The commutator pairing of an extraspecial $p$-group relative to a chosen generator of its centre

Let $P$ be an extraspecial $p$-group (Special and extraspecial $p$-groups) and fix a generator $z$ of its centre, so that $Z(P)=⟨z⟩$ has order $p$. Write

$V:=P\mathrm{/}Z(P),$

which is elementary abelian (Three equivalent descriptions of an extraspecial $p$-group, Elementary abelian $p$-groups, The quotient group $G\mathrm{/}N$ and coset product $(gN)(hN)=ghN$), and give it its canonical ${\mathbb{F}}_p$-vector-space structure (An elementary abelian $p$-group has a canonical ${\mathbb{F}}_p$-vector-space structure), where ${\mathbb{F}}_p=\mathbb{Z}\mathrm{/}p$ is the field of For every prime $p$, the two operations on $\mathbb{Z}\mathrm{/}p$ make it a field and For every natural $n$, $(\mathbb{Z}\mathrm{/}n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold, the group operation of $V$ is vector addition and the scalar action is $\bar{a}\cdot \bar{x}={\bar{x}}^a$. Elements of $V$ are written multiplicatively, $\bar{x}$ denoting the coset $xZ(P)$; scalars are written additively.

The commutator pairing of $P$ relative to $z$ is the map

$b_z:V\times V\to {\mathbb{F}}_p,[x,y]=z^{b_z(\bar{x},\bar{y})}\mathrm{.}$

Why the exponent exists and is unique. Every commutator of $P$ lies in $[P,P]=Z(P)=⟨z⟩$ (An extraspecial $p$-group is nilpotent of class exactly two and its derived subgroup has order $p$, Commutators $[g,h]=gh{g}^{-1}{h}^{-1}$ and the commutator subgroup $[G,G]$, The center $Z(G)$ of a group), so $[x,y]=z^k$ for some integer $k$. Since $z$ has order $p$ (The order $\mathrm{\mid }G\mathrm{\mid }$ of a finite group and the order $\mathrm{ord}(g)$ of an element, with $\mathrm{ord}(g)=\mathrm{\infty }$ when no positive power of $g$ is the identity, A finite group of prime order is cyclic and every nonidentity element generates it), $z^k$ depends only on the class of $k$ in $\mathbb{Z}\mathrm{/}p$, and $z^k=z^{k^{\mathrm{\prime }}}$ forces $k\equiv k^{\mathrm{\prime }}$; so the class $\bar{k}\in {\mathbb{F}}_p$ is determined by $[x,y]$, and that class is $b_z(\bar{x},\bar{y})$.

That the value depends only on the cosets $\bar{x}$ and $\bar{y}$, and not on the representatives $x$ and $y$, is proved in The commutator pairing is well defined on the central quotient, is bilinear over ${\mathbb{F}}_p$, and is alternating ↗.

The pairing depends on the choice of $z$, and only on it: replacing $z$ by $z^c$ with $c\mathrm{\ne }0$ replaces $b_z$ by $c^{-1}{b}_z$. So the radical, the orthogonality relation and every statement about a subspace being self-orthogonal are independent of the choice, while the individual values are not. The companion page records the false statement that no choice is needed.

Nothing here is imported from a theory of bilinear forms. The target ${\mathbb{F}}_p$ is the field $\mathbb{Z}\mathrm{/}p$, the vector-space structure on $V$ is the canonical scalar action of an elementary abelian $p$-group, and every property of $b_z$ used below is proved from the commutator identities of a group whose derived subgroup is central.