Weakly closed implies conjugation-invariantly relatively normal in finite group

Statement
Suppose $$H \le K \le G$$ are groups such that $$H$$ is a fact about::weakly closed subgroup of $$K$$ relative to $$G$$. Then, $$H$$ is a fact about::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) uses::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).