Show that if $HK$ is an abelian group ($H$,$K$ have coprime orders), then $H$ is in $N_G(K)$

Question

My question stems form this excerpt from Dummet. I follow all of it, except the claim that $PQ$ is a subgroup of the normalizer. I feel like the solution is some simple algebraic one but i cannot seem to find it.

Answer 1

Let $h\in H$. To show $h\in N_G(K)$ we have to show that $hK=Kh$. So let $x\in hK$. Then we have $x=hk$ for some $k\in K$. Since $HK$ is abelian, also $x=hk=kh\in Kh$. So this shows $hK\subseteq Kh$. Similarly we can show the reverse inclusion.