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Let $R$ be a ring with unity, and let $M$ be a finitely generated $R$-module.
I wish to show that if every quotient of $M$ is isomorphic to a direct summand of $M$, then $M$ is a semisimple $R$-module.

Attempt at a solution:
Let $N$ be a submodule of $M$. Then, by hypothesis, $M/N\cong N'$ where $N'$ is a direct summand of $M$. This means that there exists $N''$ such that $M=N'\oplus N''$. If we show that $N=N''$, then we are done, as this means every submodule of $M$ is a direct summand of $M$, which is equivalent to $M$ being semisimple.
I thought this would be easy to do using equivalence relations or quotient mappings, but it turns out this is not the case. I'm also running into problems because the submodules of a finitely generated module aren't necessarily finitely generated. If anyone could help me out it would be very much appreciated! Thanks.

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