Why is $M/\operatorname{rad}(M)$ semisimple?

Let $$M$$ be an $$A$$-module for $$A$$ a finite dimensional algebra. Let $$\operatorname{rad}(M)=\bigcap\{N\subsetneq M\ \text{maximal}\}$$. Clearly, $$M/N$$ is simple for any maximal submodule $$N$$. It seems to be standard that $$M/\operatorname{rad}(M)$$ is semisimple; however, I struggle to see this.

My attempt: In case of finitely many maximal submodules, this works out easily. Indeed, let $$N, N'\subseteq M$$ be submodules such that $$M/N, M/N'$$ is semisimple. Then $$N/N\cap N'\cong (N+N')/N'\subseteq M/N'$$, so as a submodule of a semisimple module, $$N/N\cap N'$$ must be semisimple itself, and the inclusion has a retraction $$M/N'\to N/N\cap N'$$. The module $$M/N\cap N'$$ in question fits into an extension $$0\to N/N\cap N'\to M/N\cap N'\to M/N\to 0$$ of two semisimple modules. Additionally, there is a retraction $$M/N\cap N'\to M/N\to N/N\cap N'$$. Hence this extension splits, and $$M/N\cap N'$$ is a direct sum of semisimple modules.

Question: How to prove this for an arbitrary number of maximal submodules?

Let $$P:=\bigoplus M/N$$ where the sum is taken over all (nontrivial) simple quotients $$M/N$$, and consider the natural map $$M\rightarrow P$$. The kernel of this map is exactly $$\mathrm{rad}(M)$$, so there is an embedding $$M/\mathrm{rad}(M)\hookrightarrow P$$.
Clearly $$P$$ is semisimple. Can you show that semisimplicity is preserved for submodules?
(There are several equivalent characterizations of semisimplicity: $$P$$ is a sum of simple submodules, $$P$$ is a direct sum of simple modules, and every submodules of $$P$$ is a direct summand.
• Ah, thanks. Sure, if $M/\operatorname{rad}(M)\subseteq P$ for $P$ semisimple, then any submodule $N$ of $M/\operatorname{rad}(M)$ is also a submodule of $P$ and thus has a retraction. The composition with $M/\operatorname{rad}(M)\to P$ then gives a retraction of $N\to M/\operatorname{rad}(M)$. – Bubaya May 8 at 16:31