Proving that any submodule of a semisimple module has a simple submodule. Proving that any submodule of a semisimple module has a simple submodule.
To prove the above statement is it necessary and sufficient to show that any semisimple module has a simple submodule. That is because submodule of a semisimple module is semisimple
or we cannot do that because after proving the above statement only we are able to show that submodule of semisimple is semisimple
 A: Let $A$ be a semisimple module, i.e. every submodule is a direct summand. We first show that every submodule $B\subset A$ is semisimple. Suppose $C\subset B$ and $A=B\bigoplus D=C\bigoplus E$. We claim that $B=C\bigoplus B\cap E$. First, take any $b\in B\subset C\bigoplus E$ and write $b=c+e$, $c\in C, e\in E$. $e=b-c\in B$ so every $b\in B$ is the sum of two elements respectively from $C$ and $B\cap E$. We then prove that $B\cap E\cap C=(0)$
. This is obvious since $C\cap E=(0)$. Therefore, any submodule of $B$ is a direct summand. $B$ is semisimple.
Let's now show that $A$ always has a simple submodule. Take any $a\in A$. By Zorn's lemma, find a maximal $B\subset A$ where $a\notin B$. By definition, write $A=B\bigoplus C$.
We claim that $C$ is simple. Suppose $C$ has a proper submodule $D$. $C$ is also semisimple so $C=D\bigoplus E$. $A=B\bigoplus D\bigoplus E$. Note that $a\notin B\bigoplus D\cap B\bigoplus E$. Otherwise, write $a=b+d=b'+e$ so $(b-b')-e+d=0$. This relation is not possible since $B\bigoplus D\bigoplus E$ is a direct sum. We assume WLOG that $a\notin B\bigoplus D$. This is absurd since $B$ is maximal among submodules not containing $a$. So $C$ is indeed a proper simple submodule of $A$.
A: Seems simplest to directly prove submodules are semisimple.
Let $N$ be a submodule and $A$ be a submodule of $N$, and we will show $A$ has a complement in $N$. Of course, $A$ has a complement in $M$, say $A\oplus B=M$. Then (it is easy to check that) $M\cap N=(A\oplus B)\cap N=A\oplus(B\cap N)$, and since $B\cap N$ is a submodule of $N$, we have our complement.
Now we can go further and show a nonzero semisimple module necessarily contains a simple submodule.
Now if $m$ is a nonzero element of $M$, we can examine its cyclic submodule $mR$, which is necessarily isomorphic to $R/T$ for some right ideal $T$.  We know by Zorn that $R$ has a maximal right ideal $T_{max}$ containing $T$, and therefore that $mR$ has a maximal submodule $C=T_{max}/T$.  We know $mR$ is semisimple (because it's a submodule of $M$), so we can write $mR=C\oplus D$. Owing to $C$'s maximality, we know that $mR/C\cong D$ is a simple $R$ module.  So $D\subseteq mR\subseteq M$ is a simple submodule of $M$.
A: 
Assume $M$ is a semi-simple module and $M=\oplus _AT_{\alpha}$.If
$$0\rightarrow K\overset{f}{\rightarrow}M\overset{g}{\rightarrow}N\rightarrow 0$$ is a
$R$-module exact sequence, then it is  a split exact sequence. In fact, there exists $B\subset  A$ such that $N\cong \oplus _BT_{\beta}$ and $K\cong \oplus _{A\backslash B}T_{\alpha}$.

Proof:

$\mathrm{Im}~f$ is a submodule of $M$, so there exists $B\subset A$
such that $M=\mathrm{Im}~f~\oplus (\oplus _BT_{\beta})$. So the exact
sequence is split and $N\cong M/\mathrm{Im}~f\cong \oplus
> _BT_{\beta}$.
Furthermore, we can get $K\cong \mathrm{Im}~f\cong \oplus _{A\backslash B}T_{\alpha}$

Use the theorem, we can get every submodule and every factor module  of a semi-simple module are semisimple.
You may also refer to Anderson and Fuller's book: Rings and Categories of Modules.
