I am trying to count the number of $p$-Sylow subgroups in $A_n$. Particularly, I want to count the number of $5$-Sylow subgroups in $A_6$. I know that to do this, I need to count the number of $5$-cycles that exist in $A_6$. This website https://groupprops.subwiki.org/wiki/Alternating_group:A6 claims we have 36 such groups, but I am having trouble finding this count. Can anyone help me out with this computation?
The way I have been thinking about counting these $5$ cycles is by considering that I could have $6*5*4*3*2$ different 5 cycles using 6 elements, without considering repetitions (meaning without considering that $(1 2 3 4 5)$ is the same as $(2 3 4 5 1)$). Thus if we divide the count we have by 5, I think we would be taking into consideration this symmetry, so we would have $6*4*3*2$ but I'm not sure if this is a good approach to counting these cycles. Thanks for your help!
This is simple combinatorics. In $A_6$, a $5$-Sylow is a cyclic group of order $5$. Any two of these intersect trivially, as nontrivial elements of a cyclic group of order $5$ are all generating elements. Every cyclic group of order $5$ contains $4$ nontrivial elements.
Putting all together: the number of $5$-Sylows is the number of elements of order $5$ divided by $4$.
By the way, you miscalculated the number of elements of order $5$. Think about it, it should be $6\cdot 4!$.