TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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A nonabelian group of order $p^{3}$ is extraspecial

Statement

Let $p$ be a prime and let $P$ be a nonabelian group of order $p^{3}$ . Then $P$ is extraspecial: $Z (P) = [P, P] = Φ (P)$ has order $p$ and $P / Z (P)$ is elementary abelian of order $p^{2}$ .

Facts & Assumptions

Given: A prime $p$ and a nonabelian group $P$ with $∣ P ∣ = p^{3}$ .

[F1]

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

[F2]

A finite $p$ -group is a finite group whose order has the form $∣ P ∣ = p^{n}$ (A finite $p$ -group has order $p^{n}$ for a prime $p$ and some $n \in N$ ).

[L1]

If $P$ is a nontrivial finite $p$ -group then $p$ divides $∣ Z (P) ∣$ (Every nontrivial finite $p$ -group has nontrivial center, in fact $p$ divides $∣ Z (P) ∣$ ).

[L2]

If the quotient group $G / Z (G)$ is cyclic, then $G$ is abelian (If $G / Z (G)$ is cyclic, then $G$ is abelian).

[L3]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L4]

If $p$ is prime and $G$ is a group of order $p^{2}$ , then $G$ is abelian (Every group of order $p^{2}$ , for prime $p$ , is abelian).

[L5]

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

[L6]

For a finite $p$ -group $P$ the following are equivalent: $P$ is extraspecial; $P$ is nonabelian, $∣ Z (P) ∣ = p$ and $P / Z (P)$ is elementary abelian; $P$ is nonabelian and $Z (P) = P^{'} = Φ (P)$ has order $p$ (Three equivalent descriptions of an extraspecial $p$ -group).

[L7]

If $P$ is a finite $p$ -group and $H \leq P$ then $∣ H ∣ = p^{k}$ for some $k$ (Every subgroup of a finite $p$ -group has order a power of $p$ ).

Proof

technique · direct

1.1

$P$ is a nontrivial finite $p$ -group, so its centre has order a power of $p$ divisible by $p$ ; and $Z (P) \neq P$ because $P$ is nonabelian. So $∣ Z (P) ∣$ is $p$ or $p^{2}$ .

F1F2L1L3L7

2.1

If $∣ Z (P) ∣ = p^{2}$ then $P / Z (P)$ has order $p$ by Lagrange, hence is cyclic, and $P$ would be abelian. So $∣ Z (P) ∣ = p$ .

L2L3step 1.1

3.1

By Lagrange $P / Z (P)$ has order $p^{2}$ , so it is abelian; it is not cyclic, since that would again force $P$ abelian; and an abelian group of order $p^{2}$ that is not cyclic has every nonidentity element of order $p$ , so it is elementary abelian.

L2L3L4L5step 2.1

4.1

So $P$ is a nonabelian finite $p$ -group with centre of order $p$ and elementary abelian central quotient, which is the second description in the characterisation; hence $P$ is extraspecial and $Z (P) = [P, P] = Φ (P)$ has order $p$ .

L6step 2.1step 3.1∎

Remarks

Order $p^{3}$ is the smallest order at which a nonabelian $p$ -group exists, and the argument shows the extraspecial condition is automatic there. At larger orders it is not: a direct product of two nonabelian groups of order $p^{3}$ is nonabelian of order $p^{6}$ with centre of order $p^{2}$ .

Depends on

Used by

Dependency tree · two levels

45 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

D. A. Craven, The Theory of p-Groups, Lemma 2.7 (standard reference, not scraped)

A nonabelian group of order p3p3 is extraspecial