Procharacteristicity is normalizer-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., procharacteristic subgroup) satisfying a subgroup metaproperty (i.e., normalizer-closed subgroup property)
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Statement

The Normalizer (?) of a procharacteristic subgroup of a group is again a procharacteristic subgroup.

Definitions used

Procharacteristic subgroup

Further information: 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 $⟨H,H^{\sigma }⟩$.

Proof

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 $\sigma \in \mathrm{A}\mathrm{u}\mathrm{t}(G)$. Then, there exists $g\in ⟨H,H^{\sigma }⟩$ 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 $g\in ⟨H,H^{\sigma }⟩$ 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, $g\in ⟨N_G(H),N_G(H{)}^{\sigma }⟩$ is such that $N_G(H{)}^{\sigma }=N_G(H{)}^g$, completing the proof.