Let $F$ be the free group on finitely many generators $x_1,\dotsc,x_n$. Let $[F,F]\subseteq F$ be its commutator, so we know $F/[F,F]\cong \mathbb{Z}^n$, and let $N\subseteq F$ be a normal subgroup with the property $N\subseteq [F,F]$. Is it true that $F/N$ is torsion-free?
It was just a conjecture. I have a family of finite aspherical cell complexes whose fundamental groups are of this form, and I thought that the torsion-freeness of them has an easier, purely algebraic reason.
Consider the multiplicative group of matrices of the form $$\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1 \end{array}\right),$$ where $a$ and $b$ are integers, and $c$ is an integer modulo $n\gt 1$.
This group is nilpotent of class $2$; it is generated by the element with $a=1$, $b=c=0$ and the element with $b=1$, $a=c=0$. The commutator of these two elements is the element with $a=b=0$, $c=1$, which is a torsion element. If you mod out by the subgroup generated by $c$, we get $\mathbb{Z}\times\mathbb{Z}$.
If we look at this group as a quotient of the free group in two elements $F$ under the obvious map, the kernel $N$ is contained in the commutator subgroup $[F,F]$. But the group is torsion (the commutator subgroup is nontrivial and finite). Thus, $F/N$ is not torsionfree.
More generally, in your free group, if you let $N$ be the normal closure of $[x_i,x_j]^k$, $i\neq j$, $k\gt 1$, then the quotient will have torsion. That the normal closure is contained in $[F,F]$ is, I hope, clear.