Integer Heisenberg group ($H(\mathbb{Z})$)

The integer Heisenberg group consists of upper-unitriangular three-by-three integer matrices. In coordinates $(x,y,z)$ its multiplication isIt is the semidirect product ${\mathbb{Z}}^2{⋊}_A\mathbb{Z}$ forand its abelianization is ${\mathbb{Z}}^2$.

Write $M(a,b,c)=\left(\begin{array}{ccc}1& a& b\\ 0& 1& c\\ 0& 0& 1\end{array}\right)$ with integer coordinates. Multiplication in this parametrization is $M(a,b,c)M(A,B,C)=M(a+A,b+B+aC,c+C)$. The subgroup $H=\{M(0,0,t):t\in \mathbb{Z}\}$ has right cosets obtained by multiplying on the left by its elements, so its right cosets are exactly the displayed sets with fixed $(a,b)$. In particular, right cosets are indexed by ${\mathbb{Z}}^2$. Left cosets differ: $M(a,b,c)H$ fixes $a$ and $b-ac$, since right multiplication changes $b$ by $at$ and $c$ by $t$. This also exhibits why $H$ is not a normal subgroup of the Integer Heisenberg group.

For a positive integer $k$, in the real Heisenberg group with multiplication $(x,y,z)(x^{\prime },y^{\prime },z^{\prime })=(x+x^{\prime },y+y^{\prime },z+z^{\prime }+x{y}^{\prime })$, the displayed set is a discrete cocompact subgroup. Writing $a=(k,0,0)$, $b=(0,k,0)$, $c=(0,0,k)$ gives $[a,b]=c^k$ and $c$ central. Hence its abelianization is ${\mathbb{Z}}^2\oplus \mathbb{Z}/k$. The right quotient maps to ${\mathbb{R}}^2/k{\mathbb{Z}}^2$ by its first two coordinates; each fibre is the central circle $\mathbb{R}/k\mathbb{Z}$.

Write the Integer Heisenberg group with central generator $c=[a,b]$, so $ab=cba$. Moving $a$ successively past powers of $b$ givesFor positive exponents this follows by two inductions; inverses give the remaining signs. The representation $a=I+E_{12}$, $b=I+E_{23}$, $c=I+E_{13}$ checks the identity and shows $c$ has infinite order.