Let $G$ be a group of order $5\times 13\times 43\times 73$. Find the number of elements of order $5$.
Here is what I do:
Since $|G| = 5m$ where $(m,5) = 1$, $m = 13\times 43\times 73$, by Sylow's Theorem, there exists a Sylow $5$-subgroup with #Sylow $5$-subgroup dividing $13\times 43\times 73$ and #Sylow $5$-subgroup = $5k + 1$ for some $k \geq 0.$ So I think that the only possible choice is $1$. This yields that there is only one Sylow $5$-subgroup which yields that the all $4$ elements of this cyclic group except the identity is of order $5$. So there are only $4$ elements of order $5$ in this group.
Is my approach to this question correct?