Kernel of a characteristic action on an abelian group implies strongly image-potentially characteristic

Statement with symbols
Suppose $$H$$ is a subgroup of a group $$G$$, and there exists an abelian group $$V$$, and a homomorphism $$\alpha:G \to \operatorname{Aut}(V)$$ such that $$H$$ equals the kernel of $$\alpha$$ and $$V$$ is a characteristic subgroup of the semidirect product $$V \rtimes G$$.

Then, there exists a group $$K$$ with a surjective homomorphism $$\rho:K \to G$$ such that both the kernel of $$\rho$$ and $$\rho^{-1}(H)$$ are characteristic subgroups of $$K$$.

Facts used

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

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

To prove: There exists a group $$K$$ with a surjective homomorphism $$\rho:K \to G$$ such that both the kernel of $$\rho$$ and $$\rho^{-1}(H)$$ are characteristic subgroups of $$K$$.

Proof: Let $$K = V \rtimes G$$ and $$\rho:K \to G$$ be the quotient map.


 * 1) By assumption, $$V$$, the kernel of $$\rho$$ is characteristic in $$K$$.
 * 2) $$C_K(V)$$ is characteristic in $$K$$: This follows from fact (1).
 * 3) $$C_K(V) = V \times H = \rho^{-1}(H)$$: Since $$V$$ is abelian, the quotient $$K/V$$ acts on $$V$$ by conjugation (fact (2)). In other words, any two elements in the same coset of $$V$$ act the same way on $$V$$ by conjugation. Thus, a coset of $$V$$ centralizes $$V$$ if and only if the element of $$G$$ in the coset centralizes $$V$$, which happens if and only if the element is in $$H$$. Thus, $$C_K(V) = V \rtimes H = V \times H$$ (because the action is trivial) and is equal to $$\rho^{-1}(H)$$.

By step (1), the kernel of $$\rho$$ is characteristic in $$K$$. Combining steps (2) and (3) yields that $$\rho^{-1}(H)$$ is also characteristic in $$K$$, completing the proof.