If n∣q and A=z∈C:zn=1 B=z:zq=1 then

q=pn for some p∈Nzq−1=znp−1=zn−1z(p−1)n+⋯+zn+1Every root of zn−1 is a root of zq−1 and every root of z(p−1)n+⋯+zn+1=0 is also a root of zq−1 Hence. A⊂B∣and A∩B=A