Kernel of a characteristic action on an abelian group with which it is characteristic in the direct product implies potentially characteristic

Statement
Suppose $$H$$ is a subgroup of a group $$G$$. Suppose there exists an abelian group $$V$$ satisfying the following two conditions:


 * There is a homomorphism $$\alpha:G \to \operatorname{Aut}(V)$$ with kernel $$H$$, such that $$V$$ is a characteristic subgroup of the semidirect product $$V \rtimes G$$.
 * $$H$$ is a characteristic subgroup of $$V \times H$$.

Then, $$H$$ is a potentially characteristic subgroup of $$G$$. In fact, $$H$$ is a characteristic subgroup of $$V \rtimes G$$.

Facts used

 * 1) uses::Characteristicity is centralizer-closed
 * 2) uses::Quotient group acts on abelian normal subgroup
 * 3) uses::Characteristicity is transitive

Proof
Given: A subgroup $$H \le G$$. An abelian group $$V$$ and a homomorphism $$\alpha: G \to \operatorname{Aut}(V)$$ with kernel $$H$$. $$V$$ is characteristic in $$V \rtimes G$$, and $$H$$ is characteristic in $$V \times H$$.

To prove: $$H$$ is characteristic in the semidirect product $$K = V \rtimes G$$.

Proof:


 * 1) $$V$$ is characteristic in $$K$$: This is by assumption.
 * 2) $$C_K(V)$$ is characteristic in $$K$$: This follows from fact (1).
 * 3) $$C_K(V) = V \times H$$: Since $$V$$ is abelian, the quotient group $$K/V \cong G$$ acts on $$V$$ (fact (2)); in particular, any two elements in the same coset of $$V$$ have the same action by conjugation on $$V$$. Thus, the centralizer of $$V$$ comprises those cosets of $$V$$ for which the corresponding element of $$G$$ fixes $$V$$. This is precisely the cosets of elements of $$H$$. Thus, $$C_K(V) = V \rtimes H$$. Since the action is trivial, $$C_K(V) = V \times H$$.
 * 4) $$H$$ is characteristic in $$C_K(V) = V \times H$$: This is again by assumption.
 * 5) $$H$$ is characteristic in $$K$$: This follows from steps (2) and (4) and fact (3).