Let be a subgroup of a finite group and let act on by left multiplication. Given , let denote the orbit of under the action of . Prove that the map given by is a bijection, so that all orbits have cardinality . From this an the preceding exercise deduce Lagrange’s Theorem: If is a finite group and a subgroup of then divides .
The mapping is “contained in” the left regular action of on itself, which is known to be faithful. Thus is injective. Now suppose ; then there exists such that , so that is surjective. So is bijective.
Now recall from the previous exercise that the action of on induces an equivalence relation on whose equivalence classes are precisely the orbits of under the action. Since is finite, we have that is the finite disjoint union of -classes. By this exercise, each of these classes has the same cardinality, . Thus . Thus divides .