbits here let us get to that point.

Proof

Define the set map $f : O_{x} \to \frac{G}{G_{x}}$ by $f (g \cdot x) \mapsto g G_{x}$ .

First, is $f$ well-defined (since we are not mapping from a quotient group)? Suppose $g_{1} \cdot x = g_{2} \cdot x$ for $g_{1}, g_{2} \in G$ and $x \in X$ . Then:

\begin{aligned} g_{1} \cdot x = g_{2} \cdot x & ⟺ g_{2}^{- 1} \cdot (g_{1} \cdot x) = x \\ ⟺ (g_{2}^{- 1} g_{1}) \cdot x = x \\ ⟺ g_{2}^{- 1} g_{1} \in G_{x} \\ ⟺ g_{1} G_{x} = g_{2} G_{x} \\ ⟺ f (g_{1} \cdot x) = f (g_{2} \cdot x) \end{aligned}

Next, is $f$ surjective? Take any $g G_{x}$ in $\frac{G}{G_{x}}$ . Then $g \cdot x \in O_{x}$ and $f (g \cdot x) = g G_{x}$ .

Is $f$ injective? Suppose $g_{1} \cdot x = g_{2} \cdot x \in O_{x}$ . Then:

\begin{array}{r} f (g_{1} \cdot x) = f (g_{2} \cdot x) \Rightarrow g_{1} \cdot x = g_{2} \cdot x \end{array}

Thus this is a valid Isomorphism!

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