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Artin transfer (group theory)

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Transversals of a subgroup

Let G be a group and H≤G be a subgroup of finite index n.

Definitions. A left transversal of H in G is an ordered system (g1,…,gn) of representatives for the left cosets of H in G such that

G=⨆i=1ngiH.

Similarly a right transversal of H in G is an ordered system (d1,…,dn) of representatives for the right cosets of H in G such that

G=⨆i=1nHdi.

Remark. For any transversal of H in G, there exists a unique subscript 1≤i0≤n such that gi0∈H, resp. di0∈H. Of course, this element with subscript i0 which represents the principal coset (i.e., the subgroup H itself) may be, but need not be, replaced by the neutral element 1.

Lemma. Let G be a non-abelian group with subgroup H. Then the inverse elements (g1−1,…,gn−1) of a left transversal (g1,…,gn) of H in G form a right transversal of H in G. Moreover, if H is a normal subgroup of G, then any left transversal is also a right transversal of H in G.

Proof. Since the mapping x↦x−1 is an involution of G we see that:
G=G−1=⨆i=1n(giH)−1=⨆i=1nH−1gi−1=⨆i=1nHgi−1.
For a normal subgroup H we have xH=Hx for each x∈G.

We must check when the image of a transversal under a homomorphism is also a transversal.

Proposition. Let ϕ:G→K be a group homomorphism and (g1,…,gn) be a left transversal of a subgroup H in G with finite index n. The following two conditions are equivalent:

  • (ϕ(g1),…,ϕ(gn)) is a left transversal of the subgroup ϕ(H) in the image ϕ(G) with finite index (ϕ(G):ϕ(H))=n.
  • ker⁡(ϕ)≤H.
Proof. As a mapping of sets ϕ maps the union to another union:
ϕ(G)=ϕ(⋃i=1ngiH)=⋃i=1nϕ(giH)=⋃i=1nϕ(gi)ϕ(H),
but weakens the equality for the intersection to a trivial inclusion:
∅=ϕ(∅)=ϕ(giH∩gjH)⊆ϕ(giH)∩ϕ(gjH)=ϕ(gi)ϕ(H)∩ϕ(gj)ϕ(H),i≠j.
Suppose for some 1≤i≤j≤n:
ϕ(gi)ϕ(H)∩ϕ(gj)ϕ(H)≠∅
then there exists elements hi,hj∈H such that
ϕ(gi)ϕ(hi)=ϕ(gj)ϕ(hj)
Then we have:
ϕ(gi)ϕ(hi)=ϕ(gj)ϕ(hj)⟹ϕ(gj)−1ϕ(gi)ϕ(hi)ϕ(hj)−1=1⟹ϕ(gj−1gihihj−1)=1⟹gj−1gihihj−1∈ker⁡(ϕ)⟹gj−1gihihj−1∈Hker⁡(ϕ)≤H⟹gj−1gi∈Hhihj−1∈H⟹giH=gjH⟹i=j
Conversely if ker⁡(ϕ)⊈H then there exists x∈G∖H such that ϕ(x)=1. But the homomorphism ϕ maps the disjoint cosets x⋅H∩1⋅H=∅ to equal cosets:
ϕ(x)ϕ(H)∩ϕ(1)ϕ(H)=1⋅ϕ(H)∩1⋅ϕ(H)=ϕ(H).

Remark. We emphasize the important equivalence of the proposition in a formula:

(1)ker⁡(ϕ)≤H⟺{ϕ(G)=⨆i=1nϕ(gi)ϕ(H)(ϕ(G):ϕ(H))=n

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