Monomorphism from a group $G$ to a symmetric group $S_q$

Question

Let $p$ and $q$ be primes with $p<q$, and let $G$ be a group of order $pq$.

If $G$ is nonabelian, show that there exists a monomorphism from $G$ into $S_q$.

My attempt: Since $G$ has unique Sylow $q$-subgroup, say $H$, $H$ is normal in $G$. Therefore, $G/H$ is a cyclic group of order $p$. Let $aH$ be a generator of $G/H$. Then $a^p H = H$, so $a^p \in H$. Since $G$ is nonabelian, the order of $a$ cannot be $pq$(otherwise $G$ will be cyclic). Therefore, the order of $a$ must be $p$. i.e. $a^p = e$. Also, let $H=\langle b \rangle$, then $b^q=e$. Since $H$ is normal in $G$, $aba^{-1}=b^i$ for some $i\neq 0, 1$.

Therefore, the desired monomorphism would send $a$ to $p$-cycle $(1, 2, ..., p)$ and $b$ to $q$-cycle $(1, 2, ..., q)$. However, I cannot figure out whether this map is indeed a homomorphism. Does anyone have ideas?

Answer 1

Since G is non- abelian, let H be a subgroup of order p. Now this subgroup is not normal. Now, make G act on the set of left cosets of $G/H$ by left multiplication. This will yield you a homomorphism day $\phi:G \to S_{G/H}$ where the map that comes from the group action is as follows

$$ g\mapsto \phi_{g} , where \phi_{g}:G/H \to G/H s.t aH\mapsto gaH$$

Now you can check that $ker(\phi) \subseteq H$. But $|H|=p$. So either $ker(\phi)$ is trivial or H. But as we have seen H is not normal and hence it must be trivial. And also observe that $S_{G/H}$ is nothing but your required $S_{q}$.

Answer 2

I think yo did not yet use that $G$ is non-abelian:

The number of Sylow $p$-subgroups is either $1$ or $q$, but in the first case, we can conclude that $G$ is abelian, hence the second case applies. $G$ acts on the set of Sylow $p$-subgroups by conjugation, thus giving rise to a homomorhism $G\to S_q$. If $g$ is in the kernel of this homomorphism, then $gHg^{-1}=H$ for all Sylow $p$-subgroups. Show that this implies $g=1$.