Let $G$ be a finite group and $p$ be a prime. Suppose that every proper subgroup of $G$ of order divisible by $p$ is a $p$-group. Let $G$ be a finite group and $p$ be a prime. Suppose that every proper subgroup of $G$ of order divisible by $p$ is a $p$-group. Prove that every two distinct Sylow $p$-subgroup of $G$ intersect trivially.
Thanks in advanced.
 A: (Edited in view of Jack Schmidt's comment): Let $P \in {\rm Syl}_{p}(G)$. Then by hypothesis, $N_{G}(Q)$ is a $p$-subgroup  whenever $1 \neq Q$ is a subgroup of $P$ which is not normal in $G.$ Suppose otherwise, and let $R$ be a subgroup of $P$ which has maximal order subject to being contained in more than one conjugate of $P.$ Then $R <P,$ and $S = N_{P}(R) >R.$ If $R \lhd G,$ and $G$ is not a $p$-group, $G$ has a Sylow $q$-subgroup $U \neq 1$ for some prime $q \neq p.$ Then $RQ$ is a proper subgroup of $G$ (it does not contain $P$), contrary to the hypotheses, as $RQ$ is not a $p$-group.
Hence we may suppose that $R \not \lhd G.$ 
Let $T = N_{G}(R),$ which is a $p$-subgroup of $G.$ Then $T^{g} \leq P$ for some $g \in G.$ Hence $S^{g} \leq P.$ By the maximal choice of $R,$ we must
 have $ g^{-1} \in N_{G}(P) = P.$ Hence $N_{G}(R) \leq P.$ But $R^{x} \leq P$ for some $x \not \in P,$ so similarly $N_{G}(R^{x}) \leq P.$ Again, the maximal choice of $R$ forces $x \in P,$ a contradiction.
A: There is nice criterion you can use here:
Let $G$ be a finite group and let $p$ a prime dividing the order of $G$. The following are equivalent.
(a) $S \cap T=1$ for all $S, T \in Syl_p(G)$
(b) $N_G(P)$ has exactly one Sylow $p$-subgroup for every non-trivial $p$-subgroup $P$ of $G$.
Observe that (b) is almost trivially implied if one assumes that every proper subgroup of $G$ of order divisible by $p$ is a $p$-group: let $P$ be a non-trivial $p$-subgroup of $G$. Then $P$ cannot be normal (if so, let $H$ be a subgroup of order not divisible by $p$. Then $|HP|=|H|.|P|$, which is divisible by $p$, so $HP$ is a $p$-group and $H$ must be trivial, that is $G$ is a $p$-group.). Hence $N_G(P)$ is proper and contains $P$, so $N_G(P)$ is a $p$-group and has exactly one Sylow $p$-subgroup.  
