False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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FALSE for odd $p$ : the scalar-valued commutator pairing needs no choice of a central generator

Statement refuted

for odd $p$ , the scalar-valued commutator pairing of an extraspecial $p$ -group is defined without choosing a generator of its centre.

Facts & Assumptions

Given: The proposed claim together with the witness named in the Statement refuted.

[F1]

For an extraspecial $p$ -group $P$ with $Z (P) = ⟨ z ⟩$ , the commutator pairing is the map $b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y}) \in Z / p$ determined by $[x, y] = z^{b_{z} (\overset{ˉ}{x}, \overset{ˉ}{y})}$ (The commutator pairing of an extraspecial $p$ -group relative to a chosen generator of its centre).

[L1]

The commutator pairing is independent of the coset representatives, is $F_{p}$ -bilinear on $P / Z (P)$ , and is alternating (The commutator pairing is well defined on the central quotient, is bilinear over $F_{p}$ , and is alternating).

[L2]

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

[L3]

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).

Refutation

technique · contradiction

1.1

Let $p$ be odd. The claim asserts that the scalar-valued pairing is independent of the generator of the centre used to define it.

F1L2assume-contra

2.1

Replacing $z$ by $z^{c}$ multiplies every value by $c^{- 1}$ , so only the pairing up to that scaling is choice-free; the scalar-valued map itself changes.

F1L1L3step 1.1discharge-contradiction∎

Remarks

At $p = 2$ the centre has a unique nonidentity element and hence a unique generator, so there is no choice to make. The refuted claim is restricted to odd $p$ , where the centre has more than one generator.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

D. A. Craven, The Theory of p-Groups (Hilary Term 2008), 48 pp. (standard reference, not scraped)
M. van Beek, Topics in Finite p-Groups, 62 pp. (standard reference, not scraped)
D. Kaur and A. Kulshrestha, Characters of real special 2-groups (arXiv:1510.06583v1) (standard reference, not scraped)

FALSE for odd pp: the scalar-valued commutator pairing needs no choice of a central generator