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If $M$ is a finite dimensional module over a $k$-algebra $A$ say (for $k$ a field), then by the Jordan-Hölder Theorem we know that $M$ admits a composition series with a unique set of composition factors.

The definition of semi-simple I am using is that $M$ has a direct sum decomposistion of simple modules. Are the simple modules in the direct sum composition necessarily the composition factors that we know $M$ admits?

If so, then when one is trying to prove the alternate definition of semi-simplicity, that is that every sub-module has a direct sum complement, is the following a valid proof?

Let $M$ be a finite dimensional module over a $k$-algebra $A$. Suppose that $M$ is semi-simple with $S=\{S_{i}\}_{i=1}^{n}$ the simple modules appearing in the direct sum decomposition of $M$, counting multiplicities. Suppose that $N$ is a submodule of $M$, then since $M$ is finite dimensional, it has a composition series including $N$. If we allow this composition series to terminate at $N$ we receive a composition series for $N$, and so by the uniqueness aspect of the Jordan-Hölder theorem, the composition factors for $N$ must be a subset of $S$, $S'$ say (counting multiplicities). Then if my statement about composition factors is correct, then both $M$ and $N$ are a direct sum of their composition factors, so if we let $S'' = S - S' $ and let $V$ be the direct sum of the simple modules in $S''$, then since $S = S' \cup S''$ (counting multiplicities), we must have that the direct sum of $V$ and $N$ is $M$.

I suspect that this proof isn't valid, since the proof that I have been provided of the above fact is more complicated (in my view), but I'm unsure just how it would be invalid.

Are there examples of finite dimensional semi-simple modules over algebras, whose direct sum decomposition is not a sum of simple-modules isomorphic to the composition factors? My suspicion is that it will come down to a problem of whether or not composition factors naturally sit in your original modules as a sub-module, but I don't know how to make this rigorous.

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    $\begingroup$ Yes, a finitely generated semisimple module is the direct sum of its composition factors. $\endgroup$ – Lord Shark the Unknown May 6 '18 at 12:49
  • $\begingroup$ Thanks for the speedy reply! How would one go about proving this fact? $\endgroup$ – Adam Higgins May 6 '18 at 12:54
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    $\begingroup$ A direct sum of simple modules $S_1\oplus\cdots\oplus S_n$ has a composition series $$0\subseteq S_1\subseteq S_1\oplus S_2\subseteq S_1\oplus S_2\oplus S_3\cdots.$$ $\endgroup$ – Lord Shark the Unknown May 6 '18 at 13:13
  • $\begingroup$ Thanks! That's the rigorous step that I was missing in my head. Then this composition series witness the composition factors as isomorphic to the simple modules in the direct sum decomposition. Perfect! $\endgroup$ – Adam Higgins May 6 '18 at 13:31

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