Specific converse to Schur's Lemma Suppose I have a finite dimensional $\mathbb C$-algebra $A$, and a finite dimensional (over $\mathbb C$) $A$-module $M$.
Suppose that $\mathrm{Hom}_A(M,M) = \mathbb C$.
Under what conditions can I conclude that $M$ is simple?
 A: Anytime your algebra is semisimple, the converse will hold. For example, consider the case when $A$ arises as the group algebra of some finite group $G$. Then the converse is a corollary to Maschke's theorem. 
Suppose $M$ has some submodule $U$. By Maschke's theorem there exists a split short exact sequence of $A=\mathbb{C}[G]$ modules
$$0 \to U \overset{i}\to M \overset{p}\to V \to 0,$$
where $V \subset M$. The map $p$ is an $A$-linear map from $M$ to $M$ which must be noninvertible if $U$ is nonzero. But this means $p=0$ since $Hom_A(M,M)= \mathbb{C}$, and so $U=M$. Hence $M$ has no nontrivial submodules. 
This result holds more generally for representations of finite groups when the characteristic of the underlying field does not divide the order of the group. 
You may also be interested in complex semisimple Lie algebras, in which case Weyl's theorem on complete reducibility would allow for a similar argument. Going even further, one may prove similar results for compact Lie groups as well as reductive algebraic groups over fields of characteristic zero. In short, any object whose representations are completely reducible will give you a converse to some appropriate formulation of Schur's lemma.
