# Does the set of all nth roots of unity form a group under multiplication?

Sep 6, 2016

Yes

#### Explanation:

• Identity: $1$ is the identity.

• Inverse: If $a$ is an $n$th root of unity, then so is $\frac{1}{a}$, since:

${\left(\frac{1}{a}\right)}^{n} = \frac{1}{{a}^{n}} = \frac{1}{1} = 1$

• Closure under product: If $a$ is an $m$th root of unity and $b$ an $n$th root of unity, then $a b$ is an $m n$th root of unity:

${\left(a b\right)}^{m n} = {\left({a}^{m}\right)}^{n} {\left({b}^{n}\right)}^{m} = {1}^{n} \cdot {1}^{m} = 1 \cdot 1 = 1$

• Associativity: Inherited from the complex numbers:

$a \left(b c\right) = \left(a b\right) c \text{ }$ for any $a , b , c$

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Footnote

The elements of this group are all the numbers of the form:

$\cos \theta + i \sin \theta$

where $\theta$ is a rational multiple of $\pi$.