As the title says, I'll assume G and H are subgroups of , where (m is not zero in this case). Here is the sketch of the proof.

Let be a multiplicative group of excluding 0. Let be a surjective group homomorphism such that , where (verify this is a well-defined group homomorphism). The kernel of is a group . By the first isomorphism theorem, (The order of a multiplicative group is 4).