Sylow $3$-subgroups of an order $180$ group 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.

Since no group of order 180 has $n_5=36$, any conclusion follows vacuously.  The normal 5-complement would give a nontrivial normal subgroup of an $A_5$ composition factor, (the only candidate for a non-abelian composition factor) so if it existed the group would have to be solvable.  But then the counting portion of Philip Hall's theorem would require we could write 36 as a product of prime powers EACH congruent to 1 mod 5, which we can't.  
