Prove that the $n$th roots of unity form a group under multiplication

Question

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:

1. Clasure: if $a\in{G}$ and $b\in{g}$, then $a*b\in{G}$.
2. Associativity: $a*(b*c)=(a*b)*C$ for all $a,b,c\in{G}$.
3. There is an element $e\in{G}$ (called the identity element) such that $a*e=a=e*a$ for every $a\in{G}$.
4. 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:

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

Please help...

Answer 1

The set $G$ is the set$$\left\{\cos\left(\frac{2k\pi}n\right)+i\sin\left(\frac{2k\pi}n\right)\,\middle|\,k\in\{0,1,\ldots,n-1\}\right\},$$since $z^k=\cos\left(\frac{2k\pi}n\right)+i\sin\left(\frac{2k\pi}n\right)$. On the other hand\begin{multline}\left(\cos\left(\frac{2k\pi}n\right)+i\sin\left(\frac{2k\pi}n\right)\right)\left(\cos\left(\frac{2k'\pi}n\right)+i\sin\left(\frac{2k'\pi}n\right)\right)=\\=\cos\left(\frac{2(k+k')\pi}n\right)+i\sin\left(\frac{2(k+k')\pi}n\right).\end{multline}Can you take it from here?

Answer 2

$$G=\{1,z,z^2,...,z^{n-1}\}$$ 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.

Answer 3

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: $z^j z^k = z^{j+k} = z^{k+j} = z^k z^j$.

Answer 4

Let $\mathbb{Z}_n=\{0,1,2,\ldots,n-1\}$ and $G$ be the set of all $n$-th roots of unity. Mathematically, \begin{equation} G=\left\{w_k=\exp\left(i \frac{2k\pi}{n}\right)\bigg| ~k\in \mathbb{Z}_n\right\}. \end{equation}

Closure: Take $k,~l\in \mathbb{Z}_n$. \begin{eqnarray} w_k w_l&=&\exp\left(i \frac{2k\pi}{n}\right)\exp\left(i \frac{2l\pi}{n}\right)\nonumber\\ &=&\exp\left(i \frac{2(k+l)\pi}{n}\right) \end{eqnarray} By division algorithm, $k+l=qn+r$ with $q\in \mathbb{Z}$ and $0\leq r<n$. \begin{eqnarray} w_k w_l&=&\exp\left(i \frac{2(qn+r)\pi}{n}\right)\\ &=&\exp\left(2q\pi i\right)\exp\left(i \frac{2r\pi}{n}\right)\\ &=&w_r\\ &=&w_{(k+l)\text{mod}~n} \end{eqnarray}

Associativity: Associativity follows from $\mathbb{C}$ since $G\subset \mathbb{C}$. Or take $k,~l,~m\in \mathbb{Z}_n$. \begin{eqnarray} w_k (w_l w_m)&=&w_k w_{(l+m)\text{mod}~n}\\ &=&w_{\left(k+(l+m)\text{mod}~n\right)\text{mod}~n}\\ &=&w_{\left((k+l)+m\right)\text{mod}~n}\\ &=&w_{\left((k+l)\text{mod}~n+m\right)\text{mod}~n}\\ &=&w_{(k+l)\text{mod}~n}w_m\\ &=&(w_kw_l)w_m \end{eqnarray}

Identity: Take $k \in \mathbb{Z}_n$. \begin{eqnarray} w_k w_0&=&w_{(k+0) \text{mod}~n}\\ &=&w_{k~\text{mod}~n}\\ &=&w_k \end{eqnarray} Similarly, for all $k\in \mathbb{Z}_n$, $w_0 w_k=w_k$. Therefore, $w_0=1$ is the identity element.

Inverse: If $k=0$, then \begin{equation} w_0 w_0=1. \end{equation} Now consider $k \in \mathbb{Z}_n$ with $k\neq 0$. Since $1\leq k <n$, then $n-k\in \mathbb{Z}_n$, and \begin{eqnarray} w_k w_{n-k}&=&w_{n~\text{mod}~n}\\ &=&w_0\\ &=&1 \end{eqnarray} Therefore, the inverse of $w_k$ is $w_{n-k}$.

Moreover, since for all $k, l \in \mathbb{Z}_n$, \begin{equation} w_k w_l=w_l w_k \end{equation} this group is commutative or, abelian.

Further, since for all $k\in \mathbb{Z}_n$, $w_k=w_1^k$, \begin{equation} G=\langle w_1 \rangle, \end{equation} in other words, $G$ is cyclic.