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PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26
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Every conjugacy class of an extraspecial p-group outside the centre has exactly p elements

Statement

Every conjugacy class of an extraspecial p-group whose representative is not central has exactly p elements. That is, for an extraspecial p-group P and x∈P∖Z(P),

∣Cl⁡P(x)∣=p.

Facts & Assumptions

Given: An extraspecial p-group P (Special and extraspecial p-groups) and an element x∈P with x∉Z(P).

[F1]

Cl⁡G(x):={gxg−1:g∈G} and CG(x):={g∈G:gx=xg}={g∈G:gxg−1=x} (The conjugacy class Cl⁡G(x) and centralizer CG(x) of an element).

[F2]

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

[F3]

For g,h∈G the commutator is [g,h]:=ghg−1h−1 (Commutators [g,h]=ghg−1h−1 and the commutator subgroup [G,G]).

[F4]

For H≤G and g∈G, the right coset is Hg:={hg:h∈H} (Left and right cosets gH and Hg of a subgroup).

[L1]

An extraspecial p-group has derived subgroup P′=Z(P) of order p (An extraspecial p-group is nilpotent of class exactly two and its derived subgroup has order p).

[L3]

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).

[L4]

[G:H]:=∣G/H∣, the number of left cosets of H in G (The coset set G/H and the index [G:H] of a subgroup).

[L5]

A finite p-group is a finite group whose order has the form ∣P∣=pn for some n∈N (A finite p-group has order pn for a prime p and some n∈N).

Proof

technique · direct
1.1F3algebra

For every g∈P one has [g,x]x=(gxg−1x−1)x=gxg−1, so each conjugate of x has the form [g,x]x.

1.2L1L5

The derived subgroup of P is Z(P) and has order p, and P is a finite p-group, so ∣P∣=pn for some n.

1.3F1L2L3L4

The size of the class of x is the index [P:CP(x)], and by Lagrange that index divides ∣P∣.

2.1F4step 1.1step 1.2algebra

Each [g,x] lies in P′=Z(P), so every conjugate of x lies in the right coset Z(P)x; the map z↦zx is a bijection from Z(P) onto that coset, so the coset has p elements and the class of x has at most p.

2.2F1F2step 1.3

Since x∉Z(P), some g∈P fails to commute with x, so CP(x)≠P, its index is greater than one, and the class of x has more than one element.

3.1step 1.2step 1.3step 2.1step 2.2∎

The class size divides pn, so it is a power of p; it lies strictly between 1 and p inclusive, and the only such power of p is p itself.

Remarks

The hypothesis x∉Z(P) is used only at step 2.2, and it is used to rule out the class of size one. A central x runs through the same computation and comes out with the class {x}, which is consistent with step 2.1 and shows that the two cases exhaust the group.

Depends on

Used by

Dependency tree · two levels

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Sources