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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?

Richard Chen
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    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 https://math.stackexchange.com/questions/135444/a-nonsplit-short-exact-sequence-of-abelian-groups-with-b-cong-a-oplus-c – Eric Wofsey May 11 '22 at 14:24
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    note that it is possible to prove this without any semisimplicity assumptions by using Krull-Schmidt. – Lukas Heger May 11 '22 at 20:56

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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.

Eric Wofsey
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