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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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A noncentral element of an extraspecial p-group has centraliser of index p

Statement

Let P be an extraspecial p-group and let x∈P∖Z(P). Then

[P:CP(x)]=p,equivalently∣CP(x)∣=∣P∣p.

Facts & Assumptions

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

[F1]

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]

[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).

[L1]

Every conjugacy class of an extraspecial p-group whose representative is not central has exactly p elements (Every conjugacy class of an extraspecial p-group outside the centre has exactly p elements).

[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).

Proof

technique · direct
1.1L1

The class of x has exactly p elements, because x is not central.

1.2F1F2L2

The size of that class is the index of the centraliser of x in P.

2.1L3step 1.1step 1.2∎

Hence [P:CP(x)]=p, and Lagrange turns this into ∣P∣=p ∣CP(x)∣.

Remarks

Every centraliser named here is proper, since x is noncentral, and maximal in the order sense: index p is the smallest index a proper subgroup of a finite p-group can have.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources