False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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FALSE: isomorphic middle groups force equivalent extensions with fixed kernel and quotient

Statement

If two extensions have isomorphic middle groups, then they are equivalent as extensions with fixed kernel and fixed quotient.

Facts & Assumptions

Given: The additive cyclic group $E = Z / 9 Z$ , the kernel group $N = Z / 3 Z$ , the quotient group $Q = Z / 3 Z$ , and the inclusion $j : N \to E$ given by $j (\overset{ˉ}{a}) = \overline{3 a}$ .

[L1]

Equivalence of extensions fixes both the chosen kernel map and the chosen quotient map (Equivalence of group extensions with fixed kernel and fixed quotient).

Refutation

technique · direct

1.1

Define two quotient maps $π_{1}, π_{2} : E \to Q$ by $π_{1} (\overset{ˉ}{x}) = \overset{ˉ}{x} (m o d 3), π_{2} (\overset{ˉ}{x}) = 2 \overset{ˉ}{x} (m o d 3) .$ Both are surjective homomorphisms with kernel ${\overset{ˉ}{0}, \overset{ˉ}{3}, \overset{ˉ}{6}} = im j$ , so they give two extensions of $Q$ by $N$ with the same middle group $E$ .

givenalgebra

2.1

An equivalence in the sense of [L1] would be an automorphism $φ : E \to E$ such that $φ \circ j = j$ and $π_{2} \circ φ = π_{1}$ . Every automorphism of the additive cyclic group $Z / 9 Z$ has the form $φ (\overset{ˉ}{x}) = u \overset{ˉ}{x}$ for a unit $u \in {1, 2, 4, 5, 7, 8}$ . The condition $φ \circ j = j$ gives $\overline{3 u} = \overset{ˉ}{3}$ , so $u \equiv 1 (m o d 3)$ . But $π_{2} \circ φ = π_{1}$ forces $\overline{2 u} = \overset{ˉ}{1}$ in $Q$ , so $u \equiv 2 (m o d 3)$ . This is impossible. Therefore the two extensions are not equivalent.

L1step 1.1algebra∎

Depends on

Equivalence of group extensions with fixed kernel and fixed quotient

Used by

The cyclic group Z/9 supports inequivalent extensions of C₃ by C₃ Counterexample

Dependency tree · two levels

3 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

J. S. Milne, Group Theory (standard reference, not scraped)