Lagrange's theorem

The idea of the proof is counting. The left cosets of \(H\) partition \(G\), and each has exactly \(|H|\) elements, because \(h\mapsto ah\) is a bijection from \(H\) onto \(aH\). If there are \([G:H]\) cosets, then \(|G|=[G:H]\,|H|\).

An immediate consequence is that a group of prime order \(p\) has only the subgroups \(\{e\}\) and \(G\); in particular, every element other than the identity generates it, and the group is cyclic. It also follows that in a finite group the order of each element divides the order of the group, since that order is the order of the subgroup the element generates.

The converse is false: \(d\) dividing \(|G|\) does not guarantee a subgroup of order \(d\). The group \(A_4\), of order \(12\), has no subgroup of order \(6\). Cauchy's and Sylow's theorems give the partial converses that do hold, for divisors that are prime powers.

The theorem requires \(G\) to be finite, although the index makes sense in general. A useful variant is the tower law: if \(K\le H\le G\), then \([G:K]=[G:H]\,[H:K]\).