 ### Patrick Stevens

Former mathematics student at the University of Cambridge; now a software engineer.

# The Orbit/Stabiliser Theorem

The Orbit/Stabiliser Theorem is a simple theorem in group theory. Thanks to Tim Gowers for the proof I outline here - I find it much more intuitive than the proof that was presented in lectures, and it involves equivalence relations (which I think are wonderful things).

Theorem: $\vert \{g(x), g \in G\} \vert \times \vert \{g \in G: g(x) = x\} \vert = \vert G \vert$.

Proof: We fix an element $x \in G$, and define two equivalence relations: $g \sim h$ iff $g(x) = h(x)$, and $g \cdot h$ if $h^{-1} g \in \text{Stab}_G(x)$, where $\text{Stab}_G(k) = \{g \in G: g(k) = k\}$.

Now, these are the same relation (we will check that they are indeed equivalence relations - don’t worry!). This is because $g \sim h \iff g(x) = h(x) \iff h^{-1}g(x) = x \iff h^{-1}g \in \text{Stab}_G(x) \iff g \cdot h$.

And $\sim$ is an equivalence relation, almost trivially: it is reflexive since $g \sim g \iff g(x) = g(x)$ is obviously true; it is symmetric, since $g \sim h \iff g(x) = h(x) \iff h(x) = g(x) \iff h \sim g$; it is transitive similarly.

Now, it is clear that the number of equivalence classes of $\sim$ is just the size of the orbit $\{g(x), g \in G \}$, because for each equivalence class there is one member of the orbit (with $[g]$ representing $g(x)$), and for each member of the orbit there is one equivalence class (with $g(x)$ being represented solely by $[g]$).

It is also clear that the size of the stabiliser $\text{Stab}_G(x)$ is just the size of an equivalence class $[g]$ of $\cdot$, since for each member $s$ of the stabiliser, we have that $g \cdot (g s)$ so $\vert [g] \vert \geq \vert \text{Stab}_G(x) \vert"$, while for each for each member $h$ of $[g]$ we have that $h^{-1}g \in \text{Stab}_G(x)$ by definition of $\cdot$ - but all these $h^{-1}g$ are different (because otherwise we could cancel a $g$) so $\vert [g] \vert \leq \vert \text{Stab}_G(x) \vert$

And the equivalence classes of $\sim \ = \cdot$ partition the set $G$, so (size of equivalence class) times (number of equivalence classes) is just $\vert G \vert$ - but this is exactly what we required.