View Single Post
P: 9
 Quote by jbunniii If xH and yK have nonempty intersection, then there is an element g contained in both: $g \in xH$ and $g \in yK$. The cosets of $H \cap K$ form a partition of G, so g is contained in exactly one such coset, call it $a(H \cap K)$. If you can show that $a(H \cap K)$ is contained in both $xH$ and $yK$ then you're done. Hint: both $xH$ and $yK$ are partitioned by cosets of $H \cap K$.
Yeah...I get it. Thanks very much. In addition, how to prove part (b), that is how can I show that both $H$ and $K$ are partitioned by finite cosets of $H \cap K$... I appreciate your insightful answer!