Let $R$ be a ring, and $M$ a left $R$-module that has been decomposed into simple submodules in the following way:
\begin{align} M = L_{1,1} \bigoplus L_{1,2} \bigoplus \cdots \bigoplus L_{1,n_1} \bigoplus L_{2,1} \bigoplus L_{2,2} \bigoplus \cdots \bigoplus L_{2,n_2} \bigoplus \cdots \bigoplus L_{r,1} \bigoplus \cdots \bigoplus L_{r, n_r}, \end{align} where the $L_{i,j}$ are all simple submodules of $M$, and $L_{i,j} \cong L_{k, l}$ if and only if $i = k$.
Is it true that $$\sum_{j=1}^{n_i} L_{i,j} = \sum_{M' \leq M, M' \cong L_{i,1}} M',$$ for each $1 \leq i \leq r$? If so, how can one go about proving this?
It is fairly clear that
$$ \sum_{j=1}^{n_i} L_{i,j} \subseteq \sum_{M' \leq M, M' \cong L_{i,1}} M', $$ though I am not sure how to prove the reverse inclusion. I believe that the $\sum_{j=1}^{n_i} L_{i,j}$ are called the homogeneous components of $M$.
Though the user asking this question does not seem to be using this forum, it is still worthwhile to give an answer to this question in case some possible users would be curious about it.
Most books discussing semisimple modules would prove the following statement:
The idea for this proof is easy: Let $p_i:M\to M_i$ be the projection map. Since $N$ is nonzero, at least one $p_i(N)$ is nonzero. However, because $N$ and $M_i$ are simple, by Schur's lemma, the restriction $\left.p_i\right\vert_N$ is an isomorphism.
In fact, such argument can tell us some more information: Consider the set $I'$ of those $i$'s where $p_i(N)$ is nonzero. Then $N$ is isomorphic to $M_i$ for each $i\in I'$. For every $x\in N$, $$x=\sum_{i\in I}{p_i(x)}=\sum_{i\in I'}{p_i(x)}\in\sum_{i\in I'}{M_i}\subseteq\sum_{i\in I,M_i\cong N}{M_i},$$ hence $N\subseteq\sum_{i\in I,M_i\cong N}{M_i}$. By similar arguments and the transitivity of isomorphism, every simple submodule of $M$ isomorphic to $N$ is also contained in $\sum_{i\in I,M_i\cong N}{M_i}$. The problem is thus resolved.