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I know about the structure theorem for finitely generated abelian groups.

I'm wondering whether there exists a similar structure theorem for abelian groups that are not finitely generated. In particular, I'm interested in torsion groups. Maybe having a finite exponent helps?

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I think that after the first theorem, the Prüfer/Kulikov's result (mainly the second theorem) should be mentioned as well: en.wikipedia.org/wiki/Pr%C3%BCfer_theorems –  pepa.dvorak Jul 8 '13 at 13:23

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up vote 40 down vote accepted

A torsion abelian group can always be decomposed into its $p$-parts: given an abelian group $A$, and a prime $p$, let $$A_p = \{a\in A\mid \mbox{the order of $a$ is a power of $p$}\}.$$ Then $A_p$ is a subgroup of $A$, each has trivial intersection with the subgroup generated by all the others, so the subgroup they generate is isomorphic to their direct sum; and so $$\Bigl\langle A_p\;\Bigm|\; \mbox{$p$ prime}\Bigr\rangle = \mathop{\oplus}\limits_{p}A_p \leq A,$$ where $p$ ranges over all primes.

Given any $a\in A$, since $a$ is torsion, $a$ can be written as a sum of elements of $A$, each of which has order a prime power, so we have $$A = \mathop{\oplus}\limits_{p} A_p,$$ where $p$ ranges over all primes.

Two torsion abelian groups are isomorphic if and only if their $p$-components are isomorphic. So the problem reduces to classifying abelian $p$-groups.

You can then decompose the group into its divisible and its reduced part, $A = A_{\mathrm{div}}\oplus A_{\mathrm{red}}$ (recall that an abelian group $A$ is divisible if and only if for every $a\in A$ and every positive integer $n$, there exists an element $x\in A$ such that $nx=a$; an abelian group $A$ is reduced if its largest divisible subgroup is the trivial group). Since every divisible abelian group is a direct sum of copies of $\mathbb{Q}$ and copies of the Prüfer groups $\mathbb{Z}_{p^{\infty}}$ for different primes, in the $p$-group torsion case your divisible part will just be a direct sum of (possibly infinitely many) copies of $\mathbb{Z}_{p^{\infty}}$.

Two abelian groups are isomorphic if and only if their divisible parts are isomorphic and their reduced parts are isomorphic. So the problem is down to reduced abelian $p$-groups.

(The following can be found in Rotman's Introduction to the Theory of Groups, 4th Edition, Chapter 10).

In the case of bounded exponent, there is no problem: Prüfer (1923 for countable groups) and Baer (1934 for the general case) proved that:

Theorem. (Prüfer, Baer) If $A$ is an abelian group of bounded exponent (that is, there exists $n\gt 0$ such that $na=0$ for all $a\in A$), then $A$ is a direct sum of cyclic groups.

So if the exponent of the reduced part is bounded, it's just a sum of cyclic groups (necessarily of $p$-power order). And then, counting with cardinality, you get that two are isomorphic if and only if they have the same number of direct summands of each order.

However, things get more complicated at this stage.

Some results: for nonnegative integer $n$, define $U\{n,A\}$ to be $$U\{n,A\}= \dim_{\mathbf{F}_p}\left(\frac{p^nA\cap A[p]}{p^{n+1}A\cap A[p]}\right)$$ where $A[p]$ is the subgroup of elements such that $pa=0$. If $A$ is a sum of cyclic groups, then $U\{n,A\}$ is the number of cyclic summands of order $p^{n+1}$.

In 1933, Ulm proved that there is a transfinite version of these numbers, with $n$ ranging over ordinal numbers, called the Ulm invariants. Two countable torsion abelian $p$-groups are isomorphic if and only if their Ulm invariants are equal. (The result does not extend to uncountable groups). Prüfer proved that countable abelian $p$-group is a direct sum of cyclic groups if and only if $\cap_{n=1}^{\infty}p^nA = 0$ (again, false for uncountable groups). Kulikov characterized the groups that are direct sums of cyclic groups.

