Wikipedia:Reference desk/Archives/Mathematics/2012 August 30

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August 30

Every nonfinitely generated abelian group must contain one of... (resumed)

I asked this question, and got one useless response (that misunderstood the question). Here it is my question, slightly revised to clarify, and my attempt to answer my own question.

Is there any list of relatively simple nonfinitely generated abelian groups, such that any arbitrary nonfinitely generated abelian group must contain one of the groups in the list as a subgroup?

An attempt to answer my own question. I believe the following is such a list, except I'm unsure about the last point. Let $A$ be an abelian group:

Case 1: $A$ contains an infinite sequence of divisible elements, ie, a sequence $x_{1},x_{2},\cdots$ such that $\langle x_{i}\rangle \subsetneq \langle x_{i+1}\rangle$. Then $\langle x_{1},x_{2},\cdots \rangle$ is isomorphic to a subgroup of $\mathbb {Q}$ or $\mathbb {Q} /\langle 1\rangle$.

Case 2: $A$ does not contain such an infinite sequence, and its torsion group is not finitely generated. Then by CRT, only finitely many primes occur as the order of some element, because otherwise every finite order element is divisible. One of those primes must occur infinitely often, because the elements of order $p^{i}$ generate the torsion subgroup, again by CRT, and if, for such a prime $p$, there are only finitely many elements of order $p$, then it's not to difficult to show, using the classification of finitely generated abelian groups, that there must be an element of order $p$ which begins a sequence of the form in case 1. Then countably infinitely many elements of order $p$ generate a vector space over $\mathbb {F} _{p}$ of countably infinite dimension - ie, $\left(\mathbb {Z} /p\mathbb {Z} \right)^{\aleph _{0}}$.

Case 3: Otherwise, must $A$ necessarily contain a direct sum of countably many copies of $\mathbb {Z}$?

Thanks for anyone who happens to know, or be able to work through the math and make any additional arguments necessary! --70.116.7.160 (talk) 14:01, 30 August 2012 (UTC)

The answer to the “Case 3” question is no: for any finite n > 1, there are infinitely generated torsion-free abelian groups of rank n with no infinitely generated subgroup of smaller rank (which in particular implies that there is no infinite descending divisor chain). For a concrete example, consider the subgroup of $\mathbb {Q} ^{2}$ generated by $\mathbb {Z} ^{2}$ and by the elements $(2^{-m},\sum _{k^{2}<m}2^{k^{2}-m})$, $m\in \mathbb {N}$. (The general pattern is given below, (6).)

I don’t have the time now to write a detailed proof, but I will sketch a (hopefully) complete classification. Let A be an infinitely generated abelian group:

• First, assume that its torsion subgroup $A^{tor}$ is infinite (= infinitely generated). If the p-torsion subgroup A_p is nontrivial for infinitely many p, then A contains a subgroup isomorphic to

$(1)\qquad \bigoplus _{p\in I}\mathbb {Z} /p\mathbb {Z} ,$

where I is an infinite set of primes. Otherwise, A_p is infinite for some prime p. If there are infinitely many elements of order p, then A contains

$(2)\qquad (\mathbb {Z} /p\mathbb {Z} )^{(\omega )}$

(the direct sum of countably many copies of $\mathbb {Z} /p\mathbb {Z}$). Otherwise, for every $x\in A_{p}$ there are only finitely many $y\in A_{p}$ such that $x=py$; since A_p is infinite, König’s lemma implies that there is an infinite chain $\{x_{n}:n\in \omega \}\subseteq A_{p}$ such that $x_{0}=0$, $x_{1}\neq 0$, and $px_{n+1}=x_{n}$. Then $\langle x_{n}:n\in \omega \rangle$ is isomorphic to the Prüfer group

$(3)\qquad \mathbb {Z} (p^{\infty }).$

• The other case is that $A^{tor}$ is finite. Then $A/A^{tor}$ is an infinitely generated torsion-free group. In fact, by successively chosing suitable elements $x_{n}\in A$ such that $x_{n}$ is not in $\langle x_{i}:i<n\rangle +A^{tor}$, one can show that A itself includes an infinitely generated torsion-free subgroup. Thus, we can assume without loss of generality that A is torsion-free.

