Proof: Let H be a subgroup of a group G. A left coset of H for an element g of G is the set gH={gh: h∈H}. It needs to be shown that the left cosets constitute a partitioning of the set G; i.e., each element of G belongs to one and only one coset.

First, any element g must belong to some coset. Obviously g belongs to its own coset because g=ge, where e is the identity element of H.

Consider two left cosets xH and yH. Suppose xH and yH have a common element z. This means that there exists h₁ and h₂ belonging to H such that z=xh₁ and z=yh₂. Thus xh₁=yh₂ and hence y = xh₁h₂^-1. Now let w be any element of yH. The there exists h₃ such that w=yh₃. But this implies that w=(xh₁h₂^-1)h₃=x(h₁h₂^-1h₃). Since h₁h₂^-1h₃ necessarily belongs to H w must belong to xH. Thus yH ⊆ xH. Likewise a similar argument demonstrates that xH ⊆ yH so xH=yH.

The cardinality of all the cosets must be equal. It cannot be that there is an element z of xH such that for two different h₁ and h₂, z=xh₁ and z=xh₂ because this would imply that x=xh₁h₂^-1 and thus h₁h₂^-1=e and hence h₁=h₂.

Thus G is partitioned into cosets of equal size (which is the order of the subgroup) and hence the order of G must be a multiple of the order of the subgroup.

Since the orders of a group and a subgroup may be infinite, it is best to state the theorem in the form

The order of a group must be a multiple
of the order of any of its subgroups.

The multiple is the number of cosets of a subgroup.