= Sylow subgroups of A6
{c}
{title2=$n_5=36,\quad n_3=10$}
In the <alternating group> $A_6$, of order $360$, a Sylow subgroup of order five is generated by a five-cycle. There are $6\cdot4!=144$ 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 $\binom63/2=10$ partitions.
Back to article page
