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This question already has an answer here:

I don't know about the Sylow Theorems.

But I have been wondering about a proof of the fact that a group or order $pq$ where $p$ and $q$ are distinct primes must be cyclic. I can't quite work out the details, but here is the general idea. I would like help with filling in details. I assume that it is already known that $G$ has subgroup(s) of order $p$ and subgroup(s) of order $q$.

If $G$ is a group of order $pq$ ($p\neq q$), then I know that $G$ has a subgroup $H$ of order $p$ and a subgroup $K$ of order $q$. Then $H\simeq \mathbb{Z}_p$ and $K\simeq \mathbb{Z}_q$.

But then $H\oplus K \simeq \mathbb{Z}_{pq}$, so I would think that $H\oplus K \simeq G$. I guess one could do an internal direct product instead of an external direct product, but I don't know that $H$ and $K$ are normal subgroups.

I am asking for help completing this argument.

Edit: I see from the comments below that I might need to assume that the smaller prime does not divide the larger prime minus $1$. Or maybe it is enough to assume that the primes are greater than or equal to $3$ (Still distinct).

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marked as duplicate by Nicky Hekster abstract-algebra May 29 at 10:41

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    $\begingroup$ $S_3$ has order $6=2^*3$ (in general you need to add the condition that the smaller prime does not divide one less than the larger). $\endgroup$ – lulu Oct 28 '15 at 18:51
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    $\begingroup$ Yep, I was going to bring up the dihedral groups. $\endgroup$ – hardmath Oct 28 '15 at 18:52
  • $\begingroup$ @lulu : I think $q < p$ should indeed be distinct primes with $q$ not dividing $p-1$. $\endgroup$ – Joel Cohen Oct 28 '15 at 18:54
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    $\begingroup$ (Assuming $p$ does not divide $q-1$) Let $H=<a>$ and $K=<b>$. Then show that $G=HK$ and that G is abelian. Then take the map $H \times K \rightarrow G$ with $(a^{m},b^{n}) \mapsto a^{m}b^{n}$ and show that it is an isomorphism. Then conclude $G \cong H \times K \cong Z_{p} \times Z_{q} \cong Z_{pq}$. $\endgroup$ – vasmous Oct 28 '15 at 19:01
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    $\begingroup$ @JohnDoe: Generally, a group $G$ of order $pq$ (not necessarily distinct) is either abelian or $Z(G)=1$. In our case you have to believe (Sylow Theorems) that $H$ and $K$ are normal in $G$. (I'm sorry for not mentioning it in my previous comment.) Now, notice that $aba^{-1}b^{-1}=(aba^{-1})b^{-1} \in K$, cause $K \trianglelefteq G$ and $aba^{-1}b^{-1}=a(ba^{-1}b^{-1}) \in H$, cause $H \trianglelefteq G$. So $aba^{-1}b^{-1} \in H\cap K=1$, which means that $ab=ba$ and you are done. $\endgroup$ – vasmous Oct 29 '15 at 21:38
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For a general group of order $p$ and $q$, there are very few possibilities (though you need Sylow theorems to know this). The fact is, for $p>q$ and $G$ a group of order $pq$, we must have $$G\cong C_p\rtimes C_q$$ where the semi-direct product is defined in terms of some homomorphism $$\Phi:C_q\to\mathrm{Aut}(C_p)\cong C_{p-1}.$$

If $q$ does not divide $p-1$, this homomorphism must be trivial and you get $G\cong C_p\times C_q\cong C_{pq}$.

When $\Phi$ is nontrivial, we can write $\Phi(c_q^k)=\phi_k$. Then, the product structure on $C_p\rtimes C_q$ is given by $$(c_p^a,c_q^b)(c_p^r,c_q^s)=(c_p^a\phi_b(c_p)^r,c_q^{b+s}).$$ It is a nice exercise to check that this is a group structure, and $C_p$ is normal. It is also useful to describe the isomorphism $S_3\to C_3\rtimes C_2$ explicitly.

EDIT: As you are requesting more detail, here you go:

Let $G$ be a group of order $pq$ with $p>q$ primes. Using Cauchy's theorem there are (cyclic) subgroups $P=\langle x\mid x^p=1\rangle$ and $Q=\langle y\mid y^q=1\rangle$ of orders $p$ and $q$, respectively. It follows from the Sylow theorems that $P\lhd G$ is normal (Since all Sylow $p$-subgroups are conjugate in $G$ and the number $n_p$ of Sylow $p$ subgroups must divide $q$ and satisfies $n_p\equiv 1$ (mod $p$)).

With this taken as given, it is straightforward to prove that $G\cong P\rtimes Q$, where the semidirect product is defined in terms of a homomorphism $\phi:Q\to\mathrm{Aut}(P)$.

  1. We first note that since $|P\cap Q|$ divides both $p$ and $q$ we must have $|P\cap Q|=1$. It follows that $$|PQ|=\frac{|P||Q|}{|P\cap Q|}=pq=|G|$$ Hence, $PQ=G$.

