Prove that the $n$th roots of unity form a group under multiplication Let $n$ be a positive integer and let $z$ be the complex number $\cos\left(\frac{2\pi}{n}\right)+i\sin\left(\frac{2\pi}{n}\right)$. Prove that $G=\{1,z,z^2,...,z^{n-1}\}$ is a group under multiplication in $\mathbb{C}$.

This is a reference to Moivre's formula, because $z^n = 1$. But I do not know how to apply this result so that it complies with the group properties. I have thought about doing induction, but it becomes more extensive. 

According to the definition of Group.
A group is a nonempty set $G$ equipped with a binary operation $*$ that satisfies the following axioms:


Clasure: if $a\in{G}$ and $b\in{g}$, then $a*b\in{G}$.
Associativity: $a*(b*c)=(a*b)*C$ for all $a,b,c\in{G}$.
There is an element $e\in{G}$ (called the identity element) such that $a*e=a=e*a$ for every $a\in{G}$.
For every $a\in{G}$, there is an element $d\in{G}$ (called the inverse of $a$) such that $a*d=e$ and $d*a=e$.


A group is said to be abelian if it also satisfies this axiom:


Commutativity: $a*b=b*a$ for all $a,b\in{G}$.


Please help...
More generally, if $C$ is a group and $z \in C$ with $z^n=1$, where $1$ is the identity of $C$, then $G=\{1,z,z^2,\dots,z^{n-1}\}$ is a group (actually, a subgroup of $C$).

Indeed:


Closure: $z^j z^k = z^{j+k} = z^r \in G$, where $r = (j+k) \bmod n$.
Associativity: comes from $C$.
Identity: $1 \in G$ by definition, and $1$ is the identity of $C$.
Inverse: $z^j z^{n-j} =1$.
Commutativity: $z^j z^k = z^{j+k} = z^{k+j} = z^k z^j$.
$$G=\{1,z,z^2,...,z^{n-1}\}$$ Consists of the $n_{th}$ roots of unity.

Note that this is like rotating around the unit circle at an equal angles of $ \frac {2\pi }{n}$ 

It is closed under multiplication and every element has an inverse, and it has an identity element and the associativity is there.