This is a standard result that any beginner in the study of the representation theory would be aware of. I merely present restate the theorem a little(essentially I don’t) and prove it.

## Theorem:

Let be representations of a group . If is a map, then the following are true assuming that and are not isomorphic:

1)If both are irreducible, then is either an

isomorphismor the zero map.2)If only is irreducible, then is either

injectiveor the zero map.3)If only is irreducible, then is either

surjectiveor the zero map

If are isomorphic as representations, then is a scalar multiplication map.

#### Proof:

Let be the respective representations. Here, are vector spaces over an algebraically closed field.

We tackle the first part of the theorem.

Consider the kernel of , say , which is set of all such that . Since is a G-linear map, implies that if , then . Hence, is stable under the action of which makes it a sub-representation. Since is irreducible, this means that either must be zero or must be the zero map. This proves that is injective.

Using the same technique, we can also prove surjectivity. Consider some image of , say . Again, since the map is G-module homomorphism, , which means that is also in the image of . So the image of is a subspace of stable under the action of . Since is also irreducible, the image must be all of .

To tackle the second and third parts of the theorem, you simply use the above the two arguments.

If are isomorphic as representations, then by the Fundamental Thoerm of Algebra, the map has an eigenvalue . Let be a corresponding eigenvector.

Now, construct another G-linear homomorphism: . . is also a homomorphism of representations. By the previous results, any homomorphism must be either an isomorphism or zero. No isomorphism would map a non-zero element to zero. So, is the zero map.

So, .

This lemma is key to developing the row orthogonality relations in representation theory which I’ll not discuss here

This entire result can be translated(or rather extended) to the language of modules quite easily.

If are two **simple modules** over a ring , then any homomorphism is either invertible or the zero map. In particular, when , Schur’s Lemma can be restated as:

The **endomorphism ring** of a simple module over a ring () is a **division ring**.

All this stuff can be generalized a little further by allowing the modules/vector spaces to be over some arbitrary algebraically closed field.