How do we prove this fact about cyclic groups? Prove that an Abelian group of order 33 is cyclic.
Can we take an element a of order 3 and an element b of order 11 and say, |ab|=33?
 A: Yes, that's definitely one way to do it.  But you'll have to be explicit about how you know that you can choose $a$ and $b$ to have those orders and once they're chosen you'll have to be explicit about how you know that the order of $ab$ is $33$.  As long as you can fill in those details your proof will be correct.
A: One possibility is to use Sylow groups. One can deduce from Sylow's theorems that there is, respectively, exactly one Sylow group $H$ of order 3 and one Sylow group $N$ of order 11. That means $H \cong (\mathbb Z/3\mathbb Z, +)$ and $N \cong (\mathbb Z/11\mathbb Z, +)$. Being unique, $H$ and $N$ are both normal, and we have $N \cap H = \{1\}$ as well as $NH = G$. One can show in general that if $N$ and $H$ are normal subgroups fulfilling these two conditions, then $G \cong N \times H$ as groups. Here, that means $G \cong \mathbb Z/11\mathbb Z \times \mathbb Z/3\mathbb Z \cong \mathbb Z/33\mathbb Z$. Thus, $G$ is cyclic. (This does not a priori use the fact that $G$ is abelian.)
