# Using jordan-holder to prove semisimplicity of a module

I was asked to show the following: Let $$A$$ be a finite dimensional algebra. If $$S$$ is a finite dimensional semisimple $$A$$-module, $$M$$ is also a finite dimensional $$A$$-module and $$nM\cong nS$$ for some $$n\in \mathbb{N}$$ then $$M\cong S$$.

The hint suggests me to use Jordan-Holder series. I don't why it leads to $$M\cong S$$. I can only do the following: Let $$S=\oplus S_i$$. Since $$nM\cong nS$$, we know that the $$S_i$$ are all the composition factors of $$M$$. And I argue randomly like following: If $$M\not\cong \oplus S_i$$, then we have $$[M]\ne0\in \mathrm{Ext}(\oplus S_{i-1},S_i)$$, so $$[nM]\ne0\in\mathrm{Ext}(n(\oplus S_{i-1}),nS_i)$$ which leads to a contradiction. Is it true? Also I'm not allowed to use $$\mathrm{Ext}$$. How to prove that under this condition?

• A problem with the Ext argument is that $[nM]\neq 0$ just tells you that a specific short exact sequence does not split. it doesn't tell you that $nM$ can't be isomorphic to a direct sum of the $nS_i$ in a different way. See for instance math.stackexchange.com/questions/135444/… Commented May 11, 2022 at 14:24
• note that it is possible to prove this without any semisimplicity assumptions by using Krull-Schmidt. Commented May 11, 2022 at 20:56

Just observe that since $$S$$ is semisimple, so is $$nS\cong nM$$, and thus $$M$$ is semisimple since a submodule of a semisimple module is semisimple. So, any Jordan-Holder series for $$M$$ automatically splits.