If $n\mid q\text{ and }A=\left\{z\in C:z^n=1\right\}$ $B=\left\{z:z^q=1\right\}$ then

$q=pn$ for some $p\in \mathbf{N}$

$z^q-1={\left(z^n\right)}^p-1=\left(z^n-1\right)\left(z^{(p-1)n}+\cdots +z^n+1\right)$

Every root of $z^n-1$ is a root of $z^q-1$ and every root of $z^{(p-1)n}+\cdots +z^n+1=0$ is also a root of $z^q-1$ Hence. $A\subset B\mid$

and $A\cap B=A$