Note that Prüfer's result gives you that in the countable reduced case you get a direct sum of cyclic groups (an element of the intersection would necessarily have a $p^i$th root for every $i$, and being of order a power of $p$, it would have $n$th roots for all $n$, hence taking it and its $n$th roots you get a divisible subgroup, so since $A$ is reduced the element must be trivial). So really, the problems start appearing when you go to the uncountable case. That argument is incorrect; taking $x$ and its $p^i$th roots does not necessarily lead to a divisible subgroup, because you may not be able to pick a system of $p^i$th roots that goes "all the way to $\infty$". E.g., you could have countably many different $p$th roots, such that the $i$th $p$th root has a $p^{i-1}$th root but no $p^i$th root; you could end up with an element that has $n$th roots for all $n$, but the roots themselves don't have $n$th roots for all $n$, so the subgroup is not actually divisible. Sorry about that, and thanks to Jack for the heads up.

Added. There is also a theorem of Kulikov:

Theorem (Kulikov, 1945) Every torsion abelian group $A$ is an extension of a direct sum of cyclic groups by a divisible group.

You might be tempted to jump to the conclusion that this means that the reduced part of the $p$-component of $A$ will necessarily be a direct sum of cyclic groups (since it has no nontrivial divisible subgroup), but the problem is that a reduced group may have divisible quotients (even if it is torsion), so this is not the case. For an example of a reduced group with a divisible quotient, take $$A = \mathop{\oplus}\limits_{n=1}^{\infty}C_{p^n},$$ where $C_{p^i}$ is a cyclic group of order $p^i$, generated by $x_i$. Note that $A$ is reduced. But if we let $$B = \bigl\langle x_1-px_2,\ x_2-px_3,\ \ldots, x_n-px_{n+1},\ldots\bigr\rangle,$$ then $A/B$ is isomorphic to the Prüfer group $\mathbb{Z}_{p^{\infty}}$, which is divisible. So Kulikov's theorem gives you a different way of thinking about the reduced part, as an extension.

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+1: thanks for the short course on infinitely generated abelian groups! –  Pete L. Clark Feb 18 '11 at 14:43
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In the theorem "bounded order" should be "bounded exponent". There are some pretty graphical methods of visualizing weird reduced but not DSC abelian p-groups. The divisibility of the elements gives you a (transfinite) tree (or a forest), and there is a presentation of the group based on the tree. The Ulm invariants describe the tree relatively directly (with the limit ordinals corresponding to infinite branches). –  Jack Schmidt Feb 18 '11 at 15:24
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@Seoral: For uncountable abelian groups, there are results; there's just nothing as nice or easy as in the countable/finite case. You start running into set-theoretic problems, where certain axioms (e.g., the Generalized Continuum Hypothesis, or large cardinal axioms) start playing an important role, and certain theorems only hold under some system of axioms for set theory, and are false under others. There is a lot of work with large cardinal groups by Selah and others. Simply put: it gets very nasty, very complicated, very fast, but there are results around. –  Arturo Magidin Feb 19 '11 at 20:14
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@Arturo: I think your claims are much too strong. A countable reduced abelian p-group need not be a direct sum of cyclic groups. Zippin gives a construction of lots of counterexamples. See section 36 of Fuchs's Abelian Groups. –  Jack Schmidt Feb 19 '11 at 20:51
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No problem. I have to run, but I was hoping to find a picture of the simply presented groups. I used to draw them for hours on end, and I think they'd make a great answer. I included a survey of more recent uncountable junk for Seoral. I never developed a taste for abelian groups done after 1970 or so. –  Jack Schmidt Feb 19 '11 at 21:46

To complement Arturo's excellent survey of the foundations of abelian p-groups: A useful survey of the later (1960-2000) development of the theory of abelian p-groups is given in:

Hill, Paul. "The development of the theory of p-groups." Rocky Mountain J. Math. 32 (2002), no. 4, 1135–1151. MR1987598 DOI:10.2307/1999950.

I personally enjoyed the earlier theory a lot more (Kulakoff, Prüfer, Zippin, Ulm) which was gathered up into textbook form by Fuchs and Kaplansky. However, lots of positive and negative results were achieved after then, so if you aren't daunted by the several hundred pages of the textbooks by Fuchs, Kaplansky, and Griffith, then the survey has a lot of important papers beyond the textbook level.

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