Does $M = \oplus_i M_i = \sum_j M'_j$ with $M_i, M'_j$ simples implies $M_i \simeq M'_j$ for some i,j Let $M$ be a $R$-module.
We assume that there exists two family $(M_i)_i$ and $(M'_j)_j$ of simple submodules of $M$ such that
$$ M = \bigoplus_i M_i = \sum_j M'_j. $$
Is there some $i,j$ such that $M_i \simeq M'_j$?
 A: For notational convenience I will let $I$ and $J$ be index sets for the $M_i$ and $M_j'$.
The answer is to your question is yes, and in fact for any $j\in J$ we can find $i\in I$ with $M_i\cong M_j'$. To see this, let $f:M_j'\hookrightarrow M$ be the inclusion map, and define $f_i=\pi_i\circ f$ for every $i\in I$, where $\pi_i:M\to M_i$ is the projection map. We cannot have every $f_i$ identically zero, or else $f$ would be identically zero, contradicting that $M_j'$ is simple. Hence there is some $i$ with $f_i$ non-zero. But any non-zero map between simple modules is an isomorphism, so $f_i$ is in fact an isomorphism $M_j'\cong M_i$, as desired.
In fact, a similar statement holds for $I$ instead of $J$: for every $i\in I$, we can find $j$ with $M_i\cong M_j'$. This follows from (the proof of) lemma 1 here; indeed, since $M=\sum_{j\in J}M'_j$, and each $M'_j$ is simple, there is some $J'\subseteq J$ with $M=\bigoplus _{j\in J'}M_j'$. Now we are in a position to apply exactly the same argument as above, by considering the compositions of the projections $\pi_j:M\to M'_j$ (for all $j\in J'$) with the inclusion $M_i\hookrightarrow M$.
