This question is Question 10.20 of Abstract Algebra: A First Course by Dan Saracino.
Suppose $G$ is a group and $H<G$, $K<G$, and $g \in G$. Prove $HgK$ is the union of exactly $|H: H\cap gKg{-1}|$ of the left cosets of $K$ in $G$.
I know the usual proof:
$$|HgK|=|HgKg^{-1}g|=|HgKg^{-1}|=\frac{|H||gKg^{-1}|}{|H\cap gKg^{-1}|}=\frac{|H||K|}{|H\cap gKg^{-1}|}=[H: H\cap gKg{-1}]|K|$$
However, I want a proof using group action. Similar to the elegant proof of Order of a product of subgroups. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$. by Martin Brandenburg.
Best Answer
$H$ acts on the set $G/K$ of cosets of $K$ in $G$ by left multiplication. The stabilizer of $gK$ is given by $H \cap gKg^{-1}$ and thus the orbit length is $[H:H\cap gKg^{-1}]$.