Question

3. Let A = {1, 5, 7, 11}. Show that (A, *) is a group, where * is a binary operator defined as x * y = (x ⋅ y) mod 24. Determine the identity and the inverse of each element. 4. Consider the group A in Exercise 3, a. Prove or disprove that A has a subgroup of order 2. b. Prove or disprove that A has a subgroup of order 3.

Answer 1

To show that (A, *) is a group, we need to verify four conditions: closure, associativity, identity, and inverses.

1. Closure: For any x, y in A, x * y = (x ⋅ y) mod 24. Since the product of any two elements in A is also in A, closure is satisfied.

2. Associativity: The binary operator * is associative if (x * y) * z = x * (y * z) for all x, y, z in A. Since multiplication and modulo operations are associative, the binary operator * is also associative.

3. Identity: An identity element e in A should satisfy x * e = e * x = x for all x in A. Let's check each element in A:

1 * 1 = (1 ⋅ 1) mod 24 = 1 mod 24 = 1 (not the identity)

5 * 5 = (5 ⋅ 5) mod 24 = 25 mod 24 = 1 (not the identity)

7 * 7 = (7 ⋅ 7) mod 24 = 49 mod 24 = 1 (not the identity)

11 * 11 = (11 ⋅ 11) mod 24 = 121 mod 24 = 1 (not the identity)

None of the elements in A satisfy the condition for an identity element, so (A, *) does not have an identity element.

4. Inverses: For each element x in A, there should exist an element y in A such that x * y = y * x = e, where e is the identity element. Since (A, *) does not have an identity element, it also does not have inverses.

Therefore, (A, *) does not form a group.

For exercise 4:

a. To prove that A has a subgroup of order 2, we need to find two elements in A whose binary operation results in the identity element (which we determined does not exist). Therefore, A does not have a subgroup of order 2.

b. To prove that A has a subgroup of order 3, we need to find three elements in A whose binary operation results in the identity element (which we determined does not exist). Therefore, A does not have a subgroup of order 3.