A $p$-subgroup of a finite group is either a Sylow $p$-subgroup or properly contained in a Sylow $p$-subgroup of its normalizer This exercice is from Aschbacher's book "Finite group theory".
I am not asking for a complete solution, just for a hint.
Here is a partial solution, when the ambient group $G$ is a $p$-group.
If $X$ is a subgroup contradicting the exercise, then $X < G$ and $X = N_G(X)$.
However, one can easily that this cannot happen in a $p$-group (by induction on the order of $G$).
 A: This can be easily seen by applying the following lemma.
Lemma Let $P$ be a $p$-subgroup of $G$. Then $$P \in Syl_p(G) \iff P \in Syl_p(N_G(P)).$$
Proof If $P \in Syl_p(G)$, then, since $P \subseteq N_G(P)$, we obviously have $P \in Syl_p(N_G(P))$. So assume $P \in Syl_p(N_G(P))$. Then $|N_G(P):P|$ is not divisible by $p$. By Sylow Theory, we can find an $S \in Syl_p(G)$ with $P \subseteq S$. Hence $P \subseteq S \cap N_G(P)=N_S(P) \subseteq N_G(P)$. By comparing indices in these last inclusions and noting that $N_S(P)$ is a $p$-subgroup, we get $P=N_S(P)$. We now apply the normalizers grow principle (see I.M. Isaacs Finite Group Theory, Theorem 1.22 here): if $P \subsetneq S$, then we would have $P \subsetneq N_S(P)$. This is a contradiction, so $P=S \in Syl_p(G)$.
Now back to your/Aschbacher's problem: if $P$ is a Sylow $p$-subgroup of $G$, then we are done, If not, then by the Lemma, the $p$-subgroup $P$ is not a Sylow $p$-subgroup of $N_G(P)$. But by Sylow Theory (applied to $N_G(P)$!) $P$ is contained and thus properly contained in a Sylow $p$-subgroup of $N_G(P)$.
