# Paranormal implies weakly normal

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., paranormal subgroup) must also satisfy the second subgroup property (i.e., weakly normal subgroup)
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## Statement

### 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 $H^g \le H$.

## 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 $H^g \le H$.