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Transfer cycle decomposition formula

Statement

Let G be a finite group, H≤G, A an abelian group written multiplicatively, φ:H→A a homomorphism, and Vφ:G→A the transfer of Transfer homomorphism for a finite index subgroup, which is independent of the transversal by Transfer is independent of the transversal. Let x∈G and let

H\G=C1⊔⋯⊔Cr

be the decomposition of the finite set of right cosets into the orbits of the right multiplication action of the cyclic subgroup ⟨x⟩≤G (The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups, ⟨g⟩={ gn:n∈Z }, and every cyclic group is abelian). Put ni:=∣Ci∣ and choose ti∈G with Hti∈Ci. Then

Vφ(x)=∏i=1rφ(ti xni ti−1).

Facts & Assumptions

Given: A finite group G, a subgroup H≤G, an abelian group A, a homomorphism φ:H→A, an element x∈G, and the transfer V=Vφ of Transfer homomorphism for a finite index subgroup.

[F1]

For α∈H\G the coset αt=Htαt is defined for every t∈G, the assignment α↦αt is a right action of G on the finite set H\G, each factor tαstαs−1 (s∈G) lies in H, and products of finitely many elements of the abelian group A are independent of the order of the factors (Transfer homomorphism for a finite index subgroup).

[F2]

The transfer V(x)=∏α∈H\Gφ(tαxtαx−1) does not depend on the transversal (Transfer is independent of the transversal).

[F3]

If u,v∈G satisfy Hu=Hv then uv−1∈H, and conversely (x∈aH iff a−1x∈H, and aH=bH iff a−1b∈H).

[F7]

Every nonempty set of positive integers has a least element, and for integers k and n≥1 there are q,r∈Z with k=qn+r and 0≤r<n (The well-ordering principle, Division with remainder in Z: for a∈Z and b>0 there are unique q,r∈Z with a=qb+r and 0≤r<b).

Proof

technique · direct
1.1

Fix α∈H\G. Since H\G is finite, the elements α,αx,αx2,… cannot all be distinct, so αxi=αxj for some 0≤i<j; applying the inverse permutation α↦αx−i (which is the action of x−i) gives αxj−i=α, an equality of the form αxn=α with n=j−i≥1. By [F7] there is a least positive integer n(α) with αxn(α)=α.

F1F5F7algebra
2.1

For every k∈Z one has αxk=α if and only if n(α)∣k: if k=qn(α) then αxk=α by [F5] and step 1.1 (including q<0, since αx−n(α)=α as well), while for arbitrary k writing k=qn(α)+r with 0≤r<n(α) by [F7] gives αxr=α, so minimality forces r=0.

F5F7step 1.1
3.1

The orbit of α under ⟨x⟩ is C(α)={αxk:k∈Z}, and by step 2.1 it equals {α,αx,…,αxn(α)−1}, whose n(α) elements are pairwise distinct. In particular αxn(α)=α and ∣C(α)∣=n(α).

step 2.1
4.1

The orbit decomposition H\G=C1⊔⋯⊔Cr of [F6] is therefore a decomposition into finitely many sets Ci each of the form Ci={αi,αix,…,αixni−1} with ni=∣Ci∣ and αixni=αi; write αi=Hti.

F6step 3.1given
5.1

The rule tαixj:=tixj for 0≤j<ni defines a transversal of H\G: indeed every coset of H\G lies in exactly one Ci and hence equals exactly one αixj, and tixj represents it, because Htixj=αixj by [F1].

F1step 4.1
6.1

Because V is independent of the transversal by [F2], it may be computed with the transversal of step 5.1: V(x)=∏i=1r∏j=0ni−1φ(tαixj x t(αixj)x−1), where α:=αixj and αx=αixj+1.

F1F2step 5.1
7.1

For 0≤j<ni−1 one has t(αixj)x=tαixj+1=tixj+1=tixjx=tαixjx, so the corresponding factor equals φ(tαixjx(tαixjx)−1)=φ(e)=1 by [F4].

F4step 6.1algebra
7.2

For j=ni−1 one has αixj+1=αixni=αi by step 4.1, so t(αixni−1)x=tαi=ti and the corresponding factor equals φ(tixni−1xti−1)=φ(tixniti−1), an element of H because Htixni=Hti by step 4.1 and [F3].

F3step 4.1step 6.1algebra
8.1

Multiplying the contributions of steps 7.1 and 7.2 over all i and j, all factors with j<ni−1 are 1 and the remaining one for each i is φ(tixniti−1); hence V(x)=∏i=1rφ(tixniti−1). ∎

step 7.1step 7.2given

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