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

Just consider two facts.


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*If $a \mid b$, then $\mathbb G_a \subseteq \mathbb G_b$; this gives you the right-to-left inclusion.

*If $\alpha^{n} = 1 = \alpha^{m}$, then $\alpha^{\gcd(m,n)} = 1$; this gives you the left-to-right inclusion.


For the latter, you will need Bézout's identity.
