Procharacteristicity is normalizer-closed

Statement
The fact about::normalizer of a procharacteristic subgroup of a group is again a procharacteristic subgroup.

Procharacteristic subgroup
(This definition uses the right-action convention).

A subgroup $$H$$ of a group $$G$$ is termed a procharacteristic subgroup of $$G$$ if, for any automorphism $$\sigma$$ of $$G$$, $$H$$ and $$H^\sigma$$ are conjugate subgroups inside the subgroup $$\langle H, H^\sigma \rangle$$.

Proof
(This proof uses the right-action convention).

Given: A group $$G$$ a procharacteristic subgroup $$H$$ with normalizer $$N_G(H)$$.

To prove: $$N_G(H)$$ is also a procharacteristic subgroup of $$G$$.

Proof: Pick $$\sigma \in \operatorname{Aut}(G)$$. Then, there exists $$g \in \langle H, H^\sigma \rangle$$ such that $$H^g = H^\sigma$$. Note that $$N_G(H)^\sigma = N_G(H^\sigma) = N_G(H^g) = N_G(H)^g$$. Thus, we have $$g \in \langle H, H^\sigma \rangle$$ such that $$N_G(H)^\sigma = N_G(H)^g$$. Further, $$H \le N_G(H)$$ and so $$H^\sigma \le N_G(H)^\sigma$$. Thus, $$g \in \langle N_G(H), N_G(H)^\sigma \rangle$$ is such that $$N_G(H)^\sigma = N_G(H)^g$$, completing the proof.