False statementConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedverified 2026-09-26 (gpt-6-sol)

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FALSE: the numerator in Hopf's formula is presentation-independent

Statement

For every pair of free presentations $G = F / R = F^{'} / R^{'}$ of the same group, the numerator subgroups $R \cap [F, F]$ and $R^{'} \cap [F^{'}, F^{'}]$ in Hopf's formula are isomorphic.

Facts & Assumptions

Given: The trivial group has free presentations $1 = Z / Z$ and $1 = F (a, b) / F (a, b)$ .

[F1]

Hopf's quotient for $F / R$ has numerator $R \cap [F, F]$ and denominator $[F, R]$ (Hopf-formula quotient).

[F2]

Reduced words distinguish the nonempty commutator word $a b a^{- 1} b^{- 1}$ from the identity in $F (a, b)$ (Reduced words form the free group on an alphabet).

[L1]

With dependent choice and supplied resolution data, the quotients $(R \cap [F, F]) / [F, R]$ from two free presentations are isomorphic through $M (G)$ (Hopf formula is presentation-independent).

Refutation

technique · direct

1.1givenF1algebra

In the first presentation $F = R = Z$ , so the numerator $R \cap [F, F] = [Z, Z]$ is trivial.

2.1givenF1F2step 1.1

In the second presentation $F = R = F (a, b)$ , so the numerator is $[F (a, b), F (a, b)]$ . It contains the commutator $a b a^{- 1} b^{- 1}$ , which is a nonempty reduced word and hence nonidentity by [F2]. The two numerator groups are therefore not isomorphic.

3.1F1L1step 1.1step 2.1∎

The denominators are respectively trivial and $[F (a, b), F (a, b)]$ , so both Hopf quotients are trivial. This agrees with the qualified invariant comparison in [L1]; it does not make their numerators isomorphic.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Clara Löh, Group Cohomology (standard reference, not scraped)