how many elements are there which has order 3?

Question

if the number of sylow-3 subgroups of a group of order 96 is 4, how many element are there which has order 3?

I dont know how to start. why isnt the answer 4?

Answer 1

Say we have a group $G$ with $|G| = 96$. If a subgroup of $G$ has order $3^n$ for some $n$, then Lagrange's theorem says that $n$ must be either $0$ or $1$. We're told that there are $4$ subgroups of order $3^n$ for the largest possible $n$. Therefore we conclude that $G$ has four subgroups of order $3$. The only element any of them have in common is the identity element $e \in G$, so there are $2\cdot 4 = 8$ other elements in these four groups, and all of those elements must have order $3$.

If there were any other elements of order $3$, then such an element would generate a cyclic subgroup of order $3$, but we've accounted for all of those, so there cannot be any other elements of order $3$ than those that are in the $4$ Sylow-3 subgroups.