The definition of internal direct sum I'm working with is the following:
Let $B,C$ be subgroups of $A$, $A$ being an abelian group. We will call $A$ an internal direct sum of $B$ and $C$, if $B + C = A$ and $A \cap B = 0$. If this is the case, $A$ is isomorphic to the direct product of $B$ and $C$.
My question is this - can we reverse the conclusion? If $B+C=A$ and $A$ is isomorphic to the direct product of $B$ and $C$, does this mean that $B \cap C = 0$ ?
I think it's true for $A$ finite - if the intersection of $B$ and $C$ was nontrivial, it would mean that $|A| < |B \times C|$, so there couldn't be an isomorphism between them. But I'm not sure whether it's true generally.
This question initially occured to me in the context of modules, so I'd also be interested whether anything changes if we consider $A$ to be a module instead.
Just a quick answer from Qiaochu Yuan's comment:
For $A,B,C$ finite-dimensional vector spaces over a field it's true - this follows from the fact that two finite dimensional vector spaces over a field are isomorphic iff they both have the same dimension.
A counterexample to the general case - let $A=B=C= \mathbb{R}^\omega$. Then $B+C=B \cap C = A$, but $\varphi : B \times C \rightarrow A$, defined by $(e_i,0) \mapsto e_{2i}$ and $(0,e_j) \mapsto e_{2j+1}$ is an isomorphism.