Remark

The analog for right cosets also holds:

|H\G| = |G|/|H| = |G/H|.

But the converse of Lagrange's theorem does not hold! The group A₅ has 60 elements, but does not have a subgroup containing 30 elements. A proof will be given below, but you may want to experiment using an earlier gapplet. (Give as input permutations from A₅ and check that you never find a a subgroup of order 30.)

So here is the proof. Suppose H is a subgroup of order 30. Then there must be a 3-cycle g which is not contained in H, since the 3-cycles generate A₅. Every (left) coset of H contains 30 elements, so there are only two of them. As soon as an element a is not contained in H, then H and aH are the two left cosets. Apply this to the elements g and g²: A₅ is partitioned into the two left cosets H and gH, but also into the two left cosets H and g²H (g² is not in H, because then g⁴, which equals g, is in H). This implies that gH = g²H. But then we conclude that the element g² is in gH, which in turn implies that g is in H. This is a contradiction.