## Lagrange’s Theorem

Let $H$ be a subgroup of a finite group $G$ and let $H$ act on $G$ by left multiplication. Given $x \in G$, let $\mathcal{O}_x$ denote the orbit of $x$ under the action of $H$. Prove that the map $\varphi : H \rightarrow \mathcal{O}_x$ given by $\varphi{h} = h \cdot x$ is a bijection, so that all orbits have cardinality $|H|$. From this an the preceding exercise deduce Lagrange’s Theorem: If $G$ is a finite group and $H$ a subgroup of $G$ then $|H|$ divides $|G|$.

The mapping $\varphi$ is “contained in” the left regular action of $G$ on itself, which is known to be faithful. Thus $\varphi$ is injective. Now suppose $y \in \mathcal{O}_x$; then there exists $h \in H$ such that $y = h \cdot x$, so that $\varphi$ is surjective. So $\varphi$ is bijective.

Now recall from the previous exercise that the action of $H$ on $G$ induces an equivalence relation $\sim$ on $G$ whose equivalence classes $G/\!\sim$ are precisely the orbits of $G$ under the action. Since $G$ is finite, we have that $G$ is the finite disjoint union of $\sim$-classes. By this exercise, each of these classes has the same cardinality, $|H|$. Thus $|G| = |\cup_{i=1}^k|\mathcal{O}_i| = \sum_{i=1}^k |H|$ $= k|H|$. Thus $|H|$ divides $|G|$.