If A has infinite rank, its injective envelope is a vector space over $\mathbb {Q}$ of infinite dimension. If we choose an infinite linearly independent set, which we may assume to lie in A, we see that A contains a copy of

$(4)\qquad \mathbb {Z} ^{(\omega )}.$

If A has finite rank, let n be the minimal rank of its infinitely generated subgroup. We may assume that A itself has rank n. Its injective envelope is a $\mathbb {Q}$-vector space of dimension n, and A includes n linearly independent elements, hence we may assume that $\mathbb {Z} ^{n}\subseteq A\subseteq \mathbb {Q} ^{n}$. The quotient $A/\mathbb {Z} ^{n}$ is an infinite torsion group contained in $(\mathbb {Q} /\mathbb {Z} )^{n}$, hence by the previous part, it contains a copy of (1) or (3). In the former case, A contains a subgroup of $\mathbb {Q} ^{n}$ of the form

$(5)\qquad \mathbb {Z} ^{n}+\langle p^{-1}a_{p}:p\in I\rangle ,$

where I is an infinite set of primes, and for each $p\in I$, $a_{p}\in \mathbb {Z} ^{n}$ is such that not all coordinates of a_p are divisible by p. Moreover, all infinitely generated subgroups of this group have rank n; one can show that this condition is equivalent to the statement that for every linear function $L(x_{1},\dots ,x_{n})=\sum _{i}m_{i}x_{i}$, where $m_{i}\in \mathbb {Z}$ and $\gcd(m_{1},\dots ,m_{n})=1$, there are only finitely many $p\in I$ such that $L(a_{p})\equiv 0{\pmod {p}}$. In particular, we may assume that all coordinates of every a_p are coprime to p.

The remaining case is that $A/\mathbb {Z} ^{n}$ includes a copy of $\mathbb {Z} (p^{\infty })$. Let $\mathbb {Z} _{p}$ denote p-adic integers, and $\mathbb {Q} _{p}$ its fraction field of p-adic numbers. For $a=\sum _{i=k}^{\infty }a_{i}p^{i}\in \mathbb {Q} _{p}$, let me write $\mathrm {Frac} (a)=\sum _{i=k}^{-1}a_{i}p^{i}$, so that $a-\mathrm {Frac} (a)\in \mathbb {Z} _{p}$ and $\mathrm {Frac} (a)\in \mathbb {Z} [p^{-1}]\subseteq \mathbb {Q}$. Moreover, if $a\in \mathbb {Q} _{p}^{n}$, I’ll write $\mathrm {Frac} (a)$ for the coordinate-wise application of Frac. Using this notation, it is not hard to show that A contains a subgroup of $\mathbb {Q} ^{n}$ of the form

$(6)\qquad \mathbb {Z} ^{n}+\langle \mathrm {Frac} (p^{-k}a):k\in \omega \rangle ,$

where $a\in \mathbb {Z} _{p}^{n}$. Again, additionally all infinitely generated subgroups of this group have rank n, which is equivalent to the condition that the n coordinates of a, as elements of $\mathbb {Q} _{p}$, are linearly independent over $\mathbb {Q} \subseteq \mathbb {Q} _{p}$.

Note that for n = 1, cases (5) and (6) reduce to the groups $\mathbb {Z} +\langle p^{-1}:p\in I\rangle$ and $\mathbb {Z} [p^{-1}]$, respectively, which, as subgroups of $\mathbb {Q}$, belong to your Case 1. However, for n > 1 the groups are new.—Emil J. 13:28, 3 September 2012 (UTC)

Thanks! I think I understood the first half of your classification perfectly, and I'll need to do a bit of research to get the background to pin down the latter parts of it. --70.116.7.160 (talk) 20:22, 3 September 2012 (UTC)

Resolved