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 $\langle H,H^{\sigma }\rangle$.

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, $H\leq N_{G}(H)$ and so $H^{\sigma }\leq N_{G}(H)^{\sigma }$. 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.