The Heisenberg group of order $p^3$ over $\mathbb{Z}\mathrm{/}p$

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p>1$ and its only positive divisors are $1$ and $p$) and let $\mathbb{Z}\mathrm{/}p$ carry the addition and multiplication of For every natural $n$, $(\mathbb{Z}\mathrm{/}n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold, which make it a field (For every prime $p$, the two operations on $\mathbb{Z}\mathrm{/}p$ make it a field). The Heisenberg group of order $p^3$ is the set

$H_p:=\{(a,b,c):a,b,c\in \mathbb{Z}\mathrm{/}p\}$

with the multiplication

$(a,b,c)(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }}):=(a+a^{\mathrm{\prime }},b+b^{\mathrm{\prime }},c+c^{\mathrm{\prime }}+a{b}^{\mathrm{\prime }})\mathrm{.}$

That this is a group law (Group and abelian group), that it is not commutative, and that $H_p$ has $p^3$ elements are proved in The Heisenberg multiplication is a group law, nonabelian, on a set of $p^3$ elements ↗.

The same multiplication is written on the published example The finite Heisenberg group is the unique Sylow $p$-subgroup of its coordinate upper-triangular group, where the triple $(a,b,c)$ records the three entries above the diagonal of a unipotent upper-triangular $3\times 3$ matrix over $\mathbb{Z}\mathrm{/}p$; the two constructions produce the same group, and the notation is kept identical so that a reader meets one object rather than two.

The asymmetry of the third coordinate, $c+c^{\mathrm{\prime }}+a{b}^{\mathrm{\prime }}$ rather than $c+c^{\mathrm{\prime }}+a^{\mathrm{\prime }}b$, is a choice of convention: the opposite choice gives the group with the roles of the first two coordinates exchanged, and the map $(a,b,c)\mapsto (b,a,c)$ carries one to the other. Nothing below depends on which is taken, provided one is taken throughout.

At $p=2$ the construction does not produce a group of exponent two: the element $(1,1,0)$ squares to $(0,0,1)$, so $H_2$ has an element of order four.