Let $P\in{\rm Syl}_p(G)$ and $K$ a subgroup of $G$ containing $N_G(P)$. Show that $N_G(K)=K$ Question:

Let $P\in{\rm Syl}_p(G)$ and $K$ a subgroup of $G$ containing $N_G(P)$. Show that $N_G(K)=K$

I'm wanting to use Frattini, but it looks like I need to use Frattini on $N_G(K)$.  Certainly, $K\trianglelefteq N_G(K)$, so I have the normal subgroup.  So now I need to find a Sylow $p$-subgroup of $K$.  Since we are given that $K$ is a subgroup of $G$ containing $N_G(P)$, then we can say that $P\subseteq K$.  So by Frattini, we'd have $N_G(K)=KN_G(P)=K$, as $N_G(P)$ is contained in $K$, by hypothesis, correct?
 A: Frattini's Srgument says: if $H$ is a normal subgroup of $G$, and $P$ is a Sylow subgroup of $H$, then $G=N_G(P)H$.
Now, you are taking the big overgroup to be $M=N_G(K)$; you say: $K$ is normal in $M$, and $P$ is a Sylow subgroup of $K$ (both are correct assertions). But then you say that Frattini's Argument gives that $N_G(K)=KN_G(P)$. That's not correct (or at least, you are missing several steps), because you are taking the normalizer of $P$ in $G$, when you should be taking it in $M=N_G(K)$. That is, the conclusion you get is
$$N_G(K) = M = KN_M(P),$$
not $N_G(K) = KN_G(P)$.
Now, not all is lost, but you are missing steps. We have $N_G(K) = KN_M(P)$. What is $N_M(P)$? It is the set of all elements of $M$ that normalize $P$. Since $M$ contains $K$, and $K$ contains $N_G(P)$, then we may conclude that  $N_G(P)=N_M(P)$, so now we can replace $N_M(P)$ in the equality we get from the Frattini argument to get
$$N_G(K) = M = KN_M(P) = KN_G(P) = K,$$
the desired conclusion. I would want to see that extra step related to $N_M(P)$ explicit, to make sure there isn't a misapplication of Frattini's Argument.
