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

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.

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.

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 $$L \le N_G(H) = K$$, yielding $$L = K$$, as desired.