= Solution
Construct the sum successively, retaining a <simple module> only when it adds a new <direct summand>. Suppose $S=V_1+\cdots+V_{j-1}$ is already expressed as a <direct sum of modules> selected from these <simple modules>. The intersection $S\cap V_j$ is a <submodule> of the <simple module> $V_j$, so it is either zero or all of $V_j$. In the latter case $S+V_j=S$. In the former case
$$
S+V_j=S\oplus V_j.
$$
Starting with $S=0$ and proceeding by <mathematical induction> gives a <direct sum of modules> equal to $V_1+\cdots+V_n$. Thus the sum is a <semisimple module>. The argument proves the useful <finite sum of simple submodules> lemma without claiming that the originally listed <submodules> are themselves independent.
Back to article page
