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

Statement with symbols
Suppose $$G$$ is a group and $$p$$ is a prime number such that $$G$$ has no nontrivial abelian normal $$p$$-subgroup. Suppose $$H$$ is a $$p$$-divisible normal subgroup of $$G$$: in other words, every element of $$H$$ is the $$p^{th}$$ power of some element of $$H$$. Then, $$H$$ is a fact about::potentially characteristic subgroup of $$G$$: there exists a group $$K$$ containing $$G$$ such that $$H$$ is characteristic in $$K$$.

Similar facts

 * Finite NPC theorem
 * Finite NIPC theorem
 * No nontrivial abelian normal p-subgroup implies every normal subgroup is 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
 * 4) uses::Characteristicity is transitive

Proof
Given: A finite group $$G$$, a prime $$p$$ such that no element of $$G$$ has order $$p$$. A normal subgroup $$H$$ of $$G$$ such that every element of $$H$$ has a $$p^{th}$$ root.

To prove: There exists a group $$K$$ containing $$G$$ such that $$H$$ is 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 automorphism group of a vector space over the field of $$p$$ elements (the dimension of the vector space is $$|L|$$). Thus, $$L$$ has a faithful representation on a vector space $$V$$ 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.
 * 3) $$V$$ is characteristic in $$K$$: $$V$$ is an abelian normal $$p$$-subgroup. By assumption, $$K/V \cong G$$ has no nontrivial abelian normal $$p$$-subgroup, so $$V$$ is the unique largest abelian normal $$p$$-subgroup. Hence, $$V$$ is a characteristic subgroup of $$K$$.
 * 4) $$C_K(V)$$ is characteristic in $$K$$: This follows from the previous step and fact (2).
 * 5) $$C_K(V) = V \times H$$: Since $$V$$ is abelian, the quotient group $$K/V \cong G$$ 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, $$C_K(V) = V \times H$$.
 * 6) $$H$$ is characteristic in $$V \times H$$: $$H$$ is a normal subgroup of $$V \times H$$, on account of being a direct factor. Further, by the assumption that every element of $$H$$ is a $$p^{th}$$ power, $$H$$ is precisely the set of $$p^{th}$$ powers in $$V \times H$$.
 * 7) $$H$$ is characteristic in $$K$$: By steps (4) and (5), $$V \times H$$ is characteristic in $$K$$, and by step (6), $$H$$ is characteristic in $$V \times H$$. Thus, by fact (4), $$H$$ is characteristic in $$K$$.