Set of all n, nth roots of unity from a finite abelian group of order n with respect to multiplication

An nth root of unity is a complex number that, when raised to the power of n, equals 1. In other words, if z is an nth root of unity, then z^n = 1. The set of all nth roots of unity forms a cyclic subgroup of the group of non-zero complex numbers under multiplication.

Finite abelian group of order n

A finite abelian group of order n is a group that has n elements and satisfies the commutative property (i.e., the order in which elements are multiplied does not matter). The group operation can be addition or multiplication, but for simplicity, let's consider the group operation as multiplication.

Proof for the set of all n, nth roots of unity is a subgroup

To prove that the set of all n, nth roots of unity forms a subgroup, we need to show that it satisfies the three conditions for being a subgroup: closure, identity, and inverses.

Closure: Let z and w be two nth roots of unity. We need to show that their product zw is also an nth root of unity. Since z^n = w^n = 1, we have (zw)^n = z^n * w^n = 1 * 1 = 1. Therefore, zw is also an nth root of unity.

Identity: The identity element of the group is 1, which is also an nth root of unity since 1^n = 1.

Inverses: Let z be an nth root of unity. We need to show that there exists an inverse element for z, denoted as z^-1, such that zz^-1 = 1. Since z^n = 1, we can write z^-1 = z^(n-1). Therefore, zz^-1 = z * z^(n-1) = z^n = 1.

The set of all n, nth roots of unity forms a subgroup of the abelian group of order n.

The set of all n, nth roots of unity satisfies the closure, identity, and inverses conditions, which means it forms a subgroup of the finite abelian group of order n with respect to multiplication.

Therefore, the correct answer is option 'C' (multiplication).