Union of $n$-th roots of unity is a subgroup of the complex unit circle under multiplication (proof) I showed that $(\upsilon_n,\cdot)$ is a subgroup of $(u,\cdot)$ where 
$$
\upsilon_n=\Big\{z\in \upsilon: z^n=1,\,n\in\mathbb{N}\Big\}
\quad\text{and} \quad \upsilon=\Big\{z\in\mathbb{C}:|z|=1\Big\}.
$$
How can I show that the group 
$(\cup_n\upsilon_n,\cdot)
$ is also a subgroup of $(\upsilon_n,\cdot)$ where $\cup_n\upsilon_n$ is the union of $\upsilon_n$ for all $n \in \mathbb{N},\, n\geq 1$.
It's known that if $G$ is a group and $K$,$L$ are subgroups of $G$ then $K\cup L$ is a subgroup of $G$ if and only if  $K$ is contained in $L$ or $L$ is contained in $K$.
Given that $\upsilon_k$ is not contained in $\upsilon_m$ for $k\neq m$, how is $\cup_n\upsilon_n$ a subgroup of $\upsilon$?
 A: I will denote the $n$-th roots of unity $G_n$. First, a couple of remarks:


*

*It is not true that $G_k \not \subset G_m$ if $k \neq m$. For example $G_2 \subseteq G_4$ and more generally, if $n | m$, then we always have $G_n \subseteq G_m$ (Hint: if $m = nd$, take $\omega \in G_n$ and compute $\omega^m$).

*As sets, $G_\infty := \bigcup_n G_n$ will contain each $G_m$, so it cannot be a subgroup the latter. However, we do have $\bigcup_n G_n \subseteq S^1 := \{\omega : |\omega| = 1\}$, so there is no contradiction. 
As for the actual proof, we clearly have $1 \in G_\infty$ because $1 \in G_1$. Now, if $z,z' \in G_\infty$, by definition there exist $n,m \in \mathbb{N}$ such that $z \in G_n$ and $z' \in G_m$. Thus, 
$$
(zz')^{nm} = (z^n)^m(z'^m)^n = 1
$$
and so $zz' \in G_{mn} \subseteq G_\infty$. Finally, since $z \in G_n$, then $z^{-1} \in G_n \subseteq G_\infty$. This proves that $G_\infty$ contains the identity, and it is closed with respect to inverses and products, which by definition means that it is a subgroup.
