I was reading the proof of $(c) \implies (a)$ i.e., (Given any submodule $M \subset N,$ there exists a submodule $M' \subset N$ such that $N = M \oplus M'$) implies ($N$ is a sum of simple modules) in the answer of the following question:
using finitely generated in proof.
But I did not get the idea of the proof. Could someone explain the idea of the proof to me or give a simpler proof, please?
Here is the proof given there:
Again, how about existence in $c)$ implying $a)$? I'll simply prove that every submodule of $N$ satisfies existence in $c)$ as well. Indeed, let $M_1 \subset M$ be submodules, let $M' \subset N$ be such that $M \oplus M'=N$. Let then $M''$ be such that $M' \oplus (M'+M_1)=N$. Let $\pi:N \rightarrow M$ be the projection with kernel $M'$. It's easy to check that $M = \pi(M'') \oplus M_1$.
Note also that every submodule of $N$ is the image of $N$ under some projection so is finitely generated. In particular, every sequence of submodules $(M_n)_n$ such that $M_n \subset M_{n+1}$ is stationary.
Let $(M_n)$ be a sequence of submodules such that $M_n \supset M_{n+1}$. Let $M_n=N_n \oplus M_{n+1}$. Then $\left(\bigoplus_{p \leq n}{N_p}\right)_n$ is a nondecreasing sequence of submodules of $N$ so is stationary, which implies that for $n$ large enough $N_n=0$, ie $M_n=M_{n+1}$ so $M_n$ is stationary.
From this, it follows that the set of nonzero submodules of $N$ has minimal elements (the simple submodules). Let $S$ be their sum, we can write $N=S \oplus S_1$ for some submodule $S_1$. Assume $S_1$ is nonzero: since every non-increasing/nondecreasing sequence of submodules of $S_1$ is stationary, $S_1$ contains a simple submodule $P$. But $P \subset S$ by definition of $S$, so $P \subset S\cap S_1 = \{0\}$, a contradiction. So $N=S$ is sum of simple modules.
In all simplicity, the idea here is to keep decreasing the submodules of $N$ until it is no longer possible to simultaneously have them decrease and stay non-trivial. For starters, choose any non-trivial submodule $M$ of $N$. Since this will be a summand of $N$, it will be finitely generated whenever $N$ is. We may then apply (c) to $M$. This gives rise to those modules $M_n$ and $N_n$.
At some point both $M_1 \supseteq M_2 \supseteq \cdots$ and $N_1 \supseteq N_2 \supseteq \cdots$ become stationary, and assuming that $M_n \neq 0$ always forces some $M_n$ to be a minimal non-trivial summand of $M_{n-1}$. Transitively, this will then be a minimal non-trivial summand of $M$, and therefore of $N$. It will have as its submodules only itself and the trivial module. In other words, $M_n$ is simple. This shows that we will always find a simple summand of $N$.
The final paragraph shows that there cannot be a non-trivial submodule of $N$ that is not a sum of simple modules, as simple modules are non-trivial, but the simple $P$ would have to satisfy $P=0$.
Is there some specific part you want to ask about, or does this answer your question?