A group of order $66$ has an element of order $33$ If I wanted to show that a group of order $66$ has an element of order $33$, could I just say that it has an element of order $3$ (by Cauchy's theorem, since $3$ is a prime and $3 \mid 66$), and similarly that there must be an element of order $11$, and then multiply these to get an element of order $33$? I'm pretty sure this is wrong, but if someone could help me out I would appreciate it. Thanks. 

Alternatively:


*

*show that the Sylow $11$-subgroup is normal.

*show that a cyclic group of order $11$ has no automorphisms of order $3$.

*pick an element of order $3$ in your group and conclude that it commutes with any element of order $11$.
