Heisenberg group over a commutative ring ($H_3(R)=U{T}_3(R)$)

For a commutative ring $R$ with an identity element, use coordinatesThese matrices form a group under multiplication, withThese formulas require no division, so they remain valid over rings with zero divisors. The center of a group is $\{A(0,y,0):y\in R\}$: the commutator has only central coordinate $x{z}^{\prime }-x^{\prime }z$, and testing $x^{\prime }=0,z^{\prime }=1$ and $x^{\prime }=1,z^{\prime }=0$ forces $x=z=0$. For finite $R$, its group order is $|R{|}^3$. This includes real, integer and modular versions of the Heisenberg group; the ring $\mathbb{Z}/4\mathbb{Z}$ must not be confused with the finite field of four elements.

In the Heisenberg group over a commutative ring with $R=\mathbb{Z}/4\mathbb{Z}$, the subset $H=\{A(t,y,t):t,y\in R\}$ is an Abelian subgroup of group order 16. Its product has central coordinate $y+y^{\prime }+t{t}^{\prime }$, symmetric in the two factors. Its inverse element is $A(-t,t^2-y,-t)$. The map $A(x,y,z)\mapsto x-z$ is a surjective group homomorphism to the additive group of $R$, with kernel of a group homomorphism $H$. Thus $H$ is a normal subgroup and has four cosets in the 64-element group.

Unlike the odd-prime case, division by two is unavailable. In fact take $u=A(1,0,1)$ and $v=A(2,0,2)$. Then $u^n=A(n,(\genfrac{}{}{0ex}{}{n}{2}),n)$, where the binomial coefficient is computed as an integer before reduction modulo four. Hence $u^4=A(0,2,0)$ and $u$ has element order eight. The element $v$ has element order two and is not a power of $u$: the powers with outer coordinates two have central coordinates one or three. Their generated subgroups intersect trivially, so the 16-element Abelian group is the direct product of groups $C_8\times C_2$.