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 isomorphism or the zero map.
2)If only
is irreducible, then
is either injective or the zero map.
3)If only
is irreducible, then
is either surjective or 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.