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|.

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