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 $\mathbb G_n \bigcap \mathbb G_m = \mathbb G_{(m:n)}.$
 A: 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.
A: Just consider two facts.


*

*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.
A: $$d=(n,m)\implies \begin{cases}\exists\;a,b,\in\Bbb Z\;\;s.t.\;\;am+bn=d\\{}\\m=rd\;,\;\;n=sd\;,\;\;r,s\in\Bbb Z\end{cases}$$
$$x\in\Bbb G_n\cap \Bbb G_m\implies x^n=x^m=1\implies x^d=(x^m)^a(x^n)^b=1\implies x\in\Bbb G_d$$
OTOH
$$x\in\Bbb G_d\implies \begin{cases}x^m=(x^d)^r=1\\{}\\x^n=(x^d)^s=1\end{cases}\implies x\in\Bbb G_n\cap\Bbb G_m$$
A: Consider $\mathbb G_{mn}$, which contains both $\mathbb G_n$ and  $\mathbb G_m$ as subgroups.
$\mathbb G_{mn}$ is a cyclic group of order $mn$. There is exactly one subgroup of each order dividing $mn$. Moreover, its lattice of subgroups is isomorphic to the lattice of divisors of $mn$. The required result is that the meet in both lattices correspond.
All this comes from the lattice of subgroups of $\mathbb Z$, since $G_n$ is a homomorphic image of $\mathbb Z$ and the lattice of subgroups of $\mathbb Z$ is isomorphic to the lattice $\mathbb N$ under divisibility.
