Final answer:
To prove that the nth roots of unity form a cyclic subgroup, we show the set is closed under multiplication with an identity element and inverses, and that it has a generator. These roots are complex numbers that satisfy the equation z^n = 1 and form a subgroup under multiplication.
Step-by-step explanation:
To prove that the nth roots of unity form a cyclic subgroup of the complex numbers under multiplication, consider the complex numbers of the form e^(2πik/n), where i is the imaginary unit, k is an integer, and n is a positive integer. The nth roots of unity are the solutions to the equation z^n = 1, which can be represented as cos(2πk/n) + isin(2πk/n), for k = 0, 1, 2, ..., n-1. They form a group under multiplication because the product of any two nth roots of unity is another nth root of unity, and this group is cyclic because there is a generator element, specifically any primitive nth root of unity, that can generate all other elements by repeated multiplication.
Steps to Prove
- Show that the set of nth roots of unity is non-empty and closed under multiplication.
- Verify that the nth roots of unity have an identity element (1) and that each element has an inverse (its complex conjugate).
- Demonstrate that the nth roots of unity form a group under multiplication.
- Identify a generator of the group, which is any primitive nth root of unity.
In summary, the nth roots of unity satisfy all the properties of a cyclic subgroup within the group of complex numbers under multiplication.