On Abelian group product of a Free and a Finite group

Question

I'd need help with this problem:

Let $H$ be an abelian group. Let $T \leq H$ and $T' \leq H$ be finite subgroups of H. Let $F \leq H$ and $F' \leq H$ be free subgroups of H.

Suppose $H = T \times F = T' \times F'$. Show that $T = T'$. Does the same conclusion hold for $F$?

I don't really know how to approach the problem. I feel like I should use the Theorem of Classification of finitely generated abelian groups, but H isn't finitely generated.

Thanks in advance for any help!

Answer 1

Hint: if $t'\in T'$, then $t'\in H = T \times F$, so you can write $t' = tf$ where $t\in T$ and $f\in F$. Now $t'$ has a finite order (since it lies in a finite subgroup), and so does $t$; let $n$ be a multiple of both those orders. What happens when you raise both sides of $t'=tf$ to the $n$th power in the abelian group $H$? What do you conclude about $f$? This should allow you to show that $T' \subseteq T$, and the reverse inclusion follows by the same argument.

Answer 2

Given a direct product of Abelian groups $G=T_0\times F_0$ with $T_0$ being finite and $F_0$ free, the torsion subgroup of $G$ is $T_0\times \{0_{F_0}\}$, as every element $x\neq 0_{F_0}$ in $F_0$ has infinite order. That is how you can identify $T_0$. So indeed, $T=T'$ in your problem.

As for the free part, it is not true (not even for finitely generated Abelian groups). Say $G= (\mathbb{Z}_2,+)\times (\mathbb{Z},+)$. Then you can have $F= \langle (0,1) \rangle$ and $F'= \langle (1,1) \rangle$.