Problem statement :
Let $G=S_5$ (it actually reads $|G|=S_5$ but this must be a misprint obviously). Using Sylow , how many 5-Sylow subgroups has G got? Is this consistent with the number of 5 cycles in G?
My proof:
we have $|G|=|S_5|=120=5\cdot 24$.
By Sylow's first theorem we now know that the order of the $5$-Sylow subgroups is 5.
By Sylow's 3rd theorem we know that the number of $5$-Sylow n_5=6.$
if $n_5=1$ then we also know that $|N_G(P)|=|G|$ but then this would imply that the normaliser contains every element in $G$ But of course $S_n$ is non-abelian for $n \geq 3$ so this can't be true.
We conclude that that $n_5=6$. So there are 6 $5$-Sylow -subgroups each with order $5$. These are distinct because the intersection is itself a subgroup and by Lagrange's theorem the intersection is either $1$ or $5$ and so must be $1$. so they intersect only at the identity. so any $5$-Sylow subgroup has order $5$ and contains the identity and $4$ $5$-cycles. Overall this means there are $(6\times 4=)\:24$ $5$-cycles, the same as would be expected in $S_5.$