Normalizer of intermediately subnormal-to-normal implies self-normalizing

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., intermediately subnormal-to-normal subgroup) must also satisfy the second subgroup property (i.e., subgroup with self-normalizing normalizer)
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Statement

Suppose $H$ is an intermediately subnormal-to-normal subgroup of a group $G$ . Then, the normalizer $N_{G} (H)$ of $H$ in $G$ is a self-normalizing subgroup of $G$ .

Property-theoretic statement

The property of being an intermediately subnormal-to-normal subgroup is stronger than the property of being a subgroup whose normalizer is a self-normalizing subgroup.

Related facts

Similar facts

Normalizer of pronormal implies abnormal: The assumption of pronormality is stronger, and the conclusion of abnormality is correspondingly stronger.
Normalizer of weakly pronormal implies weakly abnormal: The assumption of weak pronormality is stronger, and the conclusion of weak abnormality is correspondingly stronger.

Converse

The converse is not true: there can exist a subnormal subgroup that is not normal but whose normalizer is self-normalizing. Further information: Abnormal normalizer and 2-subnormal not implies normal

Proof

Given: A group $G$ with an intermediately subnormal-to-normal subgroup $H$ , with normalizer $N_{G} (H)$ .

To prove: $N_{G} (H)$ is a self-normalizing subgroup of $G$ .

Proof: Suppose $K = N_{G} (H)$ and $L = N_{G} (K)$ . Our goal is to show that $L = K$ .

Note first that $H$ is normal in $K$ and $K$ is normal in $L$ . Thus, $H$ is a 2-subnormal subgroup of $L$ . By the assumption about $H$ being intermediately subnormal-to-normal, $H$ is normal in $L$ , so $L \leq N_{G} (H) = K$ , yielding $L = K$ , as desired.