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LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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The commutator pairing is well defined on the central quotient, is bilinear over Fp, and is alternating

Statement

Let P be an extraspecial p-group with Z(P)=⟨z⟩ and V=P/Z(P). The commutator pairing bz is well defined on V×V: the value of bz(xˉ,yˉ) does not depend on the representatives x and y. It is Fp-bilinear,

bz(xˉyˉ,wˉ)=bz(xˉ,wˉ)+bz(yˉ,wˉ),bz(wˉ,xˉyˉ)=bz(wˉ,xˉ)+bz(wˉ,yˉ),bz(xˉ a,yˉ)=a bz(xˉ,yˉ)=bz(xˉ,yˉ a),

and alternating: bz(xˉ,xˉ)=0 for every xˉ∈V. Consequently bz(yˉ,xˉ)=−bz(xˉ,yˉ).

Facts & Assumptions

Given: An extraspecial p-group P with Z(P)=⟨z⟩ of order p, the quotient V=P/Z(P) with its canonical Fp-structure, and the pairing bz defined by [x,y]=z bz(xˉ,yˉ).

[F1]

The commutator pairing of P relative to z is the map bz:V×V→Fp determined by [x,y]=z bz(xˉ,yˉ), where V=P/Z(P) carries the canonical scalar action aˉ⋅xˉ=xˉa (The commutator pairing of an extraspecial p-group relative to a chosen generator of its centre).

[F2]

For g,h∈G the commutator is [g,h]:=ghg−1h−1 (Commutators [g,h]=ghg−1h−1 and the commutator subgroup [G,G]).

[F3]

The rule aˉ⋅x=xa gives every elementary abelian p-group its canonical Fp-vector-space structure, with the group operation as vector addition and the identity as zero (An elementary abelian p-group has a canonical Fp-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]≤Z(G) then [xy,w]=[x,w][y,w], [x,yw]=[x,y][x,w], and [xn,y]=[x,y]n=[x,yn] for every integer n (Commutator identities in a group whose derived subgroup is central).

[L3]

Z(G):={z∈G:zg=gz for every g∈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
1.1L1L2

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

1.2F1F2

Alternation: [x,x]=xxx−1x−1=e=z0, so bz(xˉ,xˉ)=0.

2.1F1F2L3step 1.1

If xˉ=x′‾ then x′=xu with u∈Z(P), and [xu,y]=[x,y][u,y]=[x,y] because u is central makes [u,y]=uyu−1y−1=e; the second variable is the same computation with the other expansion identity. So bz(xˉ,yˉ) depends only on the two cosets.

2.2F1F3step 1.1

Additivity in the first variable: z bz(xˉyˉ,wˉ)=[xy,w]=[x,w][y,w]=z bz(xˉ,wˉ)z bz(yˉ,wˉ)=z bz(xˉ,wˉ)+bz(yˉ,wˉ), and exponents of z are determined modulo p; the second variable is symmetric.

2.3F1F3L4step 1.1

Compatibility with scalars: for an integer a, z bz(xˉ a,yˉ)=[xa,y]=[x,y]a=z a bz(xˉ,yˉ), and both sides depend only on a modulo p because z has order p; the scalar action on V is aˉ⋅xˉ=xˉa, so this is exactly bz(a⋅xˉ,yˉ)=a bz(xˉ,yˉ), and likewise in the second variable.

3.1step 2.2step 1.2∎

Expanding 0=bz(xˉyˉ,xˉyˉ) by steps 2.2 and 1.2 gives 0=bz(xˉ,yˉ)+bz(yˉ,xˉ), so bz(yˉ,xˉ)=−bz(xˉ,yˉ).

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 bz(xˉ,xˉ) 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