My task is to show that if $G$ is a group with order $180 =2^23^25$ with $36$ Sylow $5$-subgroups, then there are two Sylow $3$-subgroups $H$ and $K$ such that $|H \cap K| = 3$. The number of Sylow $3$ groups $n_3$ is either $1$, $4$, or $10$ since $ n_3 \equiv 1 \pmod 3$ and $n_3 | 2^2\cdot5$. However from here I don't know how to force two subgroups to intersect by mere pigeonholeing.
It turns out from this post that $G$ is not simple, so either $n_2=1$ or $n_3=1$, but I don't see how to use this either . In particular we get a normal $5$-complement.
Best Answer
Let $P_5$ be a Sylow 5-subgroup. Since $n_5 = 36$ and $36 \cdot 5 = 180$, we conclude that $N(P_5) = P_5$.
If $n_3 = 10$ and any two Sylow 3-subgroups intersect in the identity, then that is too many elements. So there must exist two Sylow 3-groups whose intersection has order 3.
If $n_3 = 4$, $|N(P_3)| = 45$ so $N(P_3)$ is the semidirect product of a normal subgroup of order 9 with a subgroup of order 5. But a group of order 9 cannot have automorphisms of order 5, so the semidirect product is a direct product. This contradicts $N(P_5) = P_5$.
If $n_3 = 1$, then $P_3$ is normal, and again because $P_3$ has no automorphisms of order 5, then the group $P_3 P_5$ is a direct product which contradicts $N(P_5) = P_5$.