Finite Heisenberg group H_3(F_q)

Let $q=p^f$ , with $p$ prime and $f\geq1$ an integer, and let $\mathbb F_q$ be the field of order $q$ . The finite Heisenberg group $H_3(\mathbb F_q)$ is $\mathbb F_q^3\times\mathbb F_q^3\times\mathbb F_q$ , with multiplication

(x,y,z)(x',y',z')=(x+x',y+y',z+z'+x\cdot y'),

where $x\cdot y'$ is the usual coordinate dot product. This formula works in every characteristic.

Order and group law

The order is $q^7$ , the identity is $(0,0,0)$ , and the inverse of $(x,y,z)$ is $(-x,-y,-z+x\cdot y)$ . Bilinearity of the dot product verifies associativity.

Center and nonsimplicity

Using $[g,h]=ghg^{-1}h^{-1}$ , the commutator is

[(x,y,z),(x',y',z')]=(0,0,x\cdot y'-x'\cdot y).

The center is exactly $\{(0,0,z):z\in\mathbb F_q\}$ : commuting with every $(x',y',0)$ forces both $x=0$ and $y=0$ . It is nontrivial and proper, so the group is not simple. The displayed commutators fill the center; consequently the group is nonabelian of nilpotency class two.