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Let $p, q$ prime, with $p<q$ and $G$ a group of order $pq$. Then by Cauchy's theorem $G$ contains elements $x$ and $y$ of order $p$ and $q$ respectively.

I have already proven that $\langle x, y\rangle=G$, since $\langle x, y\rangle$ is the group of all combinations of powers of $x$ and powers of $y$. (Correct?)

I now need to prove that $\langle y\rangle$ is a normal subgroup of $G$. So for every element $g\in G \mid g\langle y\rangle g^{-1}=\langle y \rangle$.

I tried some examples like $x^3y^5x^{-3}$, but I don't know how to work with this algebraically and if this is the correct course to a proof.

Can I use Sylow's theorems?

Edit: Sylow's theorems

So maybe I should show that $\langle y\rangle$ conjugated by elements of $G$ is also of order $p$ by (20.2)?

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    $\begingroup$ Do you know Sylow's theorems? $\endgroup$ – the_fox Oct 31 '18 at 12:17
  • $\begingroup$ See here for a proof (Thm. 3.8): math.uconn.edu/~kconrad/blurbs/grouptheory/cauchyapp.pdf $\endgroup$ – the_fox Oct 31 '18 at 12:22
  • $\begingroup$ @the_fox Yes, but I only thought of it because it also uses prime numbers. $\endgroup$ – The Coding Wombat Oct 31 '18 at 12:23
  • $\begingroup$ Very close to being a dupe of this. $\endgroup$ – Jyrki Lahtonen Oct 31 '18 at 12:31
  • $\begingroup$ You might want to spell that abbreviation as "dup". "Dupe" is a different word that has pejorative connotations. $\endgroup$ – C Monsour Oct 31 '18 at 14:32
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This is an instance of the following more general result.

If $G$ is a finite group of order $n$ and $p$ is the smallest prime divisor of $n$, then any subgroup $H$ of $G$ of index $p$ is normal in $G$.

Proof. Consider the left action of $G$ on the set of left cosets of $H$ in $G$, and let $\phi:G\to S_p$ be the corresponding group homomorphism. Let $K$ be the kernel of $\phi$. Then $K$ is normal in $G$ by construction, and by the first isomorphism theorem $\phi(G)\cong G/K$ is a subgroup of $S_p$. Therefore $[G:K]\big||S_p|=p!$ by Lagrange's theorem. On the other hand, $n=|G|=[G:K]\cdot|K|$ , also from Lagrange's theorem. Since $p$ is the smallest prime dividing $n$, you get that $[G:K]$ must divide $\gcd(n,p!)=p$. However, $K\supseteq H$ by construction, thus again from Lagrange's theorem $p=[G:H]=[G:K][K:H]$, which forces $[G:K]=p$ and $[H:K]=1$. Therefore $H=K$ is normal in $G$.

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  • $\begingroup$ And in my case the subgroup $\langle y \rangle$ of $G$ is of index $p$ because $\frac{|G|}{|\langle y \rangle |}=p$, correct? $\endgroup$ – The Coding Wombat Oct 31 '18 at 12:40
  • $\begingroup$ Yes, and because $p<q$. $\endgroup$ – Maurizio Moreschi Oct 31 '18 at 12:46
  • $\begingroup$ What does $S_p$ mean? $\endgroup$ – The Coding Wombat Oct 31 '18 at 12:49
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    $\begingroup$ Oh, sorry. I denote $S_p$ the group of permutations of a set with $p$ elements (in this case the set of left cosets of $H$ in $G$). $\endgroup$ – Maurizio Moreschi Oct 31 '18 at 12:51
  • $\begingroup$ So $S_p=G / H$? (the quotient group) $\endgroup$ – The Coding Wombat Oct 31 '18 at 13:03

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