False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

False: every short exact sequence of groups splits

Statement

False claim: every short exact sequence of groups has a homomorphic section and therefore splits.

Facts & Assumptions

Given: A prime $p$ and the sequence induced by multiplication by $p$ and reduction modulo $p$ ,

$1 \to C_{p} \to C_{p^{2}} \to^{π} C_{p} \to 1.$

[A1]

The false claim says that this short exact sequence has a section.

[L1]

A section is a homomorphism $s$ satisfying $π s = id$ (Group extensions, sections, complements, and split extensions).

[L2]

Every subgroup of a cyclic group is cyclic (Every subgroup of a cyclic group is cyclic; the least positive exponent in a nontrivial subgroup supplies a generator).

[L3]

The order criterion determines which multiples of a finite-order element are zero (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Refutation

technique · contradiction

1.1L3algebra

The kernel of reduction $π : C_{p^{2}} \to C_{p}$ is $p C_{p^{2}}$ , which has order $p$ by [L3], so the displayed sequence is short exact.

1.2A1L1L2L3assume-contraalgebra

Assume, for contradiction, that [A1] holds and let $s$ be a section. Since $π s$ is the identity, $s$ is injective and its image $H$ has order $p$ . By [L2], $H = ⟨ [a]_{p^{2}} ⟩$ for some $a$ . Its generator has order $p$ , so [L3] gives $p^{2} ∣ p a$ and hence $p ∣ a$ . Thus $H \subseteq p C_{p^{2}}$ ; both subgroups have $p$ elements, so $H = \ker π$ .

2.1step 1.2A1L1discharge-contradiction∎

Hence $π s$ is zero, contradicting [L1]. Therefore this sequence does not split and [A1] is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

28 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)