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.
 A: 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 $N(P_5) = P_5$.
A: The hard part is to observe that there exist two different Sylow-3 subgroups. Indirect assumption: there is only one (which is a normal subgroup, of course). Denote it by $P_9$. 
If $P_9$ is normal, then any 5-Sylow $P_5$ normalizes it: so $P_5$ acts on $P_9$ by conjugation. This action must be trivial: there are two groups of order $9$, and none of the two has an automorphism of order $5$. (Check this!) 
But then elements of $P_5$ and $P_9$ commute, so $P_9$ is contained in the normalizer of $P_5$. But this is nonsense: the normalizer of $P_5$ is $P_5$ itself, as its index is 36 (by Sylow's theorem, this index is the number of $5$-Sylows). 
The $36$ Sylow-5 subgroups provide $144$ elements of order $5$: all such subgroups contain $4$ nontrivial elements (of order $5$). 
Everything else must be squeezed into the remianing $36$ elements. 
Assume indirectly that all Sylow-3 subgroups intersect trivially. All of them have $8$ nontrivial elements.  so that would provide us with at least $8n_3\leq 36$ elements whose order is a power of $3$. Thus, $n_3\leq 4$. 
Combining this with the observation $n_3\geq 2$ and the Sylow theorems, $n_3=4$ is the only possibility. 
Thus there are $32$ nontrivial elements contained in $3$-Sylows, i.e., of order which is a power of $3$. 
There are only 4 elemets remaining: they must be the elements of a 2-Sylow. Hence, there is only one 2-Sylow, which must be a normal subgroup. 
But then we get the same contradiction as above: there are only two $4$-element grups, and none of them has an automorphism of order 5. So this Sylow-2 subgroup would normalize any Sylow-5 subgroup, which is impossible. 
A: 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.  
