Characteristic subgroup of Sylow subgroup is weakly closed iff it is normal in every Sylow subgroup containing it

Statement
Suppose $$G$$ is a finite group, $$p$$ is a prime number, $$P$$ is a $$p$$-fact about::Sylow subgroup, and $$K$$ is a fact about::characteristic subgroup of $$P$$. Then, $$K$$ is a fact about::weakly closed subgroup in $$P$$ (relative to $$G$$) if and only if $$K$$ is a normal subgroup in every $$p$$-Sylow subgroup containing it.

p-functor version

 * Equivalence of definitions of weakly closed conjugacy functor

Facts used

 * 1) uses::Weakly closed implies conjugation-invariantly relatively normal in finite group
 * 2) uses::Sylow implies order-conjugate: For a prime $$p$$, any two $$p$$-Sylow subgroups are conjugate.
 * 3) uses::Sylow implies WNSCDIN
 * 4) uses::WNSCDIN implies every normalizer-relatively normal conjugation-invariantly relatively normal subgroup is weakly closed

Proof
Given: A finite group $$G$$, a prime $$p$$, a $$p$$-Sylow subgroup $$P$$ of $$G$$. A characteristic subgroup $$K$$ of $$P$$.

Weakly closed implies normal in every Sylow subgroup containing it
To prove: Given that $$K$$ is weakly closed in $$P$$, and $$Q$$ is a $$p$$-Sylow subgroup of $$G$$ containing $$K$$, $$K$$ is normal in $$Q$$.

Proof (quick version): This follows from facts (1) and (2). By fact (2), $$P$$ and $$Q$$ are conjugate, and by fact (1), $$K$$ is normal in $$Q$$. (Note that this part does not use the assumption that $$K$$ is a characteristic subgroup of $$P$$).

Proof (hands-on):

Normal in every Sylow subgroup containing it implies weakly closed
To prove: Given that $$K$$ is normal in every $$p$$-Sylow subgroup of $$G$$ containing it, $$K$$ is weakly closed in $$P$$. In other words, if $$g \in G$$ is such that $$gKg^{-1} \le P$$, then $$gKg^{-1} = K$$.

Proof (quick version): The proof follows from facts (3) and (4), and the observation (again stemming from fact (2)) that the conjugates of a particular $$p$$-Sylow subgroup are precisely all the $$p$$-Sylow subgroups.

Proof (hands-on):