**Theorem**. A representation of a finite abelian group is irreducible if and only if

*Proof*. One side of the theorem was already proved in Remark 2. Now, suppose that is an irreducible representation of So, by definition, is a simple -module and hence

by the second part of Diximier-Schur’s lemma . So if we show that is a -module homomorphism for all then by for all and hence Note that is always a -module homomorphism but in order for to be a -module homomorphism as well, we need to be abelian, as you will see shortly. To show that is a -module endomorphism of we need to prove that for all and all This is easy to see:

**Corollary**. The number of irreducible representations of a finite abelian group is

*Proof*. It is a trivial result of the above theorem and the lemma in this post.

**Question**. Explain how to find all irreducible representations of a finite abelian group

**Answer**. First recall that, by the fundamental theorem of finite abelian groups, every finite abelian group is a finite direct product of cyclic groups. Also, we have already described all the representations of a finite cyclic group in Example 1. So, this is how we can find all irreducible (= degree one) representations of We first write where Thus So every element of can be written uniquely as Now let be an -th root of unity. Define by

for all Clearly each is a group homomorphism and hence a representation of degree one for The number of these representations is because each has possible values and thus we have possible representations of this form. Also, by the Corollary, we know that has exactly irreducible representations. So are all the irreducible representations of

**Example**. Find all irreducible representations of the Klein-four group.

**Solution**. is the direct product of two cyclic groups of order So Let be the generators of these groups. Then Now the second roots of unity are Thus the irreducible representations of are defined by

1)

2)

3)

4)

where