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.