Does the set of all $n$ th roots of unity form a group under multiplication?

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Answer 1 · 2016-09-06T21:49:31

Yes

Identity: $1$ is the identity.
Inverse: If $a$ is an $n$ th root of unity, then so is $1/a$ , since:

$(1/a)^n = 1/(a^n) = 1/1 = 1$
Closure under product: If $a$ is an $m$ th root of unity and $b$ an $n$ th root of unity, then $ab$ is an $mn$ th root of unity:

$(ab)^(mn) = (a^m)^n(b^n)^m = 1^n*1^m = 1*1 = 1$
Associativity: Inherited from the complex numbers:

$a(bc) = (ab)c " "$ 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 $pi$ .

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