8
$\begingroup$

Let $G$ be an abelian group. Let $G_T$ be the torsion part of $G$, i.e. the set of all elements of $G$ of finite order. And let $G_F$ be the torsion-free part of $G$, i.e. the set containing $0$ along with all elements of $G$ of infinite order. Then my question is, is the isomorphism type of $G$ uniquely determined by the isomorphism types of $G_T$ and $G_F$?

I think this is definitely true in the case when $G$ is finitely generated, but I'm asking about the general case.

$\endgroup$
6
  • 3
    $\begingroup$ Neither $G_T$ or $G_F$ are necessarily groups (the product of two elements of finite order can have infinite order and vice versa), so what do you mean by 'isomorphism type' here? $\endgroup$ – Steven Stadnicki Dec 3 '20 at 17:22
  • 4
    $\begingroup$ @StevenStadnicki Since we are dealing with abelian groups only, $G_T$ is guaranteed to be a (fully characteristic) subgroup, and the product of two elements of finite order necessarily has finite order. $\endgroup$ – sTertooy Dec 3 '20 at 17:35
  • 1
    $\begingroup$ @sTertooy Ahhh! I somehow missed the 'abelian' in the first sentence. That makes much more sense. $\endgroup$ – Steven Stadnicki Dec 3 '20 at 17:36
  • 4
    $\begingroup$ But, $G_F$ is definitely not guaranteed to be a subgroup, even in the finitely generated case. Take $G = \mathbb{Z} \times \mathbb{Z}_2$, then $G_F = G - \{(0,1)\}$ is not a subgroup. It would make more sense to consider $G_T$ and the quotient $G/G_T$. $\endgroup$ – sTertooy Dec 3 '20 at 17:37
  • 3
    $\begingroup$ It would make sense to ask whether $G$ is determined by $G_T$ and the torsion-free group $G/G_T$ (but I expect the answer is no). $\endgroup$ – Derek Holt Dec 3 '20 at 19:26
11
$\begingroup$

(Still not quite a complete answer.)

An abelian group $A$ does not have a canonically defined torsion-free subgroup in general (the elements of infinite order usually aren't a subgroup). What is canonically defined is a short exact sequence

$$0 \to A_T \to A \to A/A_T \to 0$$

where $A/A_T$, which we'll write $A_F$, is the universal torsion-free abelian group admitting a map from $A$; you might call this the "torsion-free part" but keep in mind that it's a quotient, not a subgroup.

$A$ is not determined by the isomorphism type of $A_T$ and $A_F$, and the reason is that the short exact sequence above does not split in general, and so in general defines a nontrivial class in $\text{Ext}^1(A_F, A_T)$. It does split if $A_F$ is free (equivalently, projective as a $\mathbb{Z}$-module), which in particular happens whenever $A$ is finitely generated; in this case we have

$$A \cong A_T \oplus A_F$$

so $A$ is in fact determined up to isomorphism by $A_T$ and $A/A_T$. So to exhibit a counterexample we need to find a torsion abelian group $A_T$ and a torsion-free abelian group $A_F$ such that $\text{Ext}^1(A_F, A_T) \neq 0$; in particular $A_F$ must be torsion-free (equivalently, flat as a $\mathbb{Z}$-module) but not free, and hence infinitely generated.

Take $A_F = \mathbb{Z} \left[ \frac{1}{p} \right]$ where $p$ is a prime. If we write $A_F$ as a filtered colimit of the diagram $\mathbb{Z} \xrightarrow{p} \mathbb{Z} \xrightarrow{p} \dots $ and try to compute $\text{Ext}^1(A_F, A_T)$ where $A_T$ is unspecified then we get a $\lim^1$ exact sequence

$$0 \to \lim^1_n A_T \to \text{Ext}^1 \left( \mathbb{Z} \left[ \frac{1}{p} \right], A_T \right) \to 0$$

so we want to find a torsion abelian group such that $\lim^1$ of the diagram $\dots \xrightarrow{p} A_T \xrightarrow{p} A_T$ is nonzero. I believe that we can take $A_T = \bigoplus_i \mathbb{Z}/p^i$ (which showed up in this previous math.SE question, also about extensions of torsion-free abelian groups by torsion abelian groups), but it seems annoying to verify that this works; I believe it does not suffice to take either $A_T = \mathbb{Z}/p$ or $A_T = \mathbb{Z} \left[ \frac{1}{p} \right]/\mathbb{Z}$ (the Prufer $p$-group), which I was hoping would work.

Edit: Apparently it's simpler to just construct $A$ directly. Supposedly the abelian group $A = \prod_i \mathbb{Z}/p^i$ has the property that its torsion sequence $0 \to A_T \to A \to A/A_T \to 0$ doesn't split, but I'm not sure how to prove it off the top of my head.

$\endgroup$
10
  • 3
    $\begingroup$ You are correct that $\mathbb Z/p$ and the Prüfer group don't work: $\mathrm{Ext}^1(\mathbb Z[1/p],\mathbb Z/p)$ is $0$ because $p$ acts invertibly on it, and as $0$ - so you need unbounded torsion; and the Prüfer $p$-group is injective so this ext is also $0$, so you need your torsion to be "spread" - so your direct sum seems like a good candidate. $\endgroup$ – Maxime Ramzi Dec 3 '20 at 20:57
  • 1
    $\begingroup$ Here is an argument for your suggested direct sum: it has an injective resolution given by $\bigoplus_i \mathbb Z/p^\infty\to\bigoplus_i\mathbb Z/p^\infty$ given by $p^i$ on the $i$th factor. Consider the map $\mathbb Z[1/p]\to \prod_i \mathbb Z/p^\infty$ given on the $i$th coordinate by $p^i$ times the canonical projection. Then this actually lands in $\bigoplus_i \mathbb Z/p^\infty$, but if it were $\bigoplus_i p^i\circ f$ for some $f$, then this $f$ would have to have infinitely many nonzero coordinates on $1/p$, which is absurd $\endgroup$ – Maxime Ramzi Dec 3 '20 at 21:07
  • 1
    $\begingroup$ It follows that $\hom(\mathbb Z[1/p],\bigoplus_i\mathbb Z/p^\infty)\to \hom(\mathbb Z[1/p],\bigoplus_i \mathbb Z/p^\infty)$ is not surjective, i.e. you have an $\mathrm{Ext}^1(\mathbb Z[1/p],\bigoplus_i \mathbb Z/p^i)$. Since this is pretty explicit, it's not unreasonable to expect that you can actually cook up the exact extension $\endgroup$ – Maxime Ramzi Dec 3 '20 at 21:09
  • 4
    $\begingroup$ @Qiaochu Yuan - You are correct that the group in your edit does not split. This fact appears as Theorem 9.2 in the second edition of Rotman's "The Theory of Groups." I'm sure it appears in newer editions (in sections dealing with infinite abelian groups), but that's the book I have in my library. The proof relies on the fact that $A$ has no nonzero element divisible by every prime, but in $A/A_T$ there is a nonzero element divisible by every prime. If $A$ were to split, we would have a contradiction. $\endgroup$ – Chris Leary Dec 3 '20 at 21:26
  • 2
    $\begingroup$ My favourite, somewhat similar, example is $A=\prod_{p}\mathbb{Z}/p\mathbb{Z}$ (the product over all primes), with torsion subgroup $A_T=\bigoplus_{p}\mathbb{Z}/p\mathbb{Z}$ and torsion free quotient $A/A_T\cong\mathbb{R}$. [Proof: It is torsion free and divisible, and it has the cardinality of the continuum.] $\endgroup$ – Jeremy Rickard Dec 4 '20 at 3:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.