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Transfer is a homomorphism

Statement

Let $G$ be a finite group, $H \leq G$ , $A$ an abelian group written multiplicatively, $φ : H \to A$ a homomorphism, and $V_{φ} : G \to A$ the transfer of Transfer homomorphism for a finite index subgroup, formed with a transversal ${t_{α}}$ of the right cosets $H \ G$ (the result is independent of the transversal by Transfer is independent of the transversal). Then

$V_{φ} (x y) = V_{φ} (x) V_{φ} (y) for all x, y \in G;$

that is, the transfer is a group homomorphism $G \to A$ .

Facts & Assumptions

Given: A finite group $G$ , a subgroup $H \leq G$ , an abelian group $A$ , a homomorphism $φ : H \to A$ , a transversal ${t_{α}}_{α \in H \ G}$ and the transfer $V (x) = \prod_{α \in H \ G} φ (t_{α} x t_{α x}^{- 1})$ of Transfer homomorphism for a finite index subgroup.

[F1]

The assignment $α \mapsto α x = H t_{α} x$ is a right action of $G$ on the finite set $H \ G$ , so $α (x y) = (α x) y$ and $α \mapsto α x$ is a permutation of $H \ G$ with inverse $α \mapsto α x^{- 1}$ ; furthermore $t_{α} x t_{α x}^{- 1} \in H$ (Transfer homomorphism for a finite index subgroup).

[F2]

The transfer does not depend on the transversal (Transfer is independent of the transversal).

[F3]

$φ (u v) = φ (u) φ (v)$ for all $u, v \in H$ , and $φ$ is defined on all of $H$ (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed, Monoid homomorphism and group homomorphism).

[F4]

The product of finitely many elements of the abelian group $A$ is independent of the order of the factors, and $H \ G$ is finite (Transfer homomorphism for a finite index subgroup, Left and right cosets $g H$ and $H g$ of a subgroup).

Proof

technique · direct

1.1

For $x, y \in G$ and $α \in H \ G$ , the identity $t_{α} x y t_{α x y}^{- 1} = (t_{α} x t_{α x}^{- 1}) (t_{α x} y t_{α x y}^{- 1})$ holds: the middle factor $t_{α x}^{- 1} t_{α x}$ cancels, and $(α x) y = α (x y)$ by [F1].

F1algebra

2.1

Both bracketed factors of step 1.1 lie in $H$ by [F1], so applying $φ$ gives $φ (t_{α} x y t_{α x y}^{- 1}) = φ (t_{α} x t_{α x}^{- 1}) φ (t_{α x} y t_{α x y}^{- 1})$ by [F3].

F3step 1.1

3.1

Hence $V (x y) = \prod_{α} φ (t_{α} x t_{α x}^{- 1}) φ (t_{α x} y t_{α x y}^{- 1})$ , the product of the two factors over $α$ ; since $A$ is abelian this equals $(\prod_{α} φ (t_{α} x t_{α x}^{- 1})) (\prod_{α} φ (t_{α x} y t_{α x y}^{- 1}))$ by [F4].

F4step 2.1

4.1

The reindexing $β : = α x$ runs over $H \ G$ as $α$ does, by the permutation property in [F1], so $\prod_{α} φ (t_{α x} y t_{α x y}^{- 1}) = \prod_{β} φ (t_{β} y t_{β y}^{- 1}) = V (y)$ .

F1step 3.1

5.1

Since $V$ does not depend on the chosen transversal by [F2], the value $V (x)$ is well defined for every $x \in G$ ; combining steps 3.1 and 4.1, $V (x y) = V (x) V (y)$ for all $x, y \in G$ , so $V$ is a homomorphism. ∎

F2step 3.1step 4.1

Depends on

Used by

Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Paul Flavell, An Introduction to Transfer and Fusion in Finite Groups, §§2–5 (standard reference, not scraped)
Hans Kurzweil and Bernd Stellmacher, The Theory of Finite Groups, §§7.1–7.2 (standard reference, not scraped)