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. 
 A: 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 $11$.
A: To compliment the other answers, I will address why commutativity is necessary. Let $G$ be your group $o(x)$ denote the order of $x\in G$. You are making use of the statement that $o(a)o(b)=o(ab)$, but this is not necessarily true. It holds when $o(a),o(b)$ are coprime and $ab=ba$ (an interesting consequence of this is that the partial sums $s_n$ of the harmonic series are never integers for $n>1$, which follows from Bertrand's postulate and $\mathbb Q/\mathbb Z$ being abelian), but there are nonabelian groups of order $66$, such as $S_6\oplus \mathbb Z_{11}$. If $ab=ba$ and $o(a),o(b)$ are coprime, then $(ab)^{o(a)o(b)}=a^{o(a)}b^{o(b)}=e\cdot e=e$, so $o(ab)\leq o(a)o(b)$. If $(ab)^n=e$ then $a^nb^n=(ab)^n=e$ so $(a^n)^{-1}=b^n$, hence $o(a^n)=o(b^n)$, and if $n<o(a)o(b)$ then since $o(a),o(b)$ coprime we have that one of $a^n,b^n\neq e$. But $o(b^n)=o(a^n)|o(a)$ since $(a^{n})^{o(a)}=(a^{o(a)})^n=e^n=e$, and similarly $o(a^n)=o(b^n)|o(b)$, so $o(a^n)=o(b^n)=1$. Thus $a^n=b^n=e$, so $n\geq o(a)o(b)$ so $o(ab)=o(a)o(b)$.
A: 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.  
