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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)
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Elementary detection at a fixed element

Statement

If ∣G∣=pnl with p∤l, then l 1G∈IEp(G), where Ep is the family of p-elementary subgroups of G.

Facts & Assumptions

[F1]
[F3]

The integral induction subgroup is an ideal by The induction subgroup is an ideal.

Proof

Given: R is a set of conjugacy-class representatives of p′-elements of G.

1.1givenconstruct

Put A=Z[ζ∣G∣]. For r∈R, choose a Sylow p-subgroup Pr of CG(r) and set Hr=⟨r⟩×Pr. On ⟨r⟩ the delta function χr(c)=∣⟨r⟩∣δc,r lies in AR(⟨r⟩) by Fourier inversion on this cyclic group. Inflate it across Pr to ψr∈AR(Hr) and form the following sum.

ψ=∑r∈RInd⁡HrGψr.

[given, construct]

2.1step 1.1algebra

This lies in the A-scalar extension of IEp(G). Each ψr is integer-valued, so the induction formula makes every value of ψ rational. On the other hand ψ is an A-linear combination of characters, hence all its values are algebraic integers. A rational algebraic integer is an integer, so ψ is integer-valued.

3.1F1step 2.1

If r0∈R, a conjugate of r0 lying in Hr lies in ⟨r⟩. The definition of χr and [F1] therefore give the following value.

(Ind⁡HrGψr)(r0)=δr,r0∣CG(r)∣∣Pr∣.

4.1F2step 2.1step 3.1

Thus ψ(r0) is an integer prime to p. For arbitrary g, its p′-part is conjugate to some r0∈R, so [F2] shows that ψ(g)≡ψ(r0)≢0(modp).

5.1F1step 1.1step 4.1construct

To use the integer-valued function from step 4.1, first let β be any integer-valued class function on G. For each conjugacy-class representative g, put Cg=⟨g⟩ and define θg(c)=∣Cg∣δc,g on Cg. The same cyclic Fourier inversion used in step 1.1 makes θg∈AR(Cg). By [F1], Ind⁡CgGθg is ∣CG(g)∣ on the conjugacy class of g and zero elsewhere. Consequently ∣G∣β=∑[g]β(g)[G:CG(g)]Ind⁡CgGθg. Every cyclic subgroup is p-elementary by splitting its order into coprime p-power and p′ factors, so this identity puts ∣G∣β in the A-scalar extension of IEp(G).

6.1F3step 1.1step 4.1step 5.1algebra

Assume n≥1 and put e=pn−1(p−1). Euler's congruence gives ψ(g)e≡1(modpn) for every g. Thus β=l(ψe−1G)/∣G∣ is an integer-valued class function, and step 5.1 puts l(ψe−1G) in the A-scalar extension of IEp(G). The same is true of lψe by [F3] and step 1.1. Subtraction puts l1G in that scalar extension. If n=0, then l=∣G∣, so step 5.1 applied to β=1G gives the same conclusion directly.

7.1step 6.1algebra∎

The cyclotomic ring A is a finite free Z-module with the power basis 1,ζ∣G∣,…,ζ∣G∣φ(∣G∣)−1. Expand the relation from step 6.1 in this basis and take its coefficient of 1. Since all inducing characters there lie in integral character rings, this yields l1G∈IEp(G), including the case n=0.

Depends on

Used by

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Sources