Let $G$ be a finite group and p a prime that divides $|G|$. Let $n_{p}$ denote the number of Sylow p-subgroups of $G$. Prove that if $n_{p}\neq 1\pmod{p^{2}}$, then for any Sylow p-subgroup $P_{1}$ of $G$ there is a Sylow p-subgroup $P_{2}$ of $G$ such that $P_{1}\cap P_{2}$ is of index $p$ in both $P_{1}$ and $P_{2}$. Hint: If P is a Sylow p-subgroup of $G$ and $Q$ is any p-subgroup of $G$, then $Q\cap N_{G}(P)=Q\cap P$.
This is an old algebra qualifying exam question. I couldn't figure out how the hint would help me with this question. I know that any two Sylow p-subgroups intersect trivially, then $|P_{1}\cap P_{2}|=1$. Since any two Sylow p-subgroups are conjugates, $|P_{1}|=|P_{2}|$. So if $P_{1}\cap P_{2}$ has index p in $P_{1}$, it means that $|P_{1}|=p$. Now the question is how $n_{p}\neq 1\pmod{p^{2}}$ leads to $|P_{1}|=|P_{2}|=p$?
We prove the contrapositive:
Suppose that for any distinct Sylow $p-$subgroups $P_i$, $P_1\cap P_2$ is of index greater than $p$ in $P_i$. Then consider the conjugation action of a fixed Sylow $p-$subgroup $P_1$ on the collection of all Sylow $p-$subgroups of $G$.
By the hint, for any $P\not=P_1$, $\operatorname{Stab}{P}=N_G(P)\cap P_1 = P_1\cap P_2$, so by the orbit stabilizer theorem, for each $P \not= P_1$, $|\operatorname{Orb}(P)|=|P_1:P_1\cap P|$. Then $p^2$ divides the size of this orbit, since the index is a power of $p$, but it isn't $1$ (because $P_1 \not= P$) and it isn't $p$ (by the hypothesis), so it must be at least $p^2$.
Thus the orbits consist of $\{P_1\}$ and then $k$ distinct orbits, each of size divisible by $p^2$. Since $n_p$ is the sum of the sizes of the orbits, we see that $n_p = 1 \pmod{p^2}$.
Remark: You should also note that Sylow $p-$subgroups need not intersect trivially, consider the Sylow $2-$subgroups of $D_{12}$, $\langle r^3, s\rangle, \langle r^3, rs\rangle, \langle r^3, r^2s\rangle$.