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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 xPZ(P). Then

[P:CP(x)]=p,equivalentlyCP(x)=Pp.

Facts & Assumptions

Given: An extraspecial p-group P and an element xP with xZ(P).

[F1]

CG(x):={gG:gx=xg}={gG:gxg1=x} (The conjugacy class ClG(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 HG, G=[G:H]H (Lagrange's theorem: G=[G:H]H for every subgroup H of a finite group G).

Proof

technique · direct
1.1

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

L1
1.2

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

F1F2L2
2.1

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

L3step 1.1step 1.2

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.

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Sources