Question

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.

Motivation:

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 $G$.

Answer 1

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

Answer 2

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 $\mathbb{Z}\subset\mathbb{Q}$?