Group of Order 33 is Always Cyclic
The number of Sylow $\;11$- subgroups in a group $\;G\;,\;\;|G|=33=3\cdot 11\;$ , has to divide $\;3\;$ and also be equal to $\;1\pmod{11}\;$ , so that means there's one single such subgroup $\;P\;$ in $\;G\;$, and this means it is normal.
With exactly the same kind of argument prove there's only one single Sylow $\;3$- subgroup $\;Q\;$ of $\;G\;$ , and it also is normal.
Now, we have that $\;|G|=|PQ|\;$ (why?) and also $\;P\cap Q=\{1\}\;$ (why?), so in fact $\;G=PQ\cong P\times Q\;$ , and since direct product of cyclic subgroups with coprime order is cyclic again, we' re done.
Response to the postscript:
Firstly, since $P\leq G, Q\leq G$, it follows that $PQ\leq G$ ($PQ$ is a subgroup because both are normal by the exercise). Furthermore, since $PQ=\{pq\mid p\in P, q\in Q\}$, and since $1\in P, Q$, setting $q=1$, $p$ arbitrary shows that every element of $P$ is in $PQ$. Similarly, by setting $p=1$ shows that $Q$ is contained in $PQ$ as well. Thus, $PQ$ is a subgroup containing both $P$ and $Q$, and thus divisible by both $3$ and $11$, and thus by $33$. Since $G$ has order $33$, $PQ$ must be all of $G$.
$P\cap Q$ is a subgroup of both $P$ and $Q$, so by Lagrange's Theorem, its order divides both $3$ and $11$. But $\gcd(3, 11)=1$, so the order of $p\cap Q$ is $1$, so $P\cap Q=\{1\}$.
Yes, you can use the map $(p, q)\mapsto pq$. This map is a homomorphism since $p_1q_1p_2q_2=p_1p_2q_1q_2$, which is true because elements of $P$ commute with elements of $Q$, by the lemma. This homomorphism is surjective since all possible products $pq\in PQ$ are obtained, and it's injective, since if $pq=1$, then $p=q^{-1}$, so $p, q\in P\cap Q=\{1\}$, so $(p, q)=(1, 1)$, and thus the kernel is trivial. Thus, it's a bijective homomorphism (the noun form of bijective is bijection) and so an isomorphism.
Since $P$ and $Q$ have prime orders, they are cyclic. This is because the order of the subgroup generated by a nonidentity element is not $1$ and divides the (prime) order of the group, so this subgroup must be the whole group. Since the group is generated by a single element, it is cyclic. Also, the fact that the product of cyclic groups of relatively prime order is cyclic is the Chinese Remainder Theorem.