Sylow subgroup

A Sylow $p$-subgroup of a finite group has the largest power of $p$ dividing the group order.

In the alternating group $A_6$, of order $360$, a Sylow subgroup of order five is generated by a five-cycle. There are $6\cdot 4!=144$ five-cycles, four in each such subgroup, giving $36$ subgroups. A Sylow subgroup of order nine is generated by disjoint three-cycles, hence is $C_3\times C_3$. The unordered partitions of six letters into two triples give all such subgroups by Sylow conjugacy, and their single three-cycles recover the partition uniquely. There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$ partitions.

Suppose distinct Sylow subgroups for $p$ in a finite group have trivial intersections. Conjugation by one of them, $P$, fixes $P$. The stabilizer of any other $Q$ is $H=P\cap N_G(Q)$. Since $H$ normalizes $Q$, the product $HQ$ is a subgroup of $p$-power order; maximality of $Q$ gives $H\le Q$. Thus $H\le P\cap Q=1$. All other orbits have size $|P|$, proving the congruence.