Let $F_2$ be the free group of rank $2$ generated by $x$ and $y$. The discrete Heisenberg group $H$ is defined to be the factor group of $F_2$ modulo the relation that $[x,[x,y]]=[y,[x,y]]=1$. Let $G$ be the group defined by $G=\langle x,y\mid [x,x^g]=[y,y^g]=1 \text{ for all } g\in F_2\rangle$ and I wish to prove that $H$ is isomorphic to $G$.
(For the notation: $[a,b]=a^{-1}b^{-1}ab$ and $a^b=b^{-1}ab$.)
It is not difficult to prove that the relation $[x,x^g]=[y,y^g]=1 \text{ for all } g\in F_2$ implies that $[x,[x,y]]=[y,[x,y]]=1$. My difficulty is to prove the other direction; that is, $[x,[x,y]]=[y,[x,y]]=1$ implies that $[x,x^g]=[y,y^g]=1 \text{ for all } g\in F_2$. I tried to prove by induction of the syllable length of $g$ but could not complete the argument. Any assistance will be greatly appreciated.
The relations $[x,[x,y]]=[y,[x,y]]=1$ tell you that $[x, y]$ is in the center of $H$.
You have $$ [x, x^{g}] = [x, x [x, g]] = [x, [x, g]] [x, x]^{[x, g]} = [x, [x, g]]. $$ Consider $g$ as a product of $x, y, x^{-1}, y^{-1}$ of some length $n+1$. We want to prove, proceeding by induction on $n$, that $[x, g] = [x, y]^{\varepsilon}$ for some $\varepsilon$. You will have then $$ [x, x^{g}] = [x, [x, g]] = [x, [x, y]^{\varepsilon}] = [x, [x, y]]^{\varepsilon} = 1. $$
If $g = x h$, with $h$ of length $n$, then $$ [x, g] = [x, x h] = [x, h] [x, x]^{h} = [x, h]. $$ If $g = y h$, with $h$ of length $n$, then $$ [x, g] = [x, y h] = [x, h] [x, y]^{h} = [x, h] [x, y]. $$ If $g = x^{-1} h$, with $h$ of length $n$, then $$ [x, g] = [x, x^{-1} h] = [x, h] [x, x^{-1}]^{h} = [x, h]. $$ If $g = y^{-1} h$, with $h$ of length $n$, then $$ [x, g] = [x, y^{-1} h] = [x, h] [x, y^{-1}]^{h} = [x, h] [x, y]^{-1}. $$
So you see that $[x, g] = [x, y]^{\varepsilon}$ for some $\epsilon$, and you're done.