Let $R$ be a ring (with unity, not necessarily commutative) and let $P$ be an irreducible $R$-module. Let $$M=\bigoplus_{i=1}^r P$$ be a direct sum of $r$ copies of $P$, for some $r\geq 1$. Then, $M$ is semi-simple by "definition". Now, let $N$ be a submodule of $M$. Is $N$ necessarily isomorphic to a direct sum of copies of $P$? If so, can you prove it? If not, could you provide an example?
My intuition tells me that the answer is yes... If $N$ is a submodule of $M$, then $N$ itself is semi-simple, so it is a direct sum of irreducible submodules of $N$, which are also submodules of $M$... This reduces the question so that we can assume $N$ to be irreducible, and then the question is whether $N\cong P$? I am not sure how to prove that. Perhaps there is some exotic $M$ such that there is another decomposition $$M=Q\oplus \bigoplus_{i=1}^{r-1} L_i$$ with $Q$ and $L_i$ irreducible but $Q\not\cong P$? In other words, if $M$ is direct sum of irreducibles... is said decomposition unique? Do we need to assume Artinian and Noetherian here, to use the Jordan-Holder theorem, or can this be proved without those assumptions?