CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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For odd $p$ , a direct product of two Heisenberg groups is special with centre of order $p^{2}$ , hence not extraspecial

Statement refuted

Every special $p$ -group is extraspecial.

Facts & Assumptions

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

[F1]

A finite $p$ -group $P$ is special when $Z (P) = P^{'} = Φ (P)$ is elementary abelian, and extraspecial when in addition $P$ is nonabelian and this common subgroup has order $p$ (Special and extraspecial $p$ -groups).

[L1]

The Heisenberg group of order $p^{3}$ is the set $(Z / p)^{3}$ with $(a, b, c) (a^{'}, b^{'}, c^{'}) = (a + a^{'}, b + b^{'}, c + c^{'} + a b^{'})$ (The Heisenberg group of order $p^{3}$ over $Z / p$ ).

[L2]

The Heisenberg multiplication makes $(Z / p)^{3}$ a nonabelian group of order $p^{3}$ (The Heisenberg multiplication is a group law, nonabelian, on a set of $p^{3}$ elements).

[L3]

The Heisenberg group of order $p^{3}$ is extraspecial, and for odd $p$ it has exponent $p$ (The Heisenberg group of order $p^{3}$ is extraspecial, and for odd $p$ it has exponent $p$ ).

[L4]

$G \times H : = {(g, h) : g \in G, h \in H}$ (The external direct product $G \times H$ with componentwise multiplication).

[L5]

An elementary abelian $p$ -group is a finite abelian $p$ -group in which every nonidentity element has order $p$ ; the trivial group is permitted (,, ). (Elementary abelian $p$ -groups).

[L6]

$Z (G) : = {z \in G : z g = g z for every g \in G} .$ (The center $Z (G)$ of a group).

[L7]

$[g, h] : = g h g^{- 1} h^{- 1} .$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[L8]

For every finite $p$ -group $P$ , the Frattini formula gives $Φ (P) = P^{'} P^{p} .$ ( $Φ (P) = P^{'} P^{p}$ for a finite $p$ -group)

[L9]

For a group $G$ and a prime $p$ , the $p$ th-power subgroup is $G^{p} = ⟨ g^{p} : g \in G ⟩ .$ (The $p$ th-power subgroup $G^{p}$ )

[L10]

For finite groups $G$ and $H$ , the direct product has order $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ . (For finite groups $G$ and $H$ , $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ )

Counterexample

technique · constructive

1.1

Fix an odd prime $p$ , let $H_{p}$ be the Heisenberg group of order $p^{3}$ , and put $G : = H_{p} \times H_{p}$ . By [L2] each factor has order $p^{3}$ , so [L10] gives $∣ G ∣ = p^{6}$ and in particular $G$ is a finite $p$ -group.

L1L2L4L10construct

1.2

Because $H_{p}$ is extraspecial, each factor has centre equal to its derived subgroup and that common subgroup has order $p$ ; hence coordinatewise multiplication in the direct product gives $Z (G) = Z (H_{p}) \times Z (H_{p})$ and $G^{'} = H_{p}^{'} \times H_{p}^{'}$ . Therefore $Z (G) = G^{'}$ is elementary abelian of order $p^{2}$ . Also every element of $H_{p}$ has $p$ th power $1$ , so every element of $G$ has $p$ th power $(1, 1)$ and therefore $G^{p} = 1$ .

F1L3L4L5L6L7L9L10algebra

2.1

Since $G$ is a finite $p$ -group, the Frattini formula gives $Φ (G) = G^{'} G^{p} = G^{'}$ . Thus $Z (G) = G^{'} = Φ (G)$ is elementary abelian, so $G$ is special.

F1L8step 1.1step 1.2

3.1

But $∣ Z (G) ∣ = p^{2}$ , not $p$ , so $G$ is not extraspecial.

F1step 1.2step 2.1discharge-construct∎

Depends on

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Sources

D. A. Craven, The Theory of p-Groups (Hilary Term 2008), 48 pp. (standard reference, not scraped)
M. van Beek, Topics in Finite p-Groups, 62 pp. (standard reference, not scraped)
D. Kaur and A. Kulshrestha, Characters of real special 2-groups (arXiv:1510.06583v1) (standard reference, not scraped)

For odd pp, a direct product of two Heisenberg groups is special with centre of order p2p2, hence not extraspecial