Let $M$ be a right module over a ring $R$ with unity. Let $K\leq A \leq M$, where $A$ is a semisimple submodule of $M$ and $K$ is a simple submodule of $M$. Assume that $\text{rad } M=0$. It is true that $K$ is a direct summand of $M$?.
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From $\mathrm {rad} \ M =0$ we can deduce that $M$ has no superfluous submodules. Hence $K$ is not a superfluous submodule of $M$. This means that $$\exists B \subsetneq M \ : \ K+B=M$$ We need to show that this is a direct sum. $K \cap B$ is a submodule of $K$. Since $K$ is simple we have either $K \cap B = 0$ or $K \cap B=K$.
However $K \cap B=K$ is equivalent to $K \subseteq B$. This leads to a contradiction $$M=K +B \subseteq B+B =B$$
This means that $K \cap B =0$ and the proof is complete.