# For $p,q$ primes such that $q|p-1$, show there is only one non-abelian group of order $pq$, up to isomorphism

This question has been answered a lot of times on this site, but I'm looking for an approach that does not use Sylow theory, since this is not covered in my syllabus. All answers I read this far used material that I did not yet learn. My level this far is up to automorphisms, group actions, and the isomorphism theorems.

My syllabus uses the following construction of a non-abelian group of order $pq$ where $q|p-1$. Let $N=C_p$ such that $\operatorname{Aut}(N)$ has order $p-1$. From Cauchy's theorem we deduce that there exists a subgroup $H\subset \mathrm{Aut}(N)$ of order $q$. Let $\tau:H\to \operatorname{Aut}(N)$ be the identity map. Then $N\rtimes_{\tau}H$ has order $pq$ and is non-abelian.

This far I can follow, but now I have to show that this group is the only non-abelian group of order $pq$. A hint for this exercise is to use that $(\mathbb{Z}/p\mathbb{Z})^*$ is cyclic if $p$ is prime.

My attempt (it is not really an attempt, I just looked what I could deduce, but it led me nowhere): Let $G$ be a finite non-abelian group of order $pq$. There exists, by Cauchy's theorem, an element $g$ of order $p$, so $\langle g\rangle$ is a subgroup of $G$ of order $p$ and index $q$, and since $q$ is the smallest prime dividing $pq$, this subgroup is normal.

Furthermore, it is easy to deduce that the center is trivial, since if it is not, $G$ is abelian because then $G/Z(G)$ is cyclic.

I have the idea that in some way I have to say that we know $G$ can be written as a semi-direct product (but how do I know that this is always the case?). Then this semi-direct product should be between $C_p$ and $C_q$, which are both abelian, so the used automorphism cannot be the identity map. From there I think I have to show there is only one map possible, thus only one possible semi-direct product. Because we already found one way to write $G$ as a semi-direct product, then the result would follow. But how exactly does this work? I think they want me prove the theorem this way, but I don't see how and why this would work.

Any help is much appreciated!

• Your link sends me to an error page. Would you mind copying the proof? Apr 23 '18 at 13:55
• It is easy to deduce that a group of order $pq$ is necessarily a semi-direct product. What ingredients do you need for that? A normal subgroup of order $p$ and one more subgroup of order $q$. Since $p, q$ are distinct primes, the intersection of those subgroups is trivial. Apr 23 '18 at 13:55
• @the_fox thank you! I see why we can write the group as a semi direct product now. We have the corresponding theorem in my syllabus but I just didn't think about using that one. Apr 23 '18 at 13:59
• @DietrichBurde Many thanks! I will look into the document now! Apr 23 '18 at 14:00
• There is a nice theorem (by Taunt, I think) we gives a necessary and sufficient condition for two semi-direct products to be isomorphic under some special circumstances. I will try to find it. Apr 23 '18 at 14:10

There is only one non-abelian group of order $$pq$$ with primes $$p and $$p\mid q-1$$, which is proved in Theorem $$3.8$$ of Keith Conrad's notes here. It is indeed given by a semidirect product; and as a subgroup of the affine group $${\rm Aff}(\mathbb{Z}/(q))$$.
Proposition: Let $$G$$ be a group of order $$pq$$ with primes $$p. If $$q\not\equiv 1 \bmod p$$, then $$G \cong C_{pq}$$. If $$q\equiv 1 \bmod p$$, then $$G$$ is isomorphic to either $$C_{pq}$$, or to the non-abelian group $$\left\{ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} \mid a\in (\Bbb{Z}/(q))^{\times}, b\in \Bbb{Z}/(q), a^p\equiv 1 \bmod q \right\}.$$
Remark: When $$q \equiv 1 \bmod p$$, so $${\rm Aut}(\mathbb Z/q\mathbb Z) = (\mathbb Z/q\mathbb Z)^\times$$ is cyclic of order $$q-1$$ and thus contains a unique subgroup of order $$p$$, the matrix $$(\begin{smallmatrix}a&b\\0&1\end{smallmatrix})$$ can be written as $$(\begin{smallmatrix}1&b\\0&1\end{smallmatrix})(\begin{smallmatrix}a&0\\0&1\end{smallmatrix})$$. Therefore the matrix group in the proposition can be described as a semidirect product $$\cong C_q\rtimes C_p$$ where $$C_p = \{a \in (\mathbb Z/q\mathbb Z)^\times : a^p \equiv 1 \bmod q\}$$ acts as automorphisms of $$C_q = \mathbb Z/q\mathbb Z$$ by multiplication
Proof: Let $$P$$ be a Sylow $$p$$-subgroup of $$G$$, and $$Q$$ be a Sylow $$q$$-subgroup of $$G$$. We have $$P\cong C_p$$, $$Q\cong C_q$$ and $$(G:Q)=p$$, which is the smallest prime dividing $$(G:1)$$. So $$Q$$ is normal. Because $$P$$ maps bijectively onto $$G/Q$$, we have that $$G=Q\rtimes P$$. Since $${\rm Aut}(Q)\cong C_{q-1}$$ we obtain $$G=Q\times P\cong C_q\times C_p\cong C_{pq}$$, unless $$p\mid (q-1)$$, i.e., $$q\equiv 1 \bmod p$$. In that case the cyclic group $${\rm Aut}(Q)$$ has a unique subgroup $$A$$ of order $$p$$. In fact, $$A$$ consists of the automorphisms $$x\mapsto x^i$$ for $$i\in \Bbb{Z}/q\Bbb{Z}$$ with $$i^p=1$$. Let $$a$$ and $$b$$ be generators of $$P$$ and $$Q$$ respectively, and let the action of $$a$$ on $$Q$$ by conjugation be $$x\mapsto x^j$$ with $$j\neq 1$$ in $$\Bbb{Z}/q\Bbb{Z}$$. Then $$G=\langle a,b\mid a^p=b^q=1,\; aba^{-1}=b^j\rangle,$$ which is the semidirect product $$Q\rtimes P$$ with this action of $$P$$ on $$Q$$ by conjugation. Choosing a different $$j$$ amounts to choosing a different generator $$a$$ for $$P$$, and so gives a group isomorphic to $$G$$. By definition, this group is non-abelian. In fact it is isomorphic to the subgroup of $${\rm Aff}(\Bbb{Z}/(q))$$ given above.