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**aman_cc** Index is 2. So there are just two right cosets - H itself and Hx where x doesn't belong to H. every element g in G belongs one of the two cosets (as they are mutually exclusive and exhaistive over G). So Hx = G - H

As there is one-to-one mapping between left and right cosets, similarly there are two left cosets H and yH. So yH = G - H = Hx

So for every x,y not in H, Hx = yH.

If x is in H, Hx = xH = H. This is very trivial prove using the fact H is a sub-group hence closed.