Roots of Unity under Multiplication form Cyclic Group

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Theorem

Let $n \in \Z$ be an integer such that $n > 0$.

The $n$th roots of unity under the operation of multiplication form the cyclic group which is isomorphic to $C_n$.


Proof

From Roots of Unity, we have:

$U_n = \left\{{e^{2 i k \pi / n}: k \in \N_n}\right\}$

where $U_n$ is the set of $n$th roots of unity.

Let $\omega = e^{2 i \pi / n}$.

Then we have:

$U_n = \left\{{\omega^k: k \in \N_n}\right\}$

that is:

$U_n = \left\{{\omega^0, \omega^1, \omega^2, \ldots, \omega^{n-1}}\right\}$

Let $\omega^a, \omega^b \in U_n$.

Then $\omega^a \omega^b = \omega^{a+b} \in U_n$.

Either $a + b < n$, in which case $\omega^{a+b} \in U_n$, or $a + b \ge n$, in which case:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \omega^a \omega^b\) \(=\) \(\displaystyle \omega^{a+b}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \omega^{n+t}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          for some $t < n$          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \omega^n \omega^t\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \omega^t\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          as $\omega^n = 1$          

So $U_n$ is closed under multiplication.

We have that $\omega_0 = 1$ is the identity and that $\omega^{n-t}$ is the inverse of $\omega^t$.

Finally we note that $U_n$ is generated by $\omega$.

Hence the result, by definition of cyclic group, and from Cyclic Groups Same Order Isomorphic:

$U_n = \left \langle \omega \right \rangle \cong C_n$.

$\blacksquare$


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