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The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating

Statement

Let $P$ be an extraspecial $p$ -group with $Z (P) = ⟨ z ⟩$ and $V = P / Z (P)$ . The commutator pairing $b_{z}$ is well defined on $V \times V$ : the value of $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ does not depend on the representatives $x$ and $y$ . It is $F_{p}$ -bilinear,

$b_{z} (\overset{ˉ}{x} \overset{ˉ}{y}, \overset{ˉ}{w}) = b_{z} (\overset{ˉ}{x}, \overset{ˉ}{w}) + b_{z} (\overset{ˉ}{y}, \overset{ˉ}{w}), b_{z} (\overset{ˉ}{w}, \overset{ˉ}{x} \overset{ˉ}{y}) = b_{z} (\overset{ˉ}{w}, \overset{ˉ}{x}) + b_{z} (\overset{ˉ}{w}, \overset{ˉ}{y}), b_{z} ({\overset{ˉ}{x}}^{a}, \overset{ˉ}{y}) = a b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) = b_{z} (\overset{ˉ}{x}, {\overset{ˉ}{y}}^{a}),$

and alternating: $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{x}) = 0$ for every $\overset{ˉ}{x} \in V$ . Consequently $b_{z} (\overset{ˉ}{y}, \overset{ˉ}{x}) = - b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ .

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ with $Z (P) = ⟨ z ⟩$ of order $p$ , the quotient $V = P / Z (P)$ with its canonical $F_{p}$ -structure, and the pairing $b_{z}$ defined by $[x, y] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ .

[F1]

The commutator pairing of $P$ relative to $z$ is the map $b_{z} : V \times V \to F_{p}$ determined by $[x, y] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ , where $V = P / Z (P)$ carries the canonical scalar action $\overset{ˉ}{a} \cdot \overset{ˉ}{x} = {\overset{ˉ}{x}}^{a}$ (The commutator pairing of an extraspecial $p$ -group relative to a chosen generator of its centre).

[F2]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F3]

The rule $\overset{ˉ}{a} \cdot x = x^{a}$ gives every elementary abelian $p$ -group its canonical $F_{p}$ -vector-space structure, with the group operation as vector addition and the identity as zero (An elementary abelian $p$ -group has a canonical $F_{p}$ -vector-space structure).

[L1]

An extraspecial $p$ -group is nilpotent of class exactly two, so its derived subgroup is central, and every nonidentity commutator has order $p$ (An extraspecial $p$ -group is nilpotent of class exactly two and its derived subgroup has order $p$ ).

[L2]

If $[G, G] \leq Z (G)$ then $[x y, w] = [x, w] [y, w]$ , $[x, y w] = [x, y] [x, w]$ , and $[x^{n}, y] = [x, y]^{n} = [x, y^{n}]$ for every integer $n$ (Commutator identities in a group whose derived subgroup is central).

[L3]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[L4]

For every prime $p$ , the operations of addition and multiplication on $Z / p$ make it a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

Proof

technique · direct

1.1

The derived subgroup of $P$ is central, so the two expansion identities and the power identity are available for all elements of $P$ .

L1L2

1.2

Alternation: $[x, x] = x x x^{- 1} x^{- 1} = e = z^{0}$ , so $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{x}) = 0$ .

F1F2

2.1

If $\overset{ˉ}{x} = \overline{x^{'}}$ then $x^{'} = x u$ with $u \in Z (P)$ , and $[x u, y] = [x, y] [u, y] = [x, y]$ because $u$ is central makes $[u, y] = u y u^{- 1} y^{- 1} = e$ ; the second variable is the same computation with the other expansion identity. So $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ depends only on the two cosets.

F1F2L3step 1.1

2.2

Additivity in the first variable: $z^{b_{z} (\overset{ˉ}{x} \overset{ˉ}{y}, \overset{ˉ}{w})} = [x y, w] = [x, w] [y, w] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{w})} z^{b_{z} (\overset{ˉ}{y}, \overset{ˉ}{w})} = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{w}) + b_{z} (\overset{ˉ}{y}, \overset{ˉ}{w})}$ , and exponents of $z$ are determined modulo $p$ ; the second variable is symmetric.

F1F3step 1.1

2.3

Compatibility with scalars: for an integer $a$ , $z^{b_{z} ({\overset{ˉ}{x}}^{a}, \overset{ˉ}{y})} = [x^{a}, y] = [x, y]^{a} = z^{a b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ , and both sides depend only on $a$ modulo $p$ because $z$ has order $p$ ; the scalar action on $V$ is $\overset{ˉ}{a} \cdot \overset{ˉ}{x} = {\overset{ˉ}{x}}^{a}$ , so this is exactly $b_{z} (a \cdot \overset{ˉ}{x}, \overset{ˉ}{y}) = a b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ , and likewise in the second variable.

F1F3L4step 1.1

3.1

Expanding $0 = b_{z} (\overset{ˉ}{x} \overset{ˉ}{y}, \overset{ˉ}{x} \overset{ˉ}{y})$ by steps 2.2 and 1.2 gives $0 = b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) + b_{z} (\overset{ˉ}{y}, \overset{ˉ}{x})$ , so $b_{z} (\overset{ˉ}{y}, \overset{ˉ}{x}) = - b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})$ .

step 2.2step 1.2∎

Remarks

Alternation is the primitive property and skew symmetry is derived from it, not the other way round. At $p = 2$ the two are not interchangeable: there $- 1 = 1$ , so skew symmetry says only that the pairing is symmetric, and it is the vanishing of $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{x})$ that carries content.

Depends on

Used by

Cited to discharge well-definedness by The commutator pairing of an extraspecial p-group relative to a chosen generator of its centre.

Dependency tree · two levels

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Sources

D. A. Craven, The Theory of p-Groups, Lemma 3.7 and §3.2 (standard reference, not scraped)
M. van Beek, Topics in Finite p-Groups, §2.4 (standard reference, not scraped)

The commutator pairing is well defined on the central quotient, is bilinear over FpFp​, and is alternating