Throughout this section is a group, is an **abelian** subgroup of and

**Definition**. Let be the partition of into left cosets of The set is called a **transversal** of in

**Lemma 1**. Let be a transversal of in and There exists a unique permutation and such that for all

*Proof.* Clearly for every there exists a unique such that Define the map by So we just need to prove that is a bijection. Since is defined on a finite set, we only need to prove that it is one-to-one. So suppose that Then and and thus and are both in the left coset Hence and so To prove that is unique, suppose that is another permutation and for all Then for some and which is nonsense.

**Lemma 2**. Let and be two transversals of in and let By Lemma 1 there exists and such that Prove that there exist permutations and such that for all

*Proof*. Applying Lemma 1 to and we find and such that

for all Let Then (1) and the fact that gives us and hence

for all Also, since we have

for all Therefore multiplying (1) by from the left and then applying (3) to the right side we get

for all Finally replcaing on the right side of (4) with what we got in (2) completes the proof.

**Remark.** Let’s take another look at the situation in Lemma 2. Well, by Lemma 1 there exist a unique and such that for all The uniqueness of and Lemma 2 gives us and for all

**Lemma 3**. Let and be two transversals of in and let By Lemma 1 there exist and such that and for all Prove that

*Proof*. By Remark 1 we know that there exist and such that and for all Thus where products are over Now, since we have and Therefore, since is abelian, we get

Now the main result of this post:

**Theorem**. Let be a transversal of in By Lemma 1, for every there exists and such that for all Define the map by Then is a well-defined group homomorphism.

*Proof.* Lemma 3 shows that is well-defined. Now let Let and be such that and Then and thus Thus, since is abelian, we have