For c: conversely given aH'=bH multiplying on both sides by b^(-1) we get b^(-1)aH'=H .....(1) since H is a sub group thus LHS is also the same subgroup Also since H' is also a subgroup then there exists an identity e' of H' such that b^(-1)a*e' = b^(-1)a Thus b^(-1)a belongs to b^(-1)aH' thus inverse also belongs to b^(-1)a Since aH = H for all a in H Thus (b^(-1)a)^(-1)*(b^(-1)a)H' = b^(-1)aH' H'=b^(-1)aH' By equation (1) H'=H