Within a proof given in my abstract algebra lecture, we conclude that for any Ring $R$ and any $R$-module $M$ any finite sum of simple submodules $(M_i)_{i\le n}$ is an inner direct sum already, if any two of those submodules intersect trivially, i.e. $\forall i\neq j:\;M_i\cap M_j=\{0\}\implies \sum^n_{i=1}M_i=\bigoplus_{i=1}^n M_i$.
I am very unsure about this 'fact' since I cannot prove it. Is this even true? Can anyone outline a proof if it is?
EDIT: My main concern is that for $n>2$ I cannot prove that $M_i\cap\sum^n_{\overset{j=1}{i\neq j}}M_j=\{0\}$ for every $i$. As far as I know that is the required condition for a sum to be direct.
We can do this inductively in the case where they are pairwise non-isomorphic.
Assume that any family of $n -1$ pairwise non-isomorphic simple sub-modules of $M$ form a direct sum.
Then we may rewrite $M_i\cap\sum^n_{\overset{j=1}{i\neq j}}M_j= M_i\cap\bigoplus^n_{\overset{j = 1}{j\neq i}} M_j$
Since $M_i$ is simple it would mean that either $ M_i\cap\bigoplus^n_{\overset{j = 1}{j\neq i}} M_j = \{0\}$ or $M_i\cap\bigoplus^n_{\overset{j = 1}{j\neq i}} M_j= M_i$.
Now $M_i\cap\bigoplus^n_{\overset{j = 1}{j\neq i}} M_j= M_i$ implies that $M_i$ has an isomorphic copy in $\bigoplus^n_{\overset{j = 1}{j\neq i}} M_j$, which is impossible by the uniqueness theorem for semisimple modules.