No nontrivial abelian normal p-subgroup for some prime p implies every normal subgroup is strongly image-potentially characteristic

Statement
Suppose $$G$$ is a group and $$p$$ is a prime number such that $$G$$ does not contain any nontrivial abelian normal $$p$$-subgroup (a group where the order of every element is a power of $$p$$). Suppose $$H$$ is a normal subgroup. Then, $$H$$ is a strongly image-potentially characteristic subgroup of $$G$$. In other words, there exists a surjective homomorphism $$\rho:K \to G$$ such that both the kernel of $$\rho$$ and the group $$\rho^{-1}(H)$$ are characteristic subgroups of $$K$$.

Related facts

 * No nontrivial abelian normal p-subgroup for some prime p implies every p-divisible normal subgroup is potentially characteristic
 * Finite NIPC theorem
 * Finite NPC theorem

Generalizations

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

Facts used

 * 1) uses::Cayley's theorem
 * 2) uses::Characteristicity is centralizer-closed
 * 3) uses::Quotient group acts on abelian normal subgroup

Proof
Given: A finite group $$G$$, a normal subgroup $$H$$ of $$G$$.

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

Proof:


 * 1) Let $$L = G/H$$. By fact (1), $$L$$ is a subgroup of the symmetric group $$\operatorname{Sym}(L)$$, which in turn can be embedded in the general linear group $$GL(n,p)$$ where $$n = |L|$$. Thus, $$L$$ has a faithful representation on a vector space $$V$$ of dimension $$|L|$$ (possibly infinite) over the prime field of order $$p$$.
 * 2) Since $$L = G/H$$, a faithful representation of $$L$$ on $$V$$ gives a representation of $$G$$ on $$V$$ whose kernel is $$H$$. Let $$K$$ be the semidirect product $$V \rtimes G$$ for this action, with $$\rho:K \to G$$ the quotient map.
 * 3) $$V$$ (the kernel of $$\rho$$) is characteristic in $$K$$: $$V$$ is an abelian normal $$p$$-subgroup. Further, since $$K/V \cong G$$ has no nontrivial abelian normal $$p$$-subgroups, $$V$$ is the unique largest abelian normal $$p$$-subgroup. Hence, it is characteristic.
 * 4) $$C_K(V)$$ is characteristic in $$K$$: This follows from the previous step and fact (2).
 * 5) $$C_K(V) = V \times H = \rho^{-1}(H)$$: Since $$V$$ is abelian, the quotient group $$K/V$$ acts on $$V$$ (fact (3)); 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, $$C_K(V) = V \times H = \rho^{-1}(H)$$.

The last two steps show that $$\rho^{-1}(H)$$ is characteristic in $$K$$, while step (3) shows that the kernel of $$\rho$$ is characteristic in $$K$$. This completes the proof.