Weakly closed implies conjugation-invariantly relatively normal in finite group

Statement

Suppose $H \le K \le G$ are groups such that $H$ is a Weakly closed subgroup (?) of $K$ relative to $G$. Then, $H$ is a Conjugation-invariantly relatively normal subgroup (?) of $K$ relative to $G$, viz., $H$ is normal in every conjugate of $K$ in $G$ containing it.

Facts used

1. Weakly closed implies normal in middle subgroup

Proof

Given: Groups $H \le K \le G$ such that $H$ is weakly closed in $K$ with respect to $G$.

To prove: If $g \in G$ is such that $H \le gKg^{-1}$, then $H$ is normal in $K$.

Proof:

1. $g^{-1}Hg \le H$: Since $H \le gKg^{-1}$, we have $g^{-1}Hg \le K$. Now, since $H$ is weakly closed in $K$, we get that $g^{-1}Hg \le H$.
2. $g^{-1}Hg = H$: Since $G$ is finite, and conjugation by $g$ is an automorphism, the sizes of $g^{-1}Hg$ and $H$ are the same. This, along with the previous step, yields $g^{-1}Hg = H$.
3. $H = gHg^{-1}$: This follows from the previous step by conjugating both sides by $g$.
4. $H$ is weakly closed in $gKg^{-1}$: Since $H$ is weakly closed in $K$, and conjugation by $g$ is an automorphism, $gHg^{-1}$ is weakly closed in $gKg^{-1}$. $H = gHg^{-1}$ by the previos step, so $H$ is weakly closed in $gKg^{-1}$.
5. $H$ is normal in $gKg^{-1}$: This follows from the previous step, and fact (1).