MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be a torsionfree subgroup of a commutative group $G$. Are there nontrivial conditions known under which there exists a torsionfree direct summand of $G$ containing $F$?

I would already be happy with such a condition in case $F$ is free or $G$ is of finite type.


We consider a property $\mathcal{P}$ of graded rings and its behaviour under coarsening. Given an epimorphism of commutative groups $\psi:G\rightarrow H$ and a $G$-graded ring $R$ with $G$-graduation $(R_g)_{g\in G}$, the $\psi$-coarsening $R_{[\psi]}$ of $R$ is the $H$-graded ring with underlying ring the ring underlying $R$ and with $H$-graduation given by $(R_{[\psi]})_h=\bigoplus_{g\in\psi^{-1}(h)}R_g$ for $h\in H$.

Suppose we know that $\mathcal{P}$ is reflected by $\psi$-coarsening for every $\psi$, i.e., if $R_{[\psi]}$ has $\mathcal{P}$ then so does $R$, and that if $\ker(\psi)$ is a torsionfree direct summand of $G$ then $\mathcal{P}$ is respected by $\psi$-coarsening, i.e., if $R$ has $\mathcal{P}$ then so does $R_{[\psi]}$.

Then, we want to know whether $\mathcal{P}$ is also respected by $\psi$-coarsening if $\ker(\psi)$ is torsionfree but not necessarily a direct summand of $G$. And this would follow immediately if $\ker(\psi)$ is contained in a torsionfree direct summand of $G$.

share|cite|improve this question
You may want to take a look at… – Misha Sep 28 '12 at 15:58
I think this is the same as the splitting of the exact sequence $0 \to T \to G/F \to G/(F,T) \to 0$ – Will Sawin Sep 28 '12 at 17:11
@Misha: Even if the torsion subgroup of $G$ is a direct summand of $G$ the answer to my question is - to me - totally unclear. @Will: No idea what you mean. – Fred Rohrer Sep 28 '12 at 17:17
Oh sorry I mistakenly had a longer thing and thought if I deleted some of it, it would still be readable. $T$ is supposed to be the torsion subgroup of $G$. $G$ and $F$ are as in the question. – Will Sawin Sep 28 '12 at 20:50
Will is correct, his condition is necessary and sufficient: The round brackets is Will's notation for "subgroup generated by". MathJax does not like inequality signs which one would ordinarily use. – Misha Sep 28 '12 at 20:51
up vote 4 down vote accepted

Here is a counter-example in the case when $G$ is 2-generated. Let $G=<a>\times <b>$ where $a$ has finite order $p>1$ and $b$ has infinite order. Let $H=<c>$, where $c=ab^p$. Then the infinite cyclic group $H$ is not contained in a free factor of $G$. Indeed, otherwise, $c$ admits $p$-th root: $c=x^p$. Then $x=a^n b$, but $x^p=(a^{pn} b^p)=b^p$. Contradiction.

Edit: Here are details for Will's comments:

Suppose that $G=T\times R$, where $T$ is the torsion subgroup of $G$. Let $F\subset G$ be a torsion-free subgroup.

Assume now that $F\subset \tilde{F}$, where $\tilde{F}$ is a direct torsion-free factor of $G$. Then it is easily seen that $G=T\times \tilde{F} \times L$. Thus, $G/F\cong T\times (\tilde{F}/F) \times L$ and the sequence $$ 0\to T\to G/F\to G/<F,T>\to 0 $$ clearly splits. Conversely, suppose that the above sequence splits. Then $G/F=T\times R$. Taking preimage of $1\times R$ under the homomorphism $G\to G/F\to T\times R$, we obtain a torsion-free subgroup $\tilde{F}\subset G$ which contains $F$. By construction $T$ maps isomorphically to $G/\tilde{F}$. Thus, $G=T\times \tilde{F}$. qed

share|cite|improve this answer
For clarity, I would emphasize the torsion free assumption and say "where Ftilde is a direct torsion free factor". It then reinforces that you are doing an iff. Gerhard "Ask Me About System Design" Paseman, 2012.09.28 – Gerhard Paseman Sep 29 '12 at 3:36
Yes, of course. Thanks, Gerhard. – Misha Sep 29 '12 at 4:26
@Misha: in your counter example, why c must admit p-root? – Wei Zhou Sep 29 '12 at 4:41
@Wei: Assume that $\tilde{H}$ is a direct torsion-free factor containing $H$. Then $\tilde{H}$ has to be infinite cyclic. Dividing $G$ by the torsion subgroup $T=<a>$, we see that $\tilde{H}$ has to map onto $G/T$. Since the image of $H$ in $G/T$ has index $p$, it follows that $|\tilde{H}:H|=p$ as well. Hence, $c$ admits a $p$-th root. – Misha Sep 29 '12 at 5:12
Misha, thanks a lot. – Wei Zhou Sep 29 '12 at 6:31

This is always true if $G$ is a finitely generated abelian group. Careful reading of the proof of the structure theorem for finitely generated abelian groups should clarify this. If it not finitely generated, then no easy condition comes to mind. For, example, how should one avoid a case like $\mathbb{Z}\subset\mathbb{Q}$?

share|cite|improve this answer
Just to inform the posterity (as well as @Mohan): This is wrong; see Misha's answer for a counterexample. – Fred Rohrer Sep 30 '12 at 15:52
Thanks. Must have been smoking the wrong thing. – Mohan Oct 2 '12 at 20:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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