Question

Show that the nth roots of unity forms an abelian group of finite order with usual multiplication

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Answer 1

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.

Answer 2

The `n`th roots of unity form a finite abelian group (under regular complex number multiplication).

It's finite because the equation `z^n=1` has exactly `n` solutions. Even without knowing there are exactly `n`, it can still be said that there are at most `n`, which is really all we need to know to say it's a finite group.

It's abelian because the `n`th roots of unity form a subset of the complex numbers, and the complex numbers are abelian with respect to multiplication.

I hope it will help you dear friend...