# 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.

• I don't want a full answer but I'd appreciate some hints. – W.Scott May 8 '17 at 21:37
• First show that $x \mapsto a$, $y \mapsto b$ extends to a surjective homomorphism $G \to H$, where $G$ is the group defined by the presentation. Then show that every element of $G$ can be written in the form $x^iy^j[x,y]^k$ for some $i,j,k \in {\mathbb Z}$, and deduce that your map is injective and hence an isomorphism. – Derek Holt May 9 '17 at 8:08
• I'm stuck in the step "show that every element of $G$ can be written as $x^{i} y^{j} [x,y]^{k}$. Specifically, I have shown $yx$ can be written in this form with $i=1, j=1, k=-1$. But what about $y^{-1} x^{-1}, y^{-1} x, yx{^-1}$ etc.? – W.Scott May 9 '17 at 10:42
• You haven't saidf what definition of $[x,y]$ you are using. But $y^{-1}x^{-1} = x^{-1}y^{-1}(yxy^{-1}x^{-1})$ and $yxy^{-1}x^{-1} = [y,x] = [x,y]^{-1}$. – Derek Holt May 9 '17 at 10:51
• Hmm... another approach if somebody wants to expand it into a full answer: the subgroup $\{ (a,b,c) \in H | a = 0 \}$ is split normal, and isomorphic to $\mathbb{Z}^2$; and the quotient is isomorphic to $\mathbb{Z}$. So, $H$ is a semidirect product of $\mathbb{Z}^2$ and $\mathbb{Z}$, and I think the induced map $\mathbb{Z} \to \mathrm{Aut}(\mathbb{Z}^2)$ sends $1$ to the shear $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. So, by math.stackexchange.com/questions/160870/… , since a presentation of $\mathbb{Z}^2$ is... – Daniel Schepler May 17 '17 at 20:54

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 $a, b \in \mathbb{Z}$.)