# Proof that $\langle y\rangle$ is a normal subgroup of $G$ with order $q$ and $|G|=pq$ with $p$,$q$ prime

Let $$p, q$$ prime, with $$p and $$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?

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

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

## 1 Answer

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 $$G$$.

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