Finding all submodules of G-modules 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 $W =
0 \oplus W.$
 A: Let $R$ be any ring, and let $M$ be any semisimple left $R$-module.  Let $\{S_i\}_{i \in I}$ be a set of representatives for the isomorphism classes of simple left $R$-modules.  For each $i \in I$, let $M_i$ be the direct sum of all simple submodules isomorphic to $S_i$, the $\bf{S_i}$-isotypic component of M.  It is immediate that $M = \bigoplus_{i \in I} M_i$: this is the isotypic decomposition of $M$.  Note that no choices have been made: this decomposition is unique.
(What is not completely obvious is that the number of independent copies of $S_i$ in $M_i$ is a well-defined invariant of $M$.  If $M$ is finitely generated this follows from the Jordan-Holder Theorem.  In the general case see $\S$ 2.3 of these notes.  But this is not needed in what follows.)
Now let $N$ be a submodule of $M$.  Then $N_i$, the $S_i$-isotypic component of $N$, is $N \cap M_i$: both are the direct sum of all simple submodules of $N$ isomorphic to $S_i$.
In the case that $M = S_1 \oplus S_2$ is a direct sum of two nonisomorphic simple modules, we get immediately that $N = (S_1 \cap N) \oplus (S_2 \cap N)$.  Since $S_1$ and $S_2$ each have exactly two submodules, this shows that the obvious submodules $0$, $S_1 \oplus 0$, $S_2 \oplus 0$, $S_1 \oplus S_2$ are indeed the only four submodules of $M$.
There is an evident generalization to semisimple modules each of whose isotypic components is simple.
A: 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 $N \cong V \cong W$.
A: Assume that V,W are irreducible G-modules.  Define M to be the direct sum of V and W.  A submodule of M has the form of a submodule of V direct sum a submodule of W.  By the assumptions that V and W are irreducible there are only two choices for each of V and W for what those submodules can be.
