Every p-subgroup is conjugate to a p-subgroup whose normalizer in the Sylow is Sylow in its normalizer

Statement
Suppose $$G$$ is a finite group and $$P$$ is a $$p$$-Sylow subgroup of $$G$$. Suppose $$H$$ is a subgroup contained inside $$P$$. Then, there exists a subgroup $$K$$ of $$P$$, such that $$H$$ and $$K$$ are conjugate subgroups inside $$G$$, and such that $$N_P(H)$$ is a Sylow subgroup of $$N_G(H)$$.

Related facts

 * Every p-subgroup is well-placed in some Sylow subgroup

Facts used

 * 1) uses::Sylow subgroups exist
 * 2) uses::Sylow implies order-dominating

Proof
Given: A finite group $$G$$, a $$p$$-Sylow subgroup $$P$$, a subgroup $$H$$ contained inside $$P$$.

To prove: There exists a subgroup $$K$$ of $$P$$ such that $$H$$ and $$K$$ are conjugate subgroups inside $$G$$, and such that ,math>N_P(H) is $$p$$-Sylow inside $$N_G(H)$$.

Proof:


 * 1) Let $$L$$ be a $$p$$-Sylow subgroup of $$N_G(H)$$. Such a $$L$$ exists by fact (1) in the subgroup $$N_G(H)$$.
 * 2) Let $$g \in G$$ be such that $$L \le g^{-1}Pg$$. Note that such a $$g$$ exists by fact (2) in the group $$G$$.
 * 3) Let $$K = gHg^{-1}$$. Then, $$gLg^{-1}$$ is a ,math>p -Sylow subgroup of $$N_G(K)$$. By construction, $$gLg^{-1} \le P$$, so a $$p$$-Sylow subgroup of $$N_G(K)$$ is contained in $$P$$. Thus, $$N_P(K)$$ is $$p$$-Sylow in $$N_G(K)$$.