Group of Order 33 is Always Cyclic

Question

I would love to get help on this problem from a chapter on Commutator of Group Theory:

Show that each group of order 33 is cyclic. (Hint: Use the result from the Exercise and Lemma below.)

Exercise: Let $p$ and $q$ be prime numbers such that $p \nmid (q-1).$ Show that each group of order $pq$ possesses a normal subgroup of order $p.$
Lemma: For any two subgroups $H$ and $K$ of $G,$ if $H \subseteq N_G(K), K \subseteq N_G(H),$ and $(|H|, |K|) = 1,$ then $[H, K] = \{ 1 \}.$

If you have to choose between elegant and down-to-earth dummy solutions, please give me the latter. I know the latter is tedious to you but you have slowpoke over here. Thanks for your time.

POST SCRIPT: ~~~~~~~~~~~~~~~~~~~~~~~~~~
Thanks for pointing out duplicate answer. Having looked it over, however, I think the 4 solutions from that 2011 posting were way, way too long to fit the mind of a beginner. (Notice especially that one of the solutions is from W. Burnside's classic book!) I think my professor has intentionally included the two hints to allow shorter solution. But unfortunately I don't know how to connect the dots, and for that reason I am asking for help here. Thanks again.

POST SCRIPT IN RESPONSE TO "TIMBUC": ~~~~~~~~~~~~~~~~
(1) On $G = PQ,$ can I prove it using $G \subseteq PQ$ and then $PQ \subseteq G$? Any shortcut?
(2) On $P \cap Q = \{1\},$ I can claim that because both are subgroups of $G$ and both have to have neutral number, therefore $P \cap Q = \{1\}.$ But why do you have to bother with it? I did not see it is useful on the next lines.
(3) On $PQ \cong P \times Q,$ can I prove it using the classic $\varphi(x, y) = \varphi(x) \varphi(y)$ and then bijectivism since they are isomorphic? (Is "bijectivism" is the correct noun of bijective?)
(4) On $P \times Q,$ how do you know that they are each cyclic in the first place? And how do you know that the direct product of cyclic subgroups with co-prime order is cyclic again? Are they the theorems?
(5) Thank you very much for your time and help.

Answer 1

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.

Answer 2

Response to the postscript:

1. 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 $G$.

2. $P\cap Q$ 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 $P\cap Q=\{1\}$.

3. 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.

4. 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.

Answer 3

There is a theorem saying that

The cyclic group with order $n$ is the only group with order $n$ if and only if $\gcd(n,\phi(n))=1$, where $\phi$ is the Euler phi function.

You can see more about it here http://www.jstor.org/stable/2324062?seq=1#page_scan_tab_contents

I know that your problems have already been solved. But I'd like to add some information for people who reach out this question in the future.