Free abelian groups. Is the following implication true? If this is the case, how can this be shown? 
If $G$ is a one-relator (neither power nor commutator relation), two-generator group, then $G/G'$ (where $G'$ is commutator subgroup) is not free abelian of rank 2.
Thanks.

Let me make this into an answer. Any 2-generator 1-relator group has the form $G = \langle a,b \mid w \rangle = F/\langle w^F \rangle$ for some $1 \ne w \in F$,
where $F$ is the free group with generators $a,b$.
So $G/G' \cong F/N$ with $N = F'\langle w^F \rangle$. Since $F/F'$ is free abelian of rank 2, and no proper quotient of $F/F'$ is free abelian of rank 2, $G/G'$ is free abelian of rank 2 if and only if $w \in F'$.
As SteveD pointed out, the condition $w \in F'$ is equivalent to the exponent sums of $a$ and $b$ in $w$ both being $0$. You can see that as follows. Any element of $F'$ is a product of commutators, so it has this property. Conversely, if $w$ has exponent sums of $a$ and $b$ both $0$ then, since $F/F'$ is commutative, the image of $w$ in $F/F'$ is trivial and hence $w \in F'$.
