Let $G$ be a finite group and $P$ be a Sylow $p$-subgroups of $G$. Let $H$ be a subgroup of $G$ such that $N_G(P)\subset H$. Prove that $N_G(H)=H$

Question

1.Let $G$ be a finite group and $P$ be a Sylow $p$-subgroups of $G$. Let $H$ be a subgroup of $G$ such that $N_G(P)\subset H$. Prove that $N_G(H)=H$
2.Let $G$ be a group of order $143$. Show that Sylow 11-subgroup of $G$ is unique. Also, show that $G$ is cyclic.

Answer 1

Hint for #2: Consider the prime factorization and use a theorem that says $G=HK$, $H$ and $K$ are normal in $G$ and $H\cap K=\{e\}$, then $G\cong H\times K$. Now use another property that says when this is the case, if the factors have relatively prime orders, the group is cyclic.

Answer 2

(2) has been solved, so here is (1).
I am going to assume known the argument of Frattini.
Since $P\subseteq H$, $P$ is a Sylow-p-subgroup of $H$. Since $H$ is normal in $N_G(H)$, we can apply Frattini argument to $N_G(H)$ and $H$, and obtain that $N_G(H)=HN_G(P)=H$.
Point it out if some error occurs, thanks.
P.S.:
The statement of Frattini argument is: Let $H$ be a normal subgroup of $G$, and $P$ a Sylow $p$-subgroup of $H$. then $G=N_G(P)H$.
Proof:
Let $G$ act on $H$ by conjugation, which is valid since $H$ is normal in $G$. Then for any $g\in G$, $gPg^{-1}$ is a Sylow $p$-subgroup of $H$, and hence, by one of the theorems of Sylow, there is $h\in H$ such that $gPg^{-1}=hPh^{-1}$, that is to say, $h^{-1}g\in N_G(P)$, so $g\in hN_G(P)\subseteq HN_G(P)$. Q.E.D.