Paranormal implies weakly normal

Verbal statement
A paranormal subgroup of a group must be a weakly normal subgroup.

Statement with symbols
Suppose $$H$$ is a subgroup of $$G$$ such that $$H$$ is contranormal in $$\langle H, H^g \rangle$$ for any $$g \in G$$. Then, if $$H^g \le N_G(H)$$, we have $$H^g \le H$$.

Related facts

 * Paranormal implies weakly closed in intermediate nilpotent

Proof
Given: A group $$G$$. A subgroup $$H$$ such that $$H$$ is contranormal in $$\langle H, H^g \rangle$$ for any $$g \in G$$.

To prove: If $$H^g \le N_G(H)$$, then $$H^g \le H$$.

Proof: Since $$H^g \le N_G(H)$$, we have $$\langle H, H^g \rangle \le N_G(H)$$. In particular, $$H$$ is normal in $$\langle H, H^g \rangle$$. On the other hand, by paranormality, we have that the normal closure of $$H$$ in $$\langle H, H^g \rangle$$ is $$\langle H, H^g \rangle$$. This forces $$H = \langle H, H^g \rangle$$, so $$H^g \le H$$.