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Every finitely generated abelian group is a direct sum of cyclic groups. Does this hold for all abelian groups in general? If not, what fails?

Shaun
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CcVHKakalLLOOPPOkKkkKk
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    See https://math.stackexchange.com/questions/313737/status-of-the-classification-of-non-finitely-generated-abelian-groups – lhf Jul 23 '22 at 23:49
  • Also this for torsion abelian groups. – Arturo Magidin Jul 25 '22 at 01:15
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    A heavily upvoted answer neither justifies the undeletion nor the reopen-votes of this pure PSQ-question. – Peter Aug 17 '22 at 10:39
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    @Peter: I consider that the context is clear with the tags Shaun added. The very first sentence, "Every finitely generated abelian group is a direct sum of cyclic groups." is context to me. Anyone who does not know/remember what an "abelian group" or "cyclic group" is should go over their textbook. What is unclear for you? Qiaochu surely thinks this post is answerable and he posted an excellent answer. In short, I do not understand what your comment is talking about. –  Aug 17 '22 at 11:45
  • It seems like this question is very nearly a duplicate of https://math.stackexchange.com/q/313737 Perhaps it can be argued otherwise because of its scope but it seems to me the heart of the question is the same. – rschwieb Aug 26 '22 at 19:41

1 Answers1

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No, it is badly false and the classification of abelian groups is much more complicated. For example $\mathbb{Q}$ is not a nontrivial direct sum in any way.

Exercise 1: Prove that if an abelian group $A$ is a nontrivial direct sum $B \oplus C$ then the endomorphism ring $\text{End}(A)$ contains a nontrivial idempotent (an element $e \neq 0, 1$ satisfying $e^2 = e$). Deduce that if $\text{End}(A)$ contains no nontrivial idempotents then $A$ is not a nontrivial direct sum.

Exercise 2: Prove that $\text{End}(\mathbb{Q}) \cong \mathbb{Q}$. Deduce that $\mathbb{Q}$ is not a nontrivial direct sum.

Exercise 3: For a prime $p$, the Prüfer $p$-group $\mu_{p^{\infty}}$ is the group of all $p$-power roots of unity, or equivalently the quotient $\mathbb{Z} \left[ \frac{1}{p} \right]/\mathbb{Z}$. Prove that $\text{End}(\mu_{p^{\infty}}) \cong \mathbb{Z}_p$ is the ring of $p$-adic integers. Deduce that $\mu_{p^{\infty}}$ is not a nontrivial direct sum.

In a positive direction you can see the Prüfer theorems, and you can also see Ulm's theorem.

In a negative direction, even the classification of countable torsion-free abelian groups is apparently hopeless in the following sense:

The classification problem for countable torsion free abelian groups is as complicated as that for arbitrary countable structures.

Qiaochu Yuan
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    +1 Ulm's theorem. We see that: even for countable abelian $p$-groups the situation is much, much more complex than the finitely-generated case. – GEdgar Jul 24 '22 at 00:27
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    I conjecture that you didn't really mean "central" in Exercise 1. – Andreas Blass Jul 24 '22 at 01:25
  • @Andreas: whoops, you're right, I got confused there. I mixed up this result and the corresponding result describing direct product decompositions of rings. – Qiaochu Yuan Jul 24 '22 at 10:10
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    I've just discovered that divisible abelian groups have a simple classification. Since every abelian groups embeds into an abelian divisible one, why is this not considered a classification of abelian groups? – CcVHKakalLLOOPPOkKkkKk Jul 24 '22 at 15:46
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    Because... it isn't one? We also know e.g. that every manifold embeds into Euclidean space but this isn't considered a classification of manifolds. A classification should tell you what the isomorphism classes are and an embedding theorem completely fails to do this. – Qiaochu Yuan Jul 24 '22 at 15:48
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    @CcVHKakalLLOOPPOkKkkKk Every group $G$ embeds as a subgroup of $\text{Sym}(G)$ through the map $\psi(g) = (h \mapsto gh)$. Now the groups $\text{Sym}(G)$ are classified simply by the cardinality of $G$. Have we discovered a classification of groups? – Charles Hudgins Jul 24 '22 at 20:13
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    The (provable) fact that abelian groups are both images of free abelian groups, and subgroups of divisible (=injective) abelian groups, does not trivialize things, but does give rise to homology and cohomology of them... which are not generally trivial. – paul garrett Jul 24 '22 at 20:24
  • @CharlesHudgins Good point, its just that divisible groups are direct sums of the rationals and Z[p^oo] and subgroups of those seem intuitively simple, and thats what lead me to feel like this was a satisfactory characterization. – CcVHKakalLLOOPPOkKkkKk Jul 24 '22 at 23:19
  • @CcVHKakalLLOOPPOkKkkKk Have you seen a complete description of nonfinitely generated subgroups of $\mathbb{Q}$? I haven't.... – Arturo Magidin Jul 25 '22 at 01:17