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An extraspecial $p$ -group is nilpotent of class exactly two and its derived subgroup has order $p$

Statement

Let $P$ be an extraspecial $p$ -group. Then $P$ is nilpotent of nilpotency class exactly two, its derived subgroup satisfies $P^{'} = Z (P)$ and has order $p$ , and every nonidentity commutator of $P$ has order $p$ .

Facts & Assumptions

Given: An extraspecial $p$ -group $P$ (Special and extraspecial $p$ -groups).

[L1]

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).

[L2]

For subgroups $A, B \leq G$ the subgroup commutator is $[A, B] = ⟨ [a, b] : a \in A, b \in B ⟩$ , and the lower central series is $γ_{1} (G) = G$ , $γ_{r + 1} (G) = [G, γ_{r} (G)]$ (Subgroup commutators and the lower central series).

[L3]

$G$ is nilpotent exactly when its lower central series reaches $1$ , and the least $c$ with $γ_{c + 1} (G) = 1$ is its nilpotency class (Nilpotence via central series, the upper central series, and the lower central series, Nilpotent groups and nilpotency class).

[L4]

In a finite group whose order is prime, every $g \neq e$ has order $∣ G ∣$ (A finite group of prime order is cyclic and every nonidentity element generates it).

[F1]

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

Proof

technique · direct

1.1

By the third description in the characterisation, $P^{'} = Z (P)$ has order $p$ and $P$ is nonabelian.

1.2

The second term of the lower central series is $γ_{2} (P) = [P, γ_{1} (P)] = [P, P] = P^{'}$ , and the third is $γ_{3} (P) = [P, P^{'}]$ .

2.1

Every element of $P^{'} = Z (P)$ commutes with every element of $P$ , so each generator $[g, z]$ of $[P, P^{'}]$ is the identity and $γ_{3} (P) = 1$ .

F1step 1.1step 1.2

3.1

Hence $P$ is nilpotent of class at most two; the class is not zero or one, since class at most one would give $γ_{2} (P) = P^{'} = 1$ and $P$ is nonabelian, so the class is exactly two.

L3step 1.1step 1.2step 2.1

4.1

Every commutator lies in $P^{'}$ , a group of order $p$ , so a nonidentity commutator has order $p$ .

L4step 1.1∎

Remarks

Both conclusions are hypotheses of the class-two commutator calculus: the derived subgroup is central, which is what class two says, and it has exponent $p$ , which is what the order- $p$ conclusion says.

Depends on

Used by

Dependency tree · two levels

32 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

M. van Beek, Topics in Finite p-Groups, Proposition 2.29 (standard reference, not scraped)
D. A. Craven, The Theory of p-Groups, §3.1 (standard reference, not scraped)

An extraspecial pp-group is nilpotent of class exactly two and its derived subgroup has order pp