As part (1) we’ll assume that is a group and is an abelian subgroup of with We will also fix a transversal of in Let be the group homomorphism introduced in the theorem we proved in part (1).

**Theorem**. There exist positive integers and elements such that and for all

*Proof.* Let By Lemma 1 in part (1), there exists and such that

for all Now let be the decomposition of into disjoint cycles. Let

for all Clearly Now by (1) we have

if and

If in (2) we let and multiply from the left by we will get

But if in (2) we let we will get

So if we replace in the right side of (4) with what I got in (5), we will get If we repeat this process, we will get If we continue in this manner, paying attention to the last step that we’ll have to use (3), we will eventually get

Thus if we let then (6) gives us

Note that since is the cycle decomposition of we have and so Thus the definition of and (7) gives us