Not sure if its proper to ask it here. I saw the following statement: if group $G$ has 6 sylow-5 subgroups then $G$ has $6$ subgroups of order $5$, when each one of them has $4$ elements of order $5$. Then $G4 has $6\cdot 4=24$ elements of order $5$.
I do understand why $G$ has $6$ subgroups of order $5$ but why each one of them has $4$ elements of order $5$?