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I faced the following problem in a book: "Give an example of a non-cyclic group of order $pq$, with $p<q$ primes such that $p\mid q-1$."

This example is supposed to talk to a previous demonstration that if $p\nmid q-1$, then $G$ is isomorphic to $\mathbb{Z}_{pq}$(which is follows easily from Sylows theorem).

The point is that I found an example, but it was through some little work and the example is a bit ugly. Namley, let $n$ be an element of order $p$ modulo $q$ (which exists as $p\mid q-1$ say by just Cauchy, or just number theory in this case), then define the group as pairs $(u,r) \in {\mathbb{Z}}_{p}\times {\mathbb{Z}}_{p}$ where you define $$ (u,r) \dot (v,s) = (u+n^rv, r+s)$$ and it can be checked that this is a non-comutative(and thus non-cyclic) group of order $pq$.

So, I would like to know if there is a more easy example.

PS: My example was found trough taking an element $t$ with order q (this element generates the only Sylow q-subgroup of $G$), looking at the $q$ p-subgroups and playing with them.

PS2: I saw some other posts with characterization of all subgroups with order pq, but I am currently unable to understand it fully. I just wanted easy examples of such order pq groups.

PS3: I mean for a general example with arbitrary p,q

sogeking_uph
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  • The simplest examples have $p = 2$. – Dermot Craddock Jun 24 '25 at 19:19
  • Think of dihedral groups. They are a semidirect product, which is the case for a non-cyclic group of order $pq$. The easiest instance of $p\mid q-1$ is $p=2$ and $q=3$. – Dietrich Burde Jun 24 '25 at 19:30
  • sorry guys, I meant for a general example case for arbitrary p,q – sogeking_uph Jun 24 '25 at 19:52
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    It looks like what you came up with is just a semidirect product of $C_p$ and $C_q$ which might be an easier way to put it. – Daniel Schepler Jun 24 '25 at 20:06
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    The nonabelian groups of order $pq$ are isomorphic. They're in their simplest form in the duplicate. – Shaun Jun 24 '25 at 20:36
  • What you found is really a concrete description of the semi-direct product of $\Bbb Z_q\rtimes \Bbb Z_q$ induced by the action $r\cdot v = n^r v$ of $\Bbb Z_p$ on $\Bbb Z_q$. Here is another more concrete description: $$ {\begin{pmatrix} a & b \ 0 & 1/a \end{pmatrix} \mid a\in G, b\in\Bbb Z_q}$$ where $G$ is the unique multiplicative subgroup of $\Bbb Z_p^{\times}$ of order $p$. – Just a user Jun 25 '25 at 04:02

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