What does $F=\sum F_{\lambda}$ mean for free abelian groups?

Question

Let $F$ be free abelian with basis $B$. If $B$ is the disjoint union $B = \bigcup B_{\lambda}$, then $F = \sum F_{\lambda}$, where $F_{\lambda}$ is free abelian with basis $B_{\lambda}$

What does $F=\sum F_{\lambda}$ mean? I'm unfamiliar with the notation of a summation sign with free abelian groups. Is it direct product (or sum), or is it supposed to be a union that was mis-written into the book?

Answer 1

The theorem is true when $\sum$ is taken to mean "direct sum of Abelian groups." Some people might use $\bigoplus$ for a direct sum, but $\sum$ also makes some sense, because you can construct a direct sum of Abelian groups as formal sums of elements from any of the individual groups.

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Note that the theorem is also true if the $F_\lambda$ are viewed as the subgroups of $F$ generated by the $B_\lambda$, and when $\sum F_\lambda$ is taken to mean the subgroup generated by all the $F_\lambda$. These results are related. Without further context, though, it's impossible to know which specific interpretation was intended. (Note that with this latter interpretation, $\sum$ means "subgroup generated by", not "set-theoretic union".)

Answer 2

I'm thinking it could have something to do with internal ($\sum$) vs external ($\bigoplus$) direct sums. That is, $F = \sum F_\lambda$ means that $F$ is a direct sum of the subgroups $F_\lambda$ (emphasis on "sub"). It could also mean the (not necessarily direct) sum or the subgroup generated by $F_\lambda$, as Mike Haskel points out in his answer.

With the (internal) direct sum interpretation, this means that every element can be written uniquely as a sum of elements of the $F_\lambda$'s. With the (not necessarily direct) sum interpretation, the claim is that every element can be written as a sum of elements of the $F_\lambda$'s but that sum may not be unique.