$\bigcup_{n=1}^{\infty}G_n $ is a multiplicative subgroup of the group $\mathbb C\setminus\{0\}$ 
If $G_n$ is the set of all $n$-th roots of unity for $n\in \mathbb{N}$ then show that $\bigcup_{n=1}^{\infty}G_n  $ is a multiplicative subgroup of the group $\mathbb{C} \setminus \{0\}$, the order of every element of which is finite.

I can prove $G_n$ is a group. I know def of subgroup. But unable to solve the problem. Please help me to solve by using elementary knowledge of group theory. 
 A: Hint: The product of an $m$-th root of unity and an $n$-th root of unity is an $mn$-th root of unity. So if $a$ is in $G_m$ and $b$ is in $G_n$, their product $ab$ is in $G_{mn}$, and therefore in $\bigcup_{i=1}^\infty G_i$.
A: Because you weren't clear about what you have a question on, I will try to add many details. Let $G=\cup_{n=1}^\infty G_n$. Recall that $|z^n|=|z|^n$ so that if $z\in G$ then there exists $n$ so that $z\in G_n$, equivalently $z^n=1$. It follows then that $|z|=1$ whenever $z\in G$. This shows that $G\subset \mathbb{S}^1\subset \mathbb{C}\backslash \{0\}$.
Now to show this is a group (under the usual multiplication operation $•$). There are four things we have to show: 1) algebraic closure (if $a,b\in G$ then $a•b\in G$); 2) associativity $a•(b•c)=(a•b)•c$ 3) the existence of an identity ($e$ such that $e•x=x=x•e$) 4) the existence of an inverse $y$ for every element $x$ such that $xy=yx=e$ we usually write this $y$ as $x^{-1}$. Below I will write $xy$ for $x•y$.
For the first requirement, the other answer suffices. Without rehashing all the details again, the point is if $x,y\in G$ then there exists $n,m$ such that $x\in G_n$ and $y\in G_m$ and therefore (from Andre's answer) $xy\in G_{nm}\subset G$. This shows $G$ is closed.
Associativity is assumed in the definition of complex multiplication, and we are viewing $G$ as a subset of $\mathbb{C}$ in such a way that it inherits this multiplication operation. Therefore, associativity is automatic.
If $e=1$ (or more precisely $e=1+0i$) then $e$ clearly satisfies $ez=z=ze$ for all $z\in G$ since $G\subset \mathbb{C}$ and again this operation is the usual operation. So we just have to check that $e\in G$; but $e^1=1$ so $e\in G_1\subset G$.
Now we check the existence of inverses. If $z\in G$ then there exists $n$ so that $z\in G_n$ equivalently $z^n=1$.
$$1=z^n=z(z^{n-1})=(z^{n-1})z$$
Therefore the inverse of $z$ is $z^{n-1}$. Now we just have to check that $z^{n-1}\in G$. But this is easy:
$$(z^{n-1})^n=(z^n)^{n-1}=1^{n-1}=1$$
Therefore, $z^{n-1}\in G_n\subset G$
I can't think of anything left to check, this is the "full painful details" approach; please be more specific in the future rather than just saying, "I'm confused explain more."
