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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?

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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.


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".)

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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.

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  • $\begingroup$ Thanks for adding an explanation for how the "direct sum" result is stronger than the "subgroup generated by" one. At the start of your second paragraph, though, did you mean to write "With the (external) direct sum interpretation...", maybe? I believe the external result gives existence and uniqueness, and the internal one gives existence only. $\endgroup$ – Mike Haskel May 30 '17 at 14:36
  • $\begingroup$ @Mike No I didn't, the point is that in this interpretation $\sum F_\lambda$ emphasizes that the sum is internal (which is functionally equivalent to an external one anyways). Internal direct sums mean that it is 1) a sum and 2) $F_\lambda \cap F_\mu \ne \emptyset$ iff $\lambda = \mu$. For an external sum you don't need this second condition because you take the sets to be "formally disjoint". $\endgroup$ – Trevor Gunn May 30 '17 at 14:40
  • $\begingroup$ @Mike Saying a sum is direct is a uniqueness claim. I made some edits to my previous comment, maybe they make the meaning clear. $\endgroup$ – Trevor Gunn May 30 '17 at 14:46
  • $\begingroup$ Okay, I think I misunderstood what you meant by "internal direct sum." You're saying that it's a stronger condition than requiring the $F_\lambda$ to generate $F$, and is equivalent to "the map from $\oplus F_\lambda$ to $F$ induced by the inclusions of the $F_\lambda$ is an isomorphism." Is that a common usage of the "$\sum$" symbol? I haven't seen it before (but it's not my field of math, really). $\endgroup$ – Mike Haskel May 30 '17 at 14:47
  • $\begingroup$ @Mike I can conceive of such a usage, though I will admit that I have only seen $\bigoplus$ or $\coprod$ being used for direct sums and $\sum$ being used for "subgroup generated by". In rare instances, I have seen $\sum$ used for both. $\endgroup$ – Trevor Gunn May 30 '17 at 14:50

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