Stormey,
I think the commutator subgroup of G is universally denoted by G'. The easiest way to answer your question is to work in the factor group G/G'. However a direct approach is:
Let h be in H and g in G. Then [g,h^{-1}]=g^{-1}hgh^{-1} is in H and so g^{-1}hg is in H.