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Let $p>q$ be primes and $G$ be a group of order $pq$. When $p\neq 1\pmod q$, $G$ must be cyclic by a couple applications of Sylow's theorems. However, when $p=1\pmod q$ is it always possible to construct a non-cyclic $G$?

So far I tried letting $P,Q$ be Sylow $p,q$-subgroups of $G$. Then both are cyclic, $P$ is normal and $n_q=p$.

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It's always possible to construct such a noncyclic group when $p \equiv 1 \pmod{q}$. Note that this means that $q$ divides $p-1$, which is the order of the automorphism group $\operatorname{Aut}(C_p)$ of a cyclic group of order $p$, so by Cauchy's theorem, $\operatorname{Aut}(C_p)$ contains an element of order $q$. Sending the generator of $C_q$ to that element in $\operatorname{Aut}(C_p)$ gives us a non-trivial homomorphism $\varphi: C_q \to \operatorname{Aut}(C_p)$. Now taking the semidirect product $C_p \rtimes_\varphi C_q$ gives us a noncyclic group of order $pq$.

It's also possible to construct this group without semidirect products.
Consider the field with $p$ elements $\Bbb F_p$. It's well known that the multiplicative group $\Bbb F_p^\times$ is cyclic of order $p-1$. So if $q \mid (p-1)$, we can find an element $a$ in $\Bbb{F}_p^\times$ of order $q$ by Cauchy's theorem. Now consider the following subgroup of $\operatorname{GL}_2(\Bbb{F_p})$:

$$\left \{\begin{pmatrix}a^n & x \\ 0 & 1 \end{pmatrix}, n \in \Bbb{F_q}, x \in \Bbb{F}_p\right \}$$ (Note that $a^n$ is well-defined because $a$ has order $q$ in $\Bbb{F}_p^\times$). It's evident that this has order $pq$ (using that $a$ has multiplicative order $q$). Checking that this is in fact a non-abelian subgroup of $\operatorname{GL}_2(\Bbb{F_p})$ requires only easy matrix computations. This construction turns out to give exactly the same group as above, but this form of presentation (which is not meant as a technical term here) is probably more accessible to those who are not familiar with semidirect products.

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  • $\begingroup$ Would it be possible to describe this more concretely, for example with permutations? $\endgroup$ – Akababa Feb 17 '18 at 2:13
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    $\begingroup$ @Akababa I hope matrices are fine, too $\endgroup$ – MatheinBoulomenos Feb 17 '18 at 2:17
  • $\begingroup$ By Cauchy's theorem????? You just need a primitive root of $p$ $\endgroup$ – Kenny Lau Jun 13 '18 at 8:40

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