TITLE: Finding all submodules of G-modules
QUESTION [5 upvotes]: Let V; W be irreducible G-modules that are not isomorphic to each other.
How to prove that the only G-submodules of M:= $V \oplus W$, other than $0$ and M itself, are $V =
V \oplus  0$
and $W =
0 \oplus W.$

REPLY [3 votes]: Pete's emphasis on isotypic components is a much better idea long term. Here is the "other" proof, for reference.
Suppose $M$ is a direct sum of two simple modules $V$ and $W$.
$N \cap V$ is a submodule of $V$, so either it is $V$ and $V \leq N$, or it is $0$ and so $N+V$ is a direct sum. In the first case, consider $N/V \leq M/V \cong W$. Since $W$ is simple, either $N=M$ or $N=V$. In the second case $(N \oplus V)/V \leq M/V \cong W$, a simple module, so either $N \oplus V = V$ (so $N=0$) or $N \oplus V = M$.
In this last case, $N \oplus V=M$ we need to consider $N \cap W$ as well. As before, we get one of $N=0$ (no), $N=M$ (no), or $N=W$ (yes, but we consider the last as well), or $N\oplus W=M$.
In this last case of the last case, we get that $N \oplus V = N \oplus W = V \oplus W$, and so quotienting both sides by $N$, we get $V \cong W$.
Hence the only possibilities are: $N=0$, $N=V$, $N=W$, $N=M$, or $N \cong V \cong W$.