Let $g$ be a positive integer, and let $G$ be a commutative group with the following constraint on its torsion subgroup: there is an injection $G[\operatorname{tors}] \hookrightarrow (\mathbb{Q}/\mathbb{Z})^{2g}$. Must there be subgroups $G\_1,\ldots,G\_g$ of $G$ such that 


(i) $G = G\_1 \times \ldots \times G\_g$ (internal direct product), and  

(ii) For all $1 \leq i \leq g$, there is an injection $G\_i[\operatorname{tors}] \hookrightarrow (\mathbb{Q}/\mathbb{Z})^2$? 


Motivation: If this is true, then it reduces the "Inverse Mordell-Weil Problem for Abelian Varieties" to the "Inverse Mordell-Weil Problem for Elliptic Curves".


Thus although the given question certain has an affirmative answer in many special cases -- e.g. it is a triviality if $G$ is finitely generated -- I am not really interested in that. But it would be "lucky for me" if the answer turns out to be affirmative in the general case, so it's worth asking. 


**Added**: [This previous question](https://mathoverflow.net/questions/60525/when-is-the-torsion-subgroup-of-an-abelian-group-a-direct-summand) contains some information on when the torsion subgroup of a commutative group is a direct summand.



Direct product decomposition for infinite abelian groups with constrained torsion
The answer is negative.


Define $S=\bigoplus\_p C\_p$, where $C\_p$ is cyclic of order $p$ and $p$ ranges over prime numbers. Fix an integer $n\ge 1$. Define $P=\prod\_p C\_p$, and consider a maximal $\mathbf{Z}$-free family in $P^n/S^n$, generating a free abelian group (of continuum rank) $L/S^n$. Fix a prime $q$ and let $U/S^n$ be (\*) a subgroup of $P^n/S^n$ isomorphic to $\mathbf{Z}\_q$ (the group of $q$-adic numbers), with $L\subset U$.


Let's check that for every direct decomposition $U=V\oplus W$, either $V$ or $W$ is finite. Since $T(U)\simeq S^n$, this implies that $U$ answers negatively your question (as soon as $n\ge 3$).


First observe that $\mathbf{Z}\_q$ is indecomposable (i.e. has no nontrivial decomposition as direct product, easy exercise (\*\*)). Since $U/T(U)\simeq\mathbf{Z}\_q$, it follows that either $V$ or $W$, say $W$, is torsion. So $V/T(V)\simeq\mathbf{Z}\_q$.


Write $T(V)=\bigoplus A\_p$ and $T(W)=\bigoplus B\_p$, so that $A\_p\oplus B\_p=C\_p^n$ for all $p$.


We have $P^n=(\prod A\_p)\times (\prod B\_p)$. So $P^n/T(V)=(\prod A\_p/\bigoplus A\_p)\times \prod B\_p$. Since there is no nonzero homomorphism $\mathbf{Z}\_q\to C\_p$ for $p\neq q$ and since the subgroup $V/T(V)$ of $P^n/T(V)$ is isomorphic to $\mathbf{Z}\_q$, the projection of $V/T(V)$ into $\prod B\_p$ is contained in $B\_q$. Thus in the above decomposition, $V\subset (\prod A\_p)\times B\_q$ and $W\subset (\bigoplus A\_p)\times (\bigoplus B\_p)$. So $U\subset (\prod A\_p)\times (\bigoplus B\_p)$. Since $P^n/U$ is torsion (because it is a quotient of $P^n/M$ which is torsion), it follows that $\prod B\_p/\bigoplus B\_p$ is torsion. This means that $B\_p=0$ for large $p$. It follows that $W$ is finite.


Here are the two easy verifications:


(\*): pick in $\mathbf{Z}\_q$ a maximal $\mathbf{Z}$-free family, generating an abelian free subgroup $Z$; since $P^n/S^n$ is a rational vector space, any isomorphism $Z\to L/S^n$ (it exists since both groups are free abelian of continuum rank) can be extended to an injective homomorphism $\mathbf{Z}\_q\to P^n/S^n$.


(\*\*): $\mathbf{Z}\_q$ is $p$-divisible for all $p\neq q$ and $\mathbf{Z}\_q/q\mathbf{Z}\_q$ has order $q$, so for every direct decomposition of $\mathbf{Z}\_q$, at least one factor is divisible. But $\mathbf{Z}\_q$ does not contain any copy of $\mathbf{Q}$.


