Throughout is a finite group.

**Theorem**. Let be a degree two representation of Then is irreducible if and only if the linear transformations have no common eigenvector.

*Proof*. Suppose first that the linear transformations have a common eigenvector, say Then and thus is a one dimensional -subspace of Also, becasue Now let Then for some because we assumed that is an eigenvector of all Thus So is a -submodule of and hence is not simple, i.e. is not irreducible. Conversely, suppose that is reducible, i.e. is not simple. Let be a -submodule of Then because So there exists such that Now, let Then because and is a -module. Therefore for some So we have proved that is an eigenvector of all

**Example**. In Example 1 in this post I gave degree two representations of the dihedral group of order In the remark, following Example 1 in that post, I showed that the matrices of the representation corresponding to an -th root of unity are

where We now have the following claim:

**Claim** . If and then is irreducible.

*Proof*. The eigenvectors of are and the eigenvectors of are So and have no common eigenvector and thus, by the above theorem, is irreducible.