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Kernel conjugation by an element of the coefficient group corresponds to a principal crossed homomorphism

Statement

Let $G$ act on an abelian group $A$ , let $z : G \to A$ be a crossed homomorphism, and let $a \in A$ . If $z_{a} : G \to A$ is defined by

$z_{a} (g) = z (g) + g \cdot a - a,$

then the graph subgroup $Γ_{z_{a}}$ is the conjugate of $Γ_{z}$ by $(- a, 1)$ in $A ⋊ G$ .

Facts & Assumptions

Given: An action of $G$ on an abelian group $A$ , a crossed homomorphism $z : G \to A$ , and an element $a \in A$ .

[L1]

The principal crossed homomorphism attached to $a$ is $g \mapsto g \cdot a - a$ (Principal crossed homomorphism for abelian coefficients).

[L2]

The graph subgroup of a map is $Γ_{z} = {(z (g), g) : g \in G}$ (The graph subgroup attached to a map into a semidirect product).

[L3]

The semidirect-product multiplication is $(x, g) (y, h) = (x + g \cdot y, g h)$ for abelian coefficients ( The semidirect-product multiplication makes $N \times H$ a group).

Proof

technique · direct

1.1

In $A ⋊ G$ , the inverse of $(- a, 1)$ is $(a, 1)$ . Therefore $(- a, 1) (z (g), g) (a, 1) = (- a + z (g) + g \cdot a, g) = (z_{a} (g), g)$ by [L3] and the definition in [L1].

givenL1L3algebra

2.1

Step 1.1 shows that conjugating each element of $Γ_{z}$ by $(- a, 1)$ produces the corresponding element of $Γ_{z_{a}}$ . Hence $(- a, 1) Γ_{z} (a, 1) = Γ_{z_{a}}$ . So kernel conjugation changes the graph exactly by a principal crossed homomorphism.

L2step 1.1∎

Depends on

Used by

Dependency tree · two levels

7 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

David A. Craven, Finite Group Theory (standard reference, not scraped)