Submodules of direct sums of simple modules I'm reading these online notes on representation theory, and I don't fully understand this:

Isn't $V\oplus(\bigoplus_{i\in I}S_i)$ a direct sum by definition, so we'd get $I=\{1,\cdots,n\}$? Can we assume the $S_i$ are pairwise disjoint submodules of $U$ based on how the lemma is phrased (or does that follow necessarily from them each being simple)? Intuitively, I feel the condition should be that $W$ as an external direct sum maps into the internal direct sum situated within $U$ in an obvious way, and we choose $I$ maximal so that this map has no kernel. Is this understanding correct?
 A: Your notes have a typo: the proof should begin "Choose a subset $I$ maximal subject to the condition that $V + \sum S_i$ is a direct sum." Of course, as you've noted, $V\oplus \bigoplus S_I$ is a direct sum by definition-but it might be considerably bigger than $U$, as Mariano's example shows. We can assume that the $S_i$'s pairwise intersections are either $0$ or an $S_i.$ If the latter case happens, $S_i \cap S_j=S_i,$ we know $S_i=S_j$, since otherwise $S_i$ would be a proper submodule of $S_j.$
Regarding your comments, the definition of direct is not that the summands have pairwise trivial intersections, but that each element of the sum can be written uniquely in terms of the summands. So $(e_1) + (e_2)=(e_1) \oplus (e_2)=V$ in some 2-dimensional vector space $V$, but $(e_1)+(e_2)+(e_1+e_2)=V$ while $(e_1)\oplus (e_2)\oplus (e_1+e_2)$ is three-dimensional and not in $V$ at all. Note, though, that these three modules have trivial intersection. On the other hand, if a sum is direct, this does imply the summands have trivial intersection: otherwise anything in any intersection can be written in at least two ways. It's just that these properties aren't equivalent.
