Alphabeta Math
CounterexampleConstruction: 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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An extraspecial group of order 32 decomposes both as two quaternion factors and as two dihedral factors

Statement refuted

The central-product decomposition of an extraspecial group into factors of order p3 is unique.

Facts & Assumptions

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

[L1]

Subgroups G1,,Gr of G form an internal central product when they generate G and [Gi,Gj]=1 for ij (Internal central products of a finite family of subgroups).

[L2]

Subgroups G1,,Gr form an internal central product of G if and only if the multiplication map G1××GrG is a surjective homomorphism each of whose factors meets its kernel trivially (Internal central products are the images of external ones).

[L3]

Q8Q8Dih(C4)Dih(C4) (Q8Q8Dih(C4)Dih(C4)).

[L4]

For each n1 there are exactly two extraspecial groups of order 21+2n up to isomorphism, with 22n+2n and 22n2n solutions of x2=1 (For each n1 there are exactly two extraspecial groups of order 21+2n).

[L5]

Q8  :=  {1,1,i,i,j,j,k,k}    H×. (The quaternion group Q8={±1,±i,±j,±k} inside the nonzero quaternions).

Counterexample

technique · constructive
1.1

Inside one extraspecial group of order thirty-two, exhibit two quaternion subgroups and two dihedral subgroups, using the explicit generators of the cited isomorphism.

L3L5L6construct
2.1

Each pair satisfies the internal central-product conditions: elementwise commuting, intersection the centre, and generating the group.

L1L2step 1.1
3.1

So the isomorphism type of the factors is not determined by the group, although the group itself is one of the two given by the classification.

L3L4step 2.1discharge-construct

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources