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 $\mathbb G_n \bigcap \mathbb G_m = \mathbb G_{(m:n)}.$

$$d=(n,m)\implies \begin{cases}\exists\;a,b,\in\Bbb Z\;\;s.t.\;\;am+bn=d\\{}\\m=rd\;,\;\;n=sd\;,\;\;r,s\in\Bbb Z\end{cases}$$
$$x\in\Bbb G_n\cap \Bbb G_m\implies x^n=x^m=1\implies x^d=(x^m)^a(x^n)^b=1\implies x\in\Bbb G_d$$
OTOH
$$x\in\Bbb G_d\implies \begin{cases}x^m=(x^d)^r=1\\{}\\x^n=(x^d)^s=1\end{cases}\implies x\in\Bbb G_n\cap\Bbb G_m$$
