Why does a p-subgroup of a group have to lie inside the normalizer of a p-sylow subgroup? 
Prove that a p-subgroup is always contained within a p-sylow subgroup of a group $G$.

Lang's Algebra mentions this fact on pg 35. However, he starts the proof by assuming that every p-subgroup at least lies within the normalizer. Of the p-sylow subgroup. I do not understand why that is. Any help would be greatly appreciated. 
 A: Let $P$ be a $p$-subgroup, let $S$ be a $p$-Sylow subgroup, and let $C(S)$ denote the set of conjugates of $S$ in $G$. Note that $|C(S)|= [G:N_G(S)]$, since $G$ acts transitively on the set of conjugates and the stabilizer of $S$ is precisely $N_G(S)$. Moreover,  $S\subset N_G(S)$, so by Lagrange's theorem $|S|$ divides $|N_G(S)|$ and $p$ does not divide $[G:N_G(S)]$. 
Consider the induced action of $P$ on $C(S)$ by conjugation. Let $F$ denote the set of fixed points of this action, and let $O_1, \dots, O_n$ denote the set of orbits containing more than one element. Note that $|O_i| = [P:\operatorname{Stab}_P(a_i)]$ for any $a_i \in O_i$, and so $|O_i|$ is a positive power of $p$. Since orbits partition the set of conjugates, we obtain:
$$[G:N_G(S)]=|F| + |O_1| + ... + |O_n|.$$
Observe that $p$ does not divide the left hand side, but divides all summands except $|F|$ on the right. Hence $F$ cannot be empty. Since elements of $F$ are precisely conjugates of $S$ containing $P$ in their normalizer, we can replace $S$ with a conjugate if necessary and assume that $P \subset N_G(S)$.
I haven't read Lang's book, nor do I have it on hand-- so it might very well be that this result follows more easily from some previous result of his.
As a side note, the proof that I recall of the fact that every $p$-subgroup is contained in a Sylow $p$-subgroup does not use the fact that every $p$-group is contained in the normalizer. Instead, one considers the action of a $P$ on the set of cosets $G/S$ by left translation, and shows (using a nearly identical argument to the one given above) that this action has a fixed point. Hence for some  $g \in G$, $h(gS)=gS$ for all $h \in P$. This implies $P \subset gSg^{-1}$.
