Just want to make sure this is true:
If $G$ is a nilpotent group such that $G/[G,G]$ is a torsion divisible abelian group (like $\mathbb{Q}/\mathbb{Z}_{(p)}$), then $G$ is abelian.
I get that $[G,G]/[G,G,G]$ is a quotient of $G/[G,G] \otimes G/[G,G] = 0$, so $[G,G]=[G,G,G]$ and since $G$ is nilpotent, $[G,G]=1$.
Yes it's true. It's enough to prove it for 2-step nilpotent groups, since any non-abelian nilpotent group has a quotient of nilpotency length exactly 2 (and the same abelianization).
The result for 2-step nilpotent groups is equivalent to showing that for any divisible torsion abelian group $A$, we have $\Lambda^2(A)=0$. More generally, for any torsion abelian group $A_1$ and divisible abelian group $A_2$, we have $A_1\otimes_{\mathbf{Z}}A_2=0$. Equivalently, any bilinear form $A_1\times A_2\to B$ where $B$ is any abelian group, is zero. Indeed, let's check that $b(a_1,a_2)=0$: then there exists $n>0$ such that $na_1=0$, and there exists $a'_2$ such that $a_2=na'_2$. Then $b(a_1,a_2)=b(a_1,na'_2)=b(na_1,a'_2)=0$.