= Abelian plane subgroup of the Heisenberg group modulo four
{title2=$H=\{A(t,y,t)\}\cong C_8\times C_2$}
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 $y+y'+tt'$, 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 $u^n=A(n,\binom n2,n)$, 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> $C_8\times C_2$.
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