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CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26
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For odd p, a direct product of two Heisenberg groups is special with centre of order p2, 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 p3 is the set (Z/p)3 with (a,b,c)(a′,b′,c′)=(a+a′,b+b′,c+c′+ab′) (The Heisenberg group of order p3 over Z/p).

[L2]

The Heisenberg multiplication makes (Z/p)3 a nonabelian group of order p3 (The Heisenberg multiplication is a group law, nonabelian, on a set of p3 elements).

[L3]

The Heisenberg group of order p3 is extraspecial, and for odd p it has exponent p (The Heisenberg group of order p3 is extraspecial, and for odd p it has exponent p).

[L4]

G×H:={(g,h):g∈G, h∈H} (The external direct product G×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∈G:zg=gz for every g∈G}. (The center Z(G) of a group).

[L8]

For every finite p-group P, the Frattini formula gives Φ(P)=P′Pp. (Φ(P)=P′Pp for a finite p-group)

[L9]

For a group G and a prime p, the pth-power subgroup is Gp=⟨gp:g∈G⟩. (The pth-power subgroup Gp)

[L10]

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

Counterexample

technique · constructive
1.1L1L2L4L10construct

Fix an odd prime p, let Hp be the Heisenberg group of order p3, and put G:=Hp×Hp. By [L2] each factor has order p3, so [L10] gives ∣G∣=p6 and in particular G is a finite p-group.

1.2F1L3L4L5L6L7L9L10algebra

Because Hp 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(Hp)×Z(Hp) and G′=Hp′×Hp′. Therefore Z(G)=G′ is elementary abelian of order p2. Also every element of Hp has pth power 1, so every element of G has pth power (1,1) and therefore Gp=1.

2.1F1L8step 1.1step 1.2

Since G is a finite p-group, the Frattini formula gives Φ(G)=G′Gp=G′. Thus Z(G)=G′=Φ(G) is elementary abelian, so G is special.

3.1F1step 1.2step 2.1discharge-construct∎

But ∣Z(G)∣=p2, not p, so G is not extraspecial.

Depends on

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