prove that the set of all nth roots of a unity forms afinite abelian group of order n with respect to multiplication
Answers
Answer:
It is a proof. Prove that the five properties of Abelian group are satisfied for the group of n th roots.
Step-by-step explanation:
1 = exp(2 pi i) = e^{2 pi i} = Cos(2 pi) + i Sin (2 pi)
Here i = square root of -1 = imaginary number.
The n roots of 1 can be given as :
1^{1/n} = ( exp(2 pi i} )^{1/n}
= Cos (2 pi k/n) + i Sin (2 pi k/n) , for k = 0, 1, 2, .. n-1
of the form A + i B
Abelian group or commutative group of numbers is a set S with an operation * satisfying five properties: Closure, commutative, associative, identity, and inverse. To prove it for the group of nth roots of 1 with the usual math multiplication operator.
Let X = Cos(2 pi k/n) + i Sin(2 pi k/n), Y = Cos(2 pi j/n)+ i Sin(2 pi j/n),
and Z = Cos(2 pi m/n) + i Sin(2 pi m/n), where k,j,m = 0,1,2,...,n-1.
1. Closure :
X * Y = Cos(2 pi (k+j)/n) + i Sin(2 pi (k+j)/n) = Cos(2 pi r/n) + i Sin(2 pi r/n)
where r = (k+j) mod n = remainder after dividing k+j with n So r = 0, 1, 2, ..., n-1,
because Cos(2 pi+ A) = Cos A and Sin(2 pi + A) = Sin A.
Result of multiplication is a complex number and is a n th root of 1.
2. Commutative property:
To prove X * Y = Y * X.
X * Y = Cos(2 pi (k+j)/n) + i Sin(2 pi (k+j)/n) = Cos(2 pi r/n) + i Sin(2 pi r/n),
Y * X = Cos(2 pi (j+k)/n) + i Sin(2 pi (j+k)/n) = Cos(2 pi r/n) + i Sin(2 pi r/n),
r = 0, 1, 2, ... , n-1.
3. Associative property:
To prove X * ( Y * Z ) = (X * Y ) * Z.
We get that product to be:
Cos(2 pi (k+j+m)/n) + i Sin(2 pi (k+j+m)/n) = Cos(2 pi r/n) + i Sin(2 pi r/n),
r = 0, 1, 2, ... , n-1.
4. Identity:
1 or unity is the multiplicative identity as 1 * X = X * 1 = X
5. Inverse:
Multiplicative inverse of X is Y where : X * Y = Y * X = identity = 1
Y = { Cos (2 pi (n-k)/n) + i Sin (2 pi (n-k)/n) }
So X * Y = Y * X = Cos(2 pi) + i Sin (2 pi) = 1.
Since all properties are satisfied, the nth roots form an Abelian group of n th order with the usual multiplication operator.