Orbit-Stabilizer Theorem

There are a lot of things about group theory that we can prove through group actions. We could take that route if we wanted, but I'm not going to do that now. For now, though, I'm going to prove a result of group actions, called the Orbit-Stabilizer Theorem. On the list of important theorems in group theory the Orbit-Stabilizer Theorem isn't very high, but it does have its usefullness and it is an interesting and unexpected result. I'm now going to state the theorem and then spend the rest of the entry proving it.

Theorem:Orbit-Stabilizer Theorem

Suppose that a group, $G$, acts on a set, $S$, and that $i \in S$. Then $\left | G \right | = \left | \text{orb}_G(i) \right | \left | \text{stab}_G(i) \right |$.

Proof:

Let $G$ be a group acting on a set, $S$, and choose $i \in S$. Since $\text{stab}_G(i)$ is a subgroup of $G$, Lagrange's Theorem gives us that $\left | G / \text{stab}_G(i) \right | = \left | G \right | / \left | \text{stab}_G(i) \right |$, or that $\left | G \right | / \left | \text{stab}_G(i) \right |$ is the number of distinct left cosets of $\text{stab}_G(i)$ in $G$. I now wish to extablish a bijection between the left cosets of $\text{stab}_G(i)$ and the elements of $\text{orb}_G(i)$.

Define $\theta : G/\text{stab}_G(i) \to \text{orb}_G(i)$ by $\theta(g \; \text{stab}_G(i)) = g \cdot i$ for $g \; \text{stab}_G(i) \in G/\text{stab}_G(i)$. It is clear that $g \cdot i$ is in $\text{orb}_G(i)$ but it is not clear that $\theta$ is well defined - that is that the image of $\theta$ is independent of the choice of coset representative meaning that if $H \in G/\text{stab}_G(i)$ and $g,h \in H$ then $\theta ( g \; \text{stab}_G (i) ) = \theta ( h \; \text{stab}_G (i) )$.

To show that $\theta$ is well-defined, suppose that $g \; \text{stab}_G(i) = h \; \text{stab}_G(i)$ for $g,h \in G$. Then from our properties of cosets we have that $g^{-1} h \in \text{stab}_G(i)$ which means that $\left ( g^{-1} h \right ) \cdot i = i$ so that $g \cdot i = h \cdot i$. We now have that $g \; \text{stab}_G(i) = h \; \text{stab}_G(i)$ implies $\theta ( g \; \text{stab}_G (i) ) = \theta ( h \; \text{stab}_G (i) )$. Thus $\theta$ is well-defined.

To show that $\theta$ is injective we can use the reverse of the previous argument. Indeed, suppose that $g \cdot i = h \cdot i$ for $g,h \in G$. We then have that $\left ( g^{-1} h \right ) \cdot i = i$ so that $g^{-1} h \in \text{stab}_G(i)$ and by the properties of cosets, $g \; \text{stab}_G(i) = h \; \text{stab}_G(i)$. This gives that $\theta$ is injective.

To show that $\theta$ is surjective, choose $j \in \text{orb}_G(i)$. By definition, $\exists g \in G$ such that $g \cdot i = j$ and so clearly $\theta (g \; \text{stab}_G(i) ) = g \cdot i = j$. This gives that $\theta$ is surjective and finally that $\theta$ is bijective.

Since $\theta$ is a bijection, we have that $\left | G/\text{stab}_G(i) \right | = \left | \text{orb}_G(i) \right |$ and so $\left | G \right | / \left | \text{stab}_G(i) \right | = \left | \text{orb}_G(i) \right |$ and finally $\left | G \right | = \left | \text{orb}_G(i) \right | \left | \text{stab}_G(i) \right |$ as desired.

This is an interesting theorem since it does not seem at first glance like orbits and stabilizers should be related to one another.

References

Previous Related Post: Group Action Attributes

Text Reference: Gallain Chapter 7

Wikipedia: Group Action - Orbits and Stabilizers