About Maschke and Schur's theorem Could you please tell what is the importance of Maschke's theorem and Schur's theorem in representation theory of finite groups? Thanks.
 A: Maschke's theorem is important because it tells you that you can decompose any finite-dimensional representation into a direct sum of irreducible ones. This allows us to study in particular the irreps of a given group, which is a problem more precise than just studying all the representations of the given group.
In other words, thanks to Maschke's theorem, we can focus on irreducible representations.

An important corollary of Schur's lemma is the following:

Let $k$ be an algebraically closed field, and $A$ be a finite-dimensional algebra over $k$. Let $M$ be a simple $A$-module. Then any $A$-module morphism $f:M\to M$ is an homothety, i.e. $f = a \;\text{id}_M$ where $a \in k$.

With this result, we can prove that any finite-dimensional $\Bbb C$-irrep of a finite abelian group is one dimensional. It follows that for any finite dimensional representation $\rho : G \to \text{GL}(\Bbb C^n)$ and for any $g \in G$, there is a basis $B$ such that the matrix of $\rho(g)$ w.r.t. $B$ is a diagonal matrix, with roots of unity on the diagonal $(*)$.
As you see, Schur's lemma has many consequences.
For instance, $(*)$ can be used also to prove that a simple group has no irrep of dimension $2$.
A very important result that use Schur's lemma is the orthonormality relations between irreducible characters.
The fact that Schur's lemma (or the corollary stated above) appears frequently in proofs in (basic) representation theory shows that it is an important result.

To sum up, when you are working with some representation of a group $G$, you can decompose it into a direct sum of irreducible ones by Maschke's theorem.
Then the main basic result on $\Bbb C$-irreps is Schur's lemma: the irrep is associated with a simple $\Bbb C[G]$-module. Then you can try to "build" an endomorphism of $\Bbb C[G]$-module, which will have to be a scalar multiple of the identity! If you work with non-isomorphic $\Bbb C[G]$-modules, then any $\Bbb C[G]$-module morphism between them has to be $0$.
