Showing the socle of a module exists and is unique I have been set the following exercise:

For every $R$-module $M$, show that there exists a unique semi-simple submodule $sM$ $\subset$ $M$ which contains every semi-simple submodule of $M$.

After looking around online, it seems that this is called the socle of $M$?  And this question is essentially asking me to show that it exists and is unique?  
Am I right in thinking that this is how to the view the problem?  And if so, how would I go about showing this?  
Thanks in advance!
 A: The key is the fact that any semisimple module is the sum of its simple submodules and, conversely, the sum of simple submodules is semisimple.
Modules will be left $R$-modules.

Theorem. For a module $M$ the following conditions are equivalent:  
  
  
*
  
*$M$ is a direct sum of simple submodules,
  
*$M$ is a sum of simple submodules,
  
*$M$ is the sum of its simple submodules,
  
*every submodule of $M$ is a direct summand.
  

The implications $(1)\implies(2)\implies(3)$ are obvious. The proof of $(3)\implies(4)$ relies on the following lemma.

Lemma. If $M=\sum_{i\in I}S_i$ with $(S_i)_{i\in I}$ a family of simple submodules of $M$ and $K$ is a submodule of $M$, then there exists $J\subseteq I$ such that
a. $(S_i)_{i\in J}$ is independent (that is, the sum $\sum_{i\in J}S_i$ is direct), and
   b. $M=K\oplus\sum_{i\in J}S_i$.

The proof is an application of Zorn's lemma. Take a subset $J\subseteq I$ maximal with respect to $(S_i)_{i\in J}$ being independent and $K\cap\sum_{i\in J}S_i=\{0\}$ (here Zorn's lemma is applied, work it out). Set $N=K+\sum_{i\in J}S_i$: we want to show that $N=M$. Let $k\in I$; since $S_k$ is simple, either $S_i\cap N=\{0\}$ or $S_i\subseteq N$. In the first case $J'=J\cup\{k\}$ would contradict the maximality of $J$, so we have that $S_i\subseteq N$ for all $i\in I$ and so $N=M$.
The lemma also shows that $(2)\implies(1)$, by taking $K=0$, so we only need to prove $(4)\implies(2)$. So, assume every submodule of $M$ is a direct summand. Let $x\in M$, $x\ne0$. Then $Rx$ an has a maximal submodule $H$ (just take a maximal left ideal of $R$ containing $\{r\in R:rx=0\}$). So $M=H\oplus H'$. Then
$$
Rx=Rx\cap M=Rx\cap (H\oplus H')=H\oplus(Rx\cap H')
$$
(verify it), so $Rx\cap H'\cong Rx/H$ is a simple submodule of $Rx$.
Let now $N$ be the sum of the simple submodules of $M$; then $M=N\oplus N'$; if $N'\ne\{0\}$, then $N'$ would have a simple submodule: contradiction.

As a consequence of this, if $M$ is any module, the sum of its simple submodules is a semisimple submodule of $M$ containing all simple submodules and so all semisimple submodules.
