False statementConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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All finite Schur covers are unique

Statement

Every finite group has a unique Schur covering group up to isomorphism.

Facts & Assumptions

Given: Let $V = C_{2} \times C_{2}$ .

Refutation

technique · direct

1.1

Since $M (V) ≅ ⋀^{2} V ≅ C_{2}$ , both $C_{2} \to D_{8} \to V$ and $C_{2} \to Q_{8} \to V$ are stem extensions with multiplier kernel. Thus both are Schur covers of $V$ .

givenalgebra

2.1

The group $D_{8}$ has five involutions whereas $Q_{8}$ has one, so the two covers are not isomorphic. Schur covers are therefore not unique in general.

step 1.1contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

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