Prove that the intersection of two internal direct sums of simples is zero Let $R$ be a ring with identity and $M$ a right unitary $R$-module. Let $I$ be an index set and $\lbrace S_i \rbrace$ and $\lbrace T_i \rbrace$ be two families of simple submodules of $M$ such that $S_i \cap T_i = 0$ and $S_i \cong T_i$ for all $i\in I$. I need to prove two things: 1) each of the direct sums $\bigoplus_{i\in I}S_i$ and $\bigoplus_{i\in I}T_i$ are internal and 2) ($\bigoplus_{i\in I}S_i)  \cap  (\bigoplus_{i\in I}T_i$ ) = 0.
I proved 1 after a great effort. But I'm still stuck with 2. Clearly, the intersection is $0$ if we regarded the direct sums as external direct sums but I couldn't prove that the intersection is $0$ when we regard the direct sums to be internal.
I appreciate any help.
Thanks in advance.
 A: Both propositions are incorrect, as far as I can tell.
Take any ring $R$ and let $S$ be one of its simple right modules. We consider submodules of $M=S\times S$.
For 1), let $S_1=T_2=S\times \{0\}$, $S_2=T_3=\{0\}\times S$ and $S_3=T_1=\{(x,x)\mid x\in S\}$. All six are mutually isomorphic and satisfy $S_i\cap T_i=\{0\}$.  But $S_1+S_2+S_3$ is not direct since $S_1\subseteq S_2+S_3$.  Similarly, $T_1+T_2+T_3$ is not direct.
For 2), let $S_1=S\times \{0\}$ and $T_2=\{0\}\times S$ and $S_2=T_1=\{(x,x)\mid x\in S\}$. All four are mutually isomorphic, and satisfy $S_i\cap T_i=\{0\}$, but $S_1\oplus S_2=T_1\oplus T_2=M\neq\{0\}$.


each of the direct sums $\bigoplus_{i\in I}S_i$ and $\bigoplus_{i\in I}T_i$ are internal

One does not go about proving "a direct sum is internal." One can prove that a collection of submodules forms an internal direct sum or not. So I've interpreted your question the only way I think is sensible, which is to view it as proving if "a particular sum of submodules forms an internal direct sum."
Now, it is true that any sum of simple modules can be reexpressed as a direct sum of a subset of those simple modules, but that does not seem to be the question here.
