Proof that $\mathbb G_n \bigcap \mathbb G_m = \mathbb G_{(m:n)}$ Being $\mathbb G_n$ the roots of unity for $n \in \mathbb N$, prove that $\mathbb G_n \bigcap \mathbb G_m = \mathbb G_{(m:n)}.$

Hint
Recall
$$z\in\mathbb G_n\iff z^n=1$$
By double inclusion 


*

*if $d|n$ prove that 
$$z\in\mathbb G_d\Rightarrow z\in\mathbb G_n$$
and hence which inclusion we can deduce?

*Use the Bézout's theorem:
$$d=\gcd(m,n)\iff \exists u,v\in\mathbb Z,\; um+vn=d$$
to get the other inclusion.
