Non Abelian group of order pq

Question

Given primes $p$ and $q$ with $q$ dividing $p-1$, construct a non-abelian group of order $pq$ as follows: Let $P$ have order $p$ and let $Q \subseteq \operatorname{Aut}(P)$ have order $q$. Let $G \subseteq \operatorname{Sym}(P)$ be the set of all maps $\phi_{a,\sigma}$ defined by $\phi_{a,\sigma}(x)=\sigma(xa)$ for $x \in P$ and $\sigma \in Q$. Show that $G$ is a non-abelian group of order $pq$.

My attempt: By Cauchy's Theorem, there are subgroups of order $p$, $H=\langle x\rangle$ and order $q$, $K=\langle y\rangle$, where $o(x)=p$ and $o(y)=q$. Where do I go from here?

Thanks!

Answer 1

The order of the group is $pq$ because its elements $\phi_{a,\sigma}$ are indexes by $P\times Q$ and $|P\times Q|=p\cdot q$. One has to be careful, however:

• Show that $\phi_{a\sigma}=\phi_{a',\sigma'}$ implies $a=a'$ and $\sigma=\sigma'$
• Show that $G$ is indeed a group, i.e., find $a,\sigma$ such that $\phi_{a,\sigma}$ is the identity in $\operatorname{Sym}(P)$, and show closure, i.e., that that $\phi_{a,\sigma}\circ \phi_{a',\sigma'}=\phi_{a'',\sigma''}$ for suitable $a'',\sigma''$, given $ a,a',\sigma,\sigma'$.
• Show that $G$ is non-abealian; one specific example using the multiplication rule obtained in the previous bullit point is sufficent.

You need Cauchy's theorem practically only to show that $Q\le \operatorname{Aut}(P)$ of order $q$ exists to begin with.