Let $M$ be a simple module. Prove that all simple submodules of $M^n$ are isomorphic.
I think this is inferred directly by Krull-Schmidt theorem which is stated as follows:
Let $M$ be a $R$-module and let $M\cong U_1\oplus\cdots\oplus U_m\cong V_1\oplus\cdots\oplus V_n$ be two decompositions of $M$ in which $U_i$'s and $V_j$'s are simple $R$-modules. Then, $m=n$ and after a rearrangement of indices we have $U_i\cong V_i$ for each $i$.
Assume $A$ is a simple submodule of $M^n$. Since $M^n=M\oplus\cdots\oplus M$ and $M$ is a simple module, $M^n$ is a semisimple module. Then, $M^n=A\oplus B_1\oplus\cdots\oplus B_k$. Therefore, $A\cong M$. I would like to ask if such an argument is right.
Any counterexample or reference or technique is very much appreciated. Thank you in advance.
In order for your proof to work, you will also need to show that every simple submodule of $M^n$ is also a direct summand. While this is true, it requires more knowledge than just Krull-Schmidt.
I suggest you consider the projection maps $\pi_i : M^n \to M$. If $S \subseteq M$ is a simple submodule, then $\pi_i(S) \neq 0$ for some $i$. Conclude that $\pi_i$ restricts to an isomorphism $S \cong M$. This way, your proof only requires the definition of simple module.