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Let $p$ and $q$ are distinct primes, $p<q$. If $p\mid(q-1)$, then there exists a unique non-abelian group of order $pq$.

I was pondering this problem for a long period of time but I was not able to come up with solution. I have found some solutions in this forum but they were unaccessible for me.

I would be very grateful if anyone can show to me not quite simple solution.

P.S. I know the Sylow theorems if it can help.

RFZ
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1 Answers1

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If $q$ is prime, then you know that there exists a group $P$ of order $q$, isomorphic to $\mathbb{Z}_q$. It's well known that $\text{Aut}(P) \cong \mathbb{Z}_q^{\times}$, where denotes the cyclic multiplicative group with the same elements as $\mathbb{Z}_q \setminus \lbrace 0 \rbrace$. Since $|\text{Aut}(P)|=q-1$ and $p|q-1$, there exists a subgroup $Q \leq \text{Aut}(P)$ of order $p$, by Cauchy's theorem. Define the $G \subset \text{Sym}(P)$ as follows:

$$G=\lbrace \varphi_{a,\sigma} \in \text{Sym}(P) :a \in P, \sigma \in Q \rbrace$$

where $\varphi_{a,\sigma}(x)=\sigma(xa)$, for all $x \in P$. It's an easy exercise to check $G$ is a non-abelian subgroup of $\text{Sym}(P)$ and that $G$ has order $pq$.

EDIT:

If there exists an other non-abelian group $G$ of order $pq$, then you can check that $G$ has a normal subgroup of order $q$ (by using Sylow's theorems) and since G also has a subgroup of order $p$ (again Cauchy), you can write $G$ as a semidirect product of these two subroups. But since the subgroup $Q$ of order $p$ was unique (up to isomorphism), the only possibility for $G$ is the one we constructed above.

  • Thanks for answer! But how to show that this group is unique up to isomorphism? – RFZ Mar 09 '18 at 19:13
  • @MorganRodgers I know it's confusing, I read $q|p-1$ in stead of $p|q-1$ at first ... –  Mar 09 '18 at 19:26
  • Is it possible to explain your EDIT without using semidirect product? Since i am not familiar with this notion – RFZ Mar 09 '18 at 19:32
  • @RFZ I see, this makes it of course difficult to understand what I'm saying. I find it strange that the book you are reading doesn't mention semidirect products, since it's such a basic concept in group theory. Is it an algebra book you are reading? –  Mar 09 '18 at 19:33
  • Yes, it is Herstein's book – RFZ Mar 09 '18 at 19:42
  • @RFZ I'm not familiar with Hernstein. On the other hand, I don't immediately see an other way to explain my edit. Perhaps if you do some reading about outer semidirect products that you might get a better understanding...it's really not difficult to grasp :) –  Mar 09 '18 at 19:51
  • Anyway thanks for answer. If my knowledge will be better could I ask you questions regarding this answer? – RFZ Mar 09 '18 at 20:05