maximum number of elements of order 5 in a group of order 80

Question

Sylow's theorem says there is either 1 Sylow 5-subgroup or 16 Sylow 5-subgroups. In each Sylow 5-subgroup, there are 4 elements of order 5. Since an element of order 5 generates a subgroup of order 5, does this mean there can be a maximum of 64 elements of order 5?

Answer 1

That is correct. Each Sylow $5$-subgroup of $G$ is disjoint, save for the identity element. That gives us $4(16) + 1 = 65$ elements in the groups of $Syl_{5}(G)$. If this is the case, then we have exactly one Sylow $2$-subgroup, containing $16$ elements.