  2. Now, since $Q=\langle y\rangle$ normalizes $P=\langle x\rangle$, the map $\phi_k:P\to P$ given by $\phi_k(x)=y^kxy^{-k}$ is well defined. Moreover, it is clearly an automorphism with inverse $\phi_{-k}$. Finally, since $\phi_{k}\phi_j=\phi_{k+j}$, the map $y^k\mapsto\phi_k$ defines a homomorphism $$\phi:Q\to \mathrm{Aut}(P).$$

  3. As above, we define $P\rtimes Q$ to be $P\times Q$ as a set, with multiplication $$(x^i,y^j)(x^k,y^l)=(x^i\phi_j(x^k),y^{j+k}).$$ Of course, one needs to verify that this is indeed a group. The identity is $(1,1)$, $(x^k,y^l)^{-1}=(\phi_{-l}(x^{-k}),y^{-l})$. Associativity is tedious but true.

  4. Define a map $\psi: P\rtimes Q\to G$ by $\psi(x^i,y^j)=x^iy^j$. The map $\psi$ is surjective since $PQ=G$, and it is injective because $|P\rtimes Q|=pq=|G|$. To see that it is a homomorphism we compute \begin{align*} \psi((x^i,y^j)(x^k,y^l))&=\psi(x^i\phi_j(x^k),y^{j+l})\\ &=x^i\phi_j(x^k)y^{j+l}\\ &=x^i(y^jx^ky^{-j})y^{j+l}\\ &=x^iy^jx^ky^l=\psi(x^i,y^j)\psi(x^k,y^l). \end{align*} Hence, $\psi$ is an isomorphism as promised.

  5. Now, either the homoorphism $\phi:Q\to\mathrm{Aut}(P)$ is trivial or it is not. If it is trivial, then $$G\cong P\rtimes Q=P\times Q\cong C_p\times C_q\cong C_{pq}.$$ If the homomorphism is nontrivial, then $G$ has the following presentation: $$G = \langle x,y\mid x^p=1=y^q, yx=x^ny\rangle$$ where $n\in\mathbb{Z}$ satisfies $n\not\equiv1$ (mod $p$), but $n^q\equiv 1$ (mod $p$). (To see this note that $yxy^{-1}=x^n$ for some $n\not\equiv_p 1$, but $x=y^qxy^{-q}=x^{n^q}$.)

  6. This works for any pair of primes with $q|(p-1)$, not just $p=3$. An example: $p=11$, $q=5$. Take $n=3$ so we have $$G=\langle x,y\mid x^{11}=1,y^5=1,yx=x^3y\rangle.$$ This group has order 55 and you can compute \begin{align*} yxy^{-1}&=x^3\\ yx^3y^{-1}&=(yxy^{-1})^3=x^9\\ yx^{9}y^{-1}&=x^{27}=x^5\\ yx^5y^{-1}&=x^{15}=x^4\\ yx^4y^{-1}&=x^{12}=x \end{align*}

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    $\begingroup$ This answer seems like it would be perfect material for a second course in Abstract Algebra. $\endgroup$ – Patrick Tam Apr 19 '18 at 20:51
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Consider $S_3$, the group of symmetries of the triangle. It's not cyclic and has order $6=2\cdot 3$.

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Let $\lvert G \rvert = pq$ for primes $p, q$ such that $q < p$ and $q \not \mid p-1$. Let $n_p$ and $n_q$ be the number of Sylow $p$-subgroups and Sylow $q$-subgroups, respectively.

By the Third Sylow Theorem, $n_p \mid q$ and $n_p \equiv 1 \pmod p$ which implies $n_p = 1$ since $q < p$.

Similarly, $n_q \mid p$ and $n_q \equiv 1 \pmod q$ implies $n_q = 1$ since $p \not\equiv 1 \pmod q$.

Let $P$ be the unique Sylow $p$-subgroup and $Q$ be the unique Sylow $q$-subgroup. Since $p$ and $q$ are prime, $P$ and $Q$ are cyclic: $P \cong \mathbb Z / p \mathbb Z$ and $Q \cong \mathbb Z / q \mathbb Z$.

Restatement of 1. from David Hill's answer: $P \cap Q \le P, Q$ so by Lagrange's theorem we have $|P\cap Q|$ divides both $p$ and $q$, and we must have $|P\cap Q|=1$. It follows that $$|PQ|=\frac{|P||Q|}{|P\cap Q|}=pq=|G|$$ Hence, $PQ=G$. Since $P$ and $Q$ are unique, by consequence of the Third Sylow Theorem, $P,Q \lhd G$. Then the internal and external direct product are isomorphic, so $G \cong P \times Q$.

An analogue to the Chinese Remainder Theorem for groups shows that if $\operatorname{gcd}(m,n)=1$ then $\mathbb Z / m \mathbb Z \times \mathbb Z / n \mathbb Z \cong \mathbb Z / mn \mathbb Z$. (prove with First Isomorphism Theorem and the standard Chinese Remainder Theorem) So by this theorem, $G \cong P \times Q \cong \mathbb Z / pq \mathbb Z$ and hence cyclic.

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