Prove that a non-abelian group of order $pq$ ($p

Question

So I've come up with a proof for the following question, and I'd like to know if it's correct (as I couldn't find anything online along the lines of what I did).

Question Let $p$ and $q$ be primes with $p<q$. Prove that a non-abelian group of order $pq$ has a nonnormal subgroup of index $q$, so there there eixists and injective homomorphism into $S_q$

First I prove a Lemma:

If $H\triangleleft G$, then for every conjugacy class $\mathcal{K} \subseteq H$ or $\mathcal{K} \cap H = \emptyset$.

proof:
If $x\in K\cap H$, then $gxg^{-1} \in gHg^{-1}$ for all $g\in G$. Since $H$ is normal $gHg^{-1}=H$, so that $H$ contains all the conjugates of $x$,i.e., $\mathcal{K} \subseteq H$ (Dummit and Foot, 4ed, pg 127).

Here's my attempt on the question:

proof:
Let $G$ be a group such that $|G|=pq$ and $p<q$. Then $G \cong \mathbb{Z}_{pq}$, and so $G$ has two subgroups, say $H$ and $K$, of orders $p$ and $q$, respectively. 'Suppose that $H$ is a subgroup in $G$ of order $p$; a subgroup $K$ of order $q$ exists in $G$ by Cauchy (Nicky Hekster).' Also, by Lagrange, $H=\langle x \rangle$ and $K=\langle y \rangle$ for some $y,x \in G$. Since $p$ is the smallest prime that divides the order of $G$ and $|G:K|=p$, then $K\triangleleft G$ (by some theorem).

Suppose, by contradiction, that $H\triangleleft G$. Then for $x^i \in H$, $gx^ig^{-1}=x^k \in H$ for all $g\in G$. So $H\subseteq \mathcal{O}_{x^i}$ (the conjugacy class of $x_i$). Since $|x^i|$ divides $p$, then $x_i\not \in K$. Thus $\mathcal{O}_{x_i} \not \subseteq K$. Also, since $e=x^p$ and $gx^pg^{-1}=e$, then $e\in \mathcal{O}_{x_i}$. Thus $\mathcal{O}_{x_i} \cap K \neq \emptyset$. Therefore $K$ is not normal in $G$, which is absurd.

Now let $\phi: G \rightarrow S_q$ be the permutations representation afforded by the action of $G$ on $A=G/H$ by left multiplication -- say $\pi_g:A\rightarrow A$ defined by $aH\mapsto gaH$ for all $g\in G, \; aH\in A$. We know that $\ker \pi_g \leq H$ and $\ker \pi_g \triangleleft G$. Since $|H|=p$, then $|\ker \pi_g|=p$ or $|\ker \pi_g|=1$. If $|\ker \pi_g|=p$, then $\ker \pi_g$ is not normal in $G$. Therefore, $|\ker \pi_g|=1$. Since $\ker \pi_g=\ker \phi$, then $\phi$ must be injective.

Sorry for any typos (and a HUGE 'thank you' to anyone who actually reads this).

Answer 1

Let $K$ be a subgroup of order $q$ of $G$. $K$ exists by Cauchy's Theorem. Then index$[G:K]=p$, the smallest prime dividing $|G|$. Now let $G$ act on the left cosets of $K$ by left-multiplication. The kernel of this action $C=core_G(K)$ is normal in $G$ and $G/C$ injects homomorphically in $S_p$. So $|G/C| \mid p!$. Since $p \lt q$, it follows that $|G/C|=p$, whence $K=C$, and $K$ is normal.

Now $G$ has a subgroup $H$ of order $p$. If it is normal, then we would have $H \cap K =1$ and $|HK|=|H||K|/|H\cap K|=pq$, so $G=HK$, and $G \cong H \times K \cong C_{pq}$, and $G$ would be abelian. So $H$ must be non-normal.

Answer 2

Since $G$ is non-Abelian the center $Z(G)$ must be trivial. Therefore, every conjugacy class except $\langle 1\rangle$ must be of order $p$ or $q$. If a subgroup of index $q$ (and therefore order $p$) is normal it must be the union of conjugacy classes. So we must have $p = 1+mp+nq$ for some non-negative integers $m$ and $n$ which is not possible.

Answer 3

By Sylow I, there is some $p$-Sylow subgroup $P$ of $G$. If $P$ is normal, then there is a homomorphism $\gamma:G\to \text{Aut}(H)$ defined by conjugation. Since $H$ is cyclic of order $p$, $|\text{Aut}(H)|=p-1$ and therefore $|\gamma(H)|$ must divide $p-1$ and $pq$ so $\gamma$ must be the trivial map. Thus, conjugation is trivial on $H$, and therefore $H\subseteq Z(G)$ (the center of $G$), but then the center is not trivial, so $G$ must be abelian (compute $|G/Z(G)|=p$ or $q$ so $G/Z(G)$ is cyclic and therefore trivial).