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.
[Math] A group of order $66$ has an element of order $33$
group-theory
Best Answer
Step 1: Show that any finite group order $4k+2$ has an index two subgroup. (Hint.)
Step 2: Show that any group of order $33$ is cyclic.