Does the set of all #n#th roots of unity form a group under multiplication?
1 Answer
Sep 6, 2016
Yes
Explanation:

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#
Footnote
The elements of this group are all the numbers of the form:
#cos theta + i sin theta#
where