The number of Sylow $5$-subgroups in $S_6$

Question

Find the number of sylow $5$-subgroups in $S_6$.

First: $ord(S_6)=6!=2^4\cdot 3^2\cdot 5=144\cdot 5$, so $n_5|144$ and $n_5\equiv 1\pmod5$, where $n_5$ is the number of sylow $5$-subgroups.

Since we are in $S_6$ the cycles for syl $5$-subgroups look like $$(a\ b\ c\ d\ e)(f)$$ This are $\frac{5!}{5}+\frac{5\cdot 5!}{5}=144$ possibilities of $5$-cycles of order 5. A syl $5$-subgroup includes the identity permutation.

What is the next step?

Answer 1

There are indeed $\frac156!=144$ different $5$-cycles in $S_6$ but note that a single $5$-Sylow accounts for 4 of them.

Indeed $c$, $c^2$, $c^3$, $c^4$ are the 4 different non-trivial elements in $\langle c\rangle$.

Thus the total number of $5$-Sylows is $\frac14144=36$.