Efficient way of explicitly writing Sylow groups of $S_5$ I am trying to write explicitly the Sylow 5-subgroups of the symmetric group $S_5$. Using the properties of the cardinal of the set of Sylow subgroups, I have come to the conclusion that there must be $6$ Sylow 5-subgroups. I know one of them is $H_1=\langle (1,2,3,4,5)\rangle$ and the other five subgruops should also be generated by a cycle of order 5. My problem is that $H_1$ contains several cycles of order 5 and I would like to write the rest of Sylow 5-subgroups making sure that I do not make the mistake of writing two subgroups that are actually equal but just "look different" because the generator is different.
The only way I can think of for approaching this problem is writing all of the elements of each subgroup and making sure that they are all different, but this is not very efficient. I was wondering if there might be some an efficient way to do this or a clever way of choosing the generator of each subgroup so as to make sure that all subgroups are different.
Any trick you can come up with would be very useful. Thanks in advance.
 A: Let's work in any general $S_p$ and look for Sylow $p$-subgroups, which must all be generated by $p$-cycles.  Any such subgroup must have a unique element that permutes $1$ to $2$.  Use that element to generate the subgroup, and (written in cycle notation) each permutation of the remaining elements gives you a unique subgroup.
Thus, $\sigma = (1~2~\tau)$ generates a unique Sylow subgroup where $\tau$ is any permutation of $\{3, 4, \ldots p \}$.  This also demonstrates that there are $(p-2)!$ such subgroups.
A: Let's generalize to $S_p$.
The map $\sigma\mapsto\langle\sigma\rangle$ from $p$-cycles to Sylow $p$-subgroups in $S_p$ is a $(p-1)$-to-$1$ map, since every such subgroup has $p-1$ generators. There are $p!/p=(p-1)!$ elements of order $p$, so there are $(p-2)!$ Sylow subgroups.
Sylow theorems says the Sylow subgroups are conjugate, which by orbit-stabilizer means they correspond to cosets of the normalizer of one of them. If we let $\sigma=(12\cdots p)$ then any $\tau$ which fixes adjacent numbers but no others cannot be in the normalizer (consider how conjugation affects cycle types), so in paticular $S_{p-2}$ is a set of coset representatives for $N_{S_p}(\langle\sigma\rangle)$. Thus, the set of $p$-cycles $\pi$ fixing $p-1$ and $p$ form an irredundant list of generators for the $p$-Sylows.
