Question

(a) Write down all abelian groups of order 360 , up to isomorphism.

(b) Let G be an abelian group of order n, where n is square-free, i.e. it is not divisible by any square >1. Prove that G is cyclic.

Answer 1

To find all the abelian groups of order 360, factorize 360 into its prime factors. List down all possible orders of abelian groups using these prime factors. To prove that a cyclic group of order n, where n is square-free, is cyclic, show that there exists an element in the group that generates the entire group.

Step-by-step explanation:

(a) To find all the abelian groups of order 360, we first need to factorize 360 into its prime factorization. 360 can be expressed as 2^3 * 3^2 * 5. Using this, we can list down all possible orders of abelian groups using these prime factors: 2^3, 2^3 * 3, 2^3 * 5, 2^3 * 3^2, 2^2 * 3^2 * 5, 2 * 3^2 * 5, 2^2 * 3 * 5, 2 * 3 * 5.

These are the different abelian groups of order 360 up to isomorphism.

(b) To prove that a cyclic group of order n, where n is square-free, is cyclic, we need to show that there exists an element in the group that generates the entire group. Since the group is abelian, every element in the group commutes with each other.

This means that for each prime factor p in the prime factorization of n, there exists an element of order p in the group. By the Chinese Remainder Theorem, we can combine these elements to form an element of order n, which generates the entire group. Therefore, the group is cyclic.