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Two elements of an extraspecial p-group with nontrivial commutator generate an extraspecial subgroup of order p3

Statement

Let P be an extraspecial p-group with Z(P)=⟨z⟩, and let x,y∈P satisfy [x,y]≠e. Then Q=⟨x,y⟩ contains Z(P), has order p3, is nonabelian, and is extraspecial with Z(Q)=Z(P).

Facts & Assumptions

Given: An extraspecial p-group P with Z(P)=⟨z⟩ of order p, the quotient V=P/Z(P) with its commutator pairing bz, and elements x,y∈P with [x,y]≠e.

[F1]

The commutator pairing of P relative to z is the map bz:V×V→Fp determined by [x,y]=z bz(xˉ,yˉ) (The commutator pairing of an extraspecial p-group relative to a chosen generator of its centre).

[F2]

Z(G):={z∈G:zg=gz for every g∈G} (The center Z(G) of a group).

[F3]

A subset S of an elementary abelian p-group is independent when ∏s∈Ssas=e with finite support forces every as=0, and it spans when every element is such a product (Fp-spanning sets, independence, and bases in an elementary abelian p-group).

[F4]

An elementary abelian p-group is a finite abelian p-group in which every nonidentity element has order p (Elementary abelian 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]

The commutator pairing is well defined, Fp-bilinear and alternating, and satisfies bz(yˉ,xˉ)=−bz(xˉ,yˉ) (The commutator pairing is well defined on the central quotient, is bilinear over Fp, and is alternating).

[L3]

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

[L4]

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

[L5]

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

[L7]

The rule aˉ⋅x=xa gives every elementary abelian p-group its canonical Fp-vector-space structure (An elementary abelian p-group has a canonical Fp-vector-space structure).

[L8]

If P is a finite p-group and H≤P, then ∣H∣=pk for some k∈N (Every subgroup of a finite p-group has order a power of p).

Proof

technique · direct
1.1F2L1L3L6

The element [x,y] lies in [P,P]=Z(P) and is not the identity, so it generates the group Z(P) of order p; since [x,y]∈Q, this gives Z(P)≤Q.

1.2F1

Write c=bz(xˉ,yˉ); then zc=[x,y]≠e, so c≠0 in Fp.

2.1F3L2L7step 1.2

The pair xˉ,yˉ is independent in V: if xˉ ayˉ d is the identity of V, pairing with yˉ gives a c=0 and hence a=0, and pairing with xˉ gives −d c=0 and hence d=0.

3.1F3L4L6L7step 1.1step 2.1

Since Z(P)≤Q, the image of Q in V is Q/Z(P), and it equals ⟨xˉ,yˉ⟩={xˉ ayˉ d:a,d∈Fp}; independence makes the p2 displayed products pairwise distinct, so ∣Q/Z(P)∣=p2 and Lagrange gives ∣Q∣=p⋅p2=p3.

4.1F2L3L4L5L8step 1.1step 3.1

Q is nonabelian because [x,y]≠e, and Z(P)≤Z(Q) because an element central in P is central in the subgroup Q containing it. The order of Z(Q) is a power of p dividing p3 and is not p3; were it p2, the quotient Q/Z(Q) would have order p and hence be cyclic, forcing Q abelian. So ∣Z(Q)∣=p and Z(Q)=Z(P).

5.1F4L1step 3.1step 4.1∎

The quotient Q/Z(Q)=Q/Z(P) is a subgroup of the elementary abelian group V, hence is itself a finite abelian p-group all of whose nonidentity elements have order p; so Q is a nonabelian finite p-group with centre of order p and elementary abelian central quotient, and the characterisation makes it extraspecial.

Remarks

The hypothesis is on the pair, not on either element separately: x and y are automatically noncentral, since a central element commutes with everything, but two noncentral elements can commute and then generate an abelian subgroup.

Depends on

Used by

Dependency tree · two levels

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Sources