Proving the Group Properties of M, the Set of Nth Roots of Unity

In summary, The conversation is about solving a set of mathematical problems involving the nth roots of unity and showing that it forms a group with complex number multiplication. The person asking for help is willing to provide explanations and is seeking understanding. They ask for the four things needed to show that something is a group and inquire about the specific part the person is stuck on.
  • #1
ustus
2
0
Hello,
Please help in solving the four set of problems, i will be very happy explaining comment as really want to understand.
The problem will spread to the extent of understanding preduduschey.

1 Problems:

The set M, M = {e^(j*2*pi*k/n) , k= 0,1,2...n-1} denotes the set of the nth roots of unity,
i.e. the solution set of z^n = 1 for fixed n.
Show that M, together with multiplication of complex numbers, forms a group.
 
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  • #2
ustus said:
Hello,
Please help in solving the four set of problems, i will be very happy explaining comment as really want to understand.
The problem will spread to the extent of understanding preduduschey.

1 Problems:

The set M, M = {e^(j*2*pi*k/n) , k= 0,1,2...n-1} denotes the set of the nth roots of unity,
i.e. the solution set of z^n = 1 for fixed n.
Show that M, together with multiplication of complex numbers, forms a group.

Ok. To show something is a group, what four things do you need to show?

Which one(s) are you stuck on?
 
  • #3
Are the following evidence to the problem?
 

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1. What is the definition of the set of Nth roots of unity?

The set of Nth roots of unity, denoted as M, is a mathematical set that contains all complex numbers that, when raised to the power of N, equal 1. In other words, M is the set of solutions to the equation x^N = 1.

2. How do you prove that M is closed under multiplication?

To prove that M is closed under multiplication, we need to show that for any two elements a and b in M, their product ab is also in M. This can be done by using the property of exponents, where (a^N)(b^N) = (ab)^N. Since a and b are both Nth roots of unity, their product ab is also an Nth root of unity, thus satisfying closure under multiplication.

3. What is the identity element of M?

The identity element of M is 1. This is because any number raised to the power of 0 is equal to 1. Therefore, 1 is an Nth root of unity and is the identity element of M.

4. How do you prove that every element in M has an inverse?

To prove that every element in M has an inverse, we need to show that for any element a in M, there exists another element b in M such that ab = 1. This can be done by taking the Nth root of 1/a, which will result in the inverse of a. Since a is an Nth root of unity, 1/a will also be an Nth root of unity, satisfying the existence of an inverse in M.

5. Can you give an example of a group that is isomorphic to M?

Yes, the group of complex numbers under multiplication, denoted as C*, is isomorphic to M. This is because both groups have the same structure and follow the same group properties, such as closure under multiplication, associativity, and the existence of an identity and inverse element. The only difference is that the elements in C* are all complex numbers, while the elements in M are only the Nth roots of unity.

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