Normalizer of pronormal implies abnormal

Verbal statement
The fact about::normalizer of a pronormal subgroup of a group is an fact about::abnormal subgroup.

Statement with symbols
Suppose $$H$$ is a pronormal subgroup of $$G$$. Then, the normalizer $$N_G(H)$$ is an abnormal subgroup of $$G$$.

Pronormal subgroup
A subgroup $$H$$ of a group $$G$$ is termed pronormal if for any $$g \in G$$, there exists $$x \in \langle H, H^g \rangle$$ such that $$H^x = H^g$$. Here, $$H^g := g^{-1}Hg$$.

Abnormal subgroup
A subgroup $$H$$ of a group $$G$$ is termed abnormal if for any $$g \in G$$, $$g \in \langle H, H^g \rangle$$.

Related facts

 * Pronormal and subnormal implies normal

Corollaries

 * Sylow-normalizer implies abnormal: The normalizer of a Sylow subgroup is an abnormal subgroup. In particular, it is self-normalizing, and any subgroup containing it is also self-normalizing.

Proof
Given: Group $$G$$, pronormal subgroup $$H$$, $$K = N_G(H)$$.

To prove: For any $$g \in G$$, $$g \in \langle K, K^g \rangle $$.

Proof: By the definition of pronormality, there exists $$x \in \langle H, H^g \rangle$$ such that $$H^g = H^x$$. Thus, $$H^{gx^{-1}} = H$$, so $$gx^{-1} \in N_G(H)$$, hence $$g \in N_G(H)x = Kx$$. We know that $$x \in \langle H, H^g \rangle \le \langle K, K^g \rangle$$, so $$g \in \langle K, K^g \rangle$$.