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Feb29-12, 01:56 AM
P: 9
Quote Quote by jbunniii View Post
If xH and yK have nonempty intersection, then there is an element g contained in both: [itex]g \in xH[/itex] and [itex]g \in yK[/itex].

The cosets of [itex]H \cap K[/itex] form a partition of G, so g is contained in exactly one such coset, call it [itex]a(H \cap K)[/itex].

If you can show that [itex]a(H \cap K)[/itex] is contained in both [itex]xH[/itex] and [itex]yK[/itex] then you're done.

Hint: both [itex]xH[/itex] and [itex]yK[/itex] are partitioned by cosets of [itex]H \cap K[/itex].
Yeah...I get it. Thanks very much. In addition, how to prove part (b), that is how can I show that both [itex]H[/itex] and [itex]K[/itex] are partitioned by finite cosets of [itex]H \cap K[/itex]... I appreciate your insightful answer!