TITLE: Intersection of cosets
QUESTION [7 upvotes]: I'm not really sure how to go about proving the following. Any help will be appreciated. 
Question:
Let $H$ and $K$ be subgroups of a group $G$. Prove that the intersection 
$$xH \cap yK$$ 
of two cosets of $H$ and $K$ is either empty or else is a coset of the subgroup $H \cap K$.

REPLY [13 votes]: Suppose $x H \cap y K \ne \emptyset$, and let $z \in x H \cap y K$. Then $z \in x H$, so that $z H = x H$, and similarly $z K = y K$, so
$$
x H \cap y K = z H \cap z K = z (H \cap K),
$$
as the map $u \mapsto z u$ on $G$ is bijective, and thus preserves intersections.