Submodule of a semisimple module is semisimple . I have a doubt in proving the above statement . 
Lets say that $M$ a semisimple module , then consider $P  \subset N$ , $P$ and $N$ are two sub-modules of $M$ . 
now since $M$ is a semisimple module , then there exists $P'$ and $N'$ such that the direct sum of $P, P'$ is $M$ and $N, N'$ is $M$. 
$N= M\cap N$ , now can i write $M$ as just $P+P'$ , i mean when can i write just sum in place of direct sum . 
EDIT
My notes say 
$N= M\cap N =(P+P') \cap N = P+(P' \cap N) $ and $P\cap (P' \cap N) = \{0\}$ 
I know that $M$ is a direct sum of $P$ and $P'$ but my notes just igonres the direct sum sign . 
I don't understand whats going on . 
 A: There are a few equivalent ways to define a semisimple module and it would be helpful to know exactly which formulation you are using.  One possible formulation is the following:
Definition. A module $M$ is semisimple if every submodule of $M$ is a direct summand of $M$.
Then to prove that every submodule of $M$ is semisimple let $N \subseteq M$ be a submodule and assume $P \subseteq N$ is a submodule of $N$.  Write $M = N \oplus N' = P \oplus P'$ for some $N'$ and $P'$.  What we need is to find $Q$ such that $N = P \oplus Q$.
Let $Q = P' \cap N$.  This is a submodule of $N$ so we need to show that $N = P + Q$ and $P \cap Q = 0$.  For the first let $n \in N$.  Then $n \in M$ so $n = a + b$ for some uniquely determined $a \in P$ and $b \in P'$.  As $P \subseteq N$ we have $n, a \in N$ therefore $b \in N$ therefore $b \in Q$.  This gives $n \in P + Q$ and consequently $N = P + Q$.
For the second assume $n \in P \cap Q$.  Then $n \in Q = P' \cap N$ so $n \in P'$.  But $n \in P$ and $n \in P'$ imply $n = 0$.  Hence $P \cap Q = 0$.
A: The simplest way to do this is based on the fact that 

a module is semisimple if it is a sum of simple submodules.

If $M$ is semisimple, and $N\subseteq M$ is a submodule, then it is easy to check that $N$ is the sum of all the simple submodules of $M$ it contains, and therefore it is semisimple.
