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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 $H \cap K$.

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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.

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