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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).

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    $\begingroup$ So why is $G$ isomorphic to $\mathbb{Z}_{pq}$? I thought the theorem was for nonabelian groups? $\endgroup$ – Bruce Zheng Oct 8 '14 at 2:59
  • $\begingroup$ 'Prove that a non-abelian group of order pq has a nonnormal subgroup of index q, so there there eixists and injective homomorphism into Sq' $\endgroup$ – pretzelman Oct 8 '14 at 5:43
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    $\begingroup$ I got that part. Line 2 of your proof states 'then $G \simeq \mathbb{Z}_{pq}$'. But $\mathbb{Z}_{pq}$ is cyclic and abelian so how can $G$ possibly be isomorphic to it? $\endgroup$ – Bruce Zheng Oct 8 '14 at 5:45
  • $\begingroup$ I'm sorry sir, you are correct. I've edited the post. Please let me know if anything else is wrong! Thanks! $\endgroup$ – pretzelman Oct 8 '14 at 6:38
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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.

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  • $\begingroup$ This is very simple and clean! Thank you for the first line; I used it in my proof (hope you don't mind). I would like to stick with mine so long as its faults are curable (thought this was VERY insightful). Thanks! $\endgroup$ – pretzelman Oct 8 '14 at 6:42
  • $\begingroup$ No problem, you are welcome! $\endgroup$ – Nicky Hekster Oct 8 '14 at 7:30
  • $\begingroup$ I don't understand how your first paragraph shows that there exists a subgroup $H$ of order $p$. Could you explain? $\endgroup$ – frito_mosquito Jul 24 '18 at 14:31
  • $\begingroup$ I think you mean "in the second paragraph"? This is again an application of Cauchy's Theorem. $p$ divides the order of $G$. There was a typo I just corrected in the first paragraph, thanks. $\endgroup$ – Nicky Hekster Jul 24 '18 at 15:05
  • $\begingroup$ Yes, of course. Thanks. $\endgroup$ – frito_mosquito Jul 24 '18 at 18:13

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