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Transfer is independent of the transversal

Statement

Let G be a finite group, let H≤G be a subgroup, let A be an abelian group written multiplicatively, and let φ:H→A be a homomorphism (Transfer homomorphism for a finite index subgroup). Let {tα}α∈H\G and {tα′}α∈H\G be two choices of representatives of the right cosets of H in G, and let

V(x):=∏α∈H\Gφ(tα x tαx−1),V′(x):=∏α∈H\Gφ(tα′ x tαx′−1)

be the two products formed from them, where αx:=Htαx. Then V(x)=V′(x) for every x∈G. In particular the transfer Vφ is a well-defined function G→A depending only on φ.

Facts & Assumptions

Given: A finite group G, a subgroup H≤G, an abelian group A, a homomorphism φ:H→A, and two transversals {tα}, {tα′} of the right cosets α∈H\G as in Transfer homomorphism for a finite index subgroup.

[F1]

For every α the coset αx=Htαx is a right coset of H, the assignment α↦αx is a permutation of H\G, each tαxtαx−1 lies in H, and the finite product ∏α∈H\Gaα of elements of the abelian group A is independent of the order of its factors (Transfer homomorphism for a finite index subgroup).

[F2]

If u,v∈G represent the same right coset of H, that is Hu=Hv, then uv−1∈H; conversely uv−1∈H implies Hu=Hv (x∈aH iff a−1x∈H, and aH=bH iff a−1b∈H, Left and right cosets gH and Hg of a subgroup).

[F4]

The map α↦αx is a bijection of the finite set H\G, so a product indexed by H\G may be reindexed along it (The coset set G/H and the index [G:H] of a subgroup, Transfer homomorphism for a finite index subgroup).

Proof

technique · direct
1.1

For every α∈H\G one has tα′=hαtα for some hα∈H: both tα′ and tα represent the coset α, so tα′tα−1∈H by [F2], and hα:=tα′tα−1 is the desired element.

F2given
2.1

For x∈G and α∈H\G the product tα′xtαx′−1 equals hα tαx tαx−1 hαx−1, because tα′=hαtα and tαx′=hαxtαx by step 1.1; hence φ(tα′xtαx′−1)=φ(hα) φ(tαxtαx−1) φ(hαx)−1 by [F3].

F3step 1.1algebra
2.2

The reindexing α↦αx is a bijection of H\G by [F4], so ∏αφ(hαx)−1=∏βφ(hβ)−1=(∏βφ(hβ))−1 by [F3].

F3F4step 1.1
3.1

Consequently V′(x)=(∏αφ(hα))(∏αφ(tαxtαx−1))(∏αφ(hαx)−1): the product over α of the three factors of step 2.1 may be rearranged because A is abelian, by [F1].

F1step 2.1
4.1

Therefore V′(x)=(∏αφ(hα)) V(x) (∏αφ(hα))−1=V(x), the two outer factors cancelling because multiplication in the abelian group A commutes.

F1step 3.1step 2.2algebra
5.1

Since x∈G was arbitrary, V′=V; the transfer is therefore independent of the choice of transversal. ∎

step 4.1given

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