Group presentation for discrete Heisenberg group I have to show that the discrete Heisenberg group H (with $a,b,c \in \mathbb{Z}$) has the presentation $H=\langle x,y\mid [[x,y],x]=[[x,y],y]=1 \rangle$.
I figured we write $H$ as a group of triples $H=\{(a,b,c) \mid a, b, c \in \mathbb{Z}\}$ with group operation defined as $(a,b,c)(a',b',c')=(a+a',b+b',ab'+c+c')$, but that is as far as I can get.
 A: Let $H = \{ (a, b, c) | a, b, c \in \mathbb{Z} \}$ and $H' = \langle x, y | [[x,y], x] = [[x,y], y] = 1 \rangle$.  Then we can define a group homomorphism $\phi : H' \to H$ such that $\phi(x) = (1, 0, 0)$ and $\phi(y) = (0, 1, 0)$ (since the images of $[[x, y], x]$ and $[[x, y], y]$ are both the identity element).  It suffices to show that $\phi$ is an isomorphism.
To show $\phi$ is surjective, note that for $a, b, c \in \mathbb{Z}$, $(a, b, c) = \phi([x, y]^c y^b x^a)$ in $H$.
To show $\phi$ is injective, we first claim that any element of $H'$ can be written in the form $[x, y]^c y^b x^a$ for some $a, b, c \in \mathbb{Z}$.  The proof of this claim is left as an exercise for the reader (hint: use induction on the number of terms of $x$, $y$, $x^{-1}$, or $y^{-1}$ making up the expansion of a preimage in the free group $\langle x, y \rangle$, along with the identities $xy = [x,y] \cdot yx$, $x y^{-1} = [x,y]^{-1} \cdot y^{-1} x$, $x^{-1} y = [x,y]^{-1} \cdot y x^{-1}$, $x^{-1} y^{-1} = [x,y] \cdot y^{-1} x^{-1}$ in $H'$, and the fact that $[x,y]$ is in the center of $H'$).  But then, $\phi([x,y]^c y^b x^a) = (a, b, c) = (0, 0, 0)$ trivially implies $[x,y]^c y^b x^a = 1$.
(Another valid approach would be to define $\psi : H \to H'$, $(a, b, c) \mapsto [x,y]^c y^b x^a$, and show $\psi$ is the inverse of $\phi$.  In this approach, the hardest step would likely be showing that $\psi$ is actually a homomorphism -- for which it would be useful to prove the identity $x^a y^b = [x,y]^{ab} y^b x^a \in H'$ for $a, b \in \mathbb{Z}$.)
