Question

Let \( A = \{0, 1, 2, 3, \dots\} \) be the set of non-negative integers. Let \( F \) be the set of functions from \( A \) to itself. For any two functions, \( f_1, f_2 \in F \), we define \[ (f_1 \circ f_2)(n) = f_1(n) + f_2(n) \] for every number \( n \in A \). Which of the following is/are CORRECT about the mathematical structure \( (F, \circ) \)?

• A. \( (F, \circ) \) is an Abelian group.
• B. \( (F, \circ) \) is an Abelian monoid.
• C. \( (F, \circ) \) is a non-Abelian group.
• D. \( (F, \circ) \) is a non-Abelian monoid.

Hint:

To check if a set with an operation forms a monoid or group, verify associativity, the existence of an identity element, and for a group, check if every element has an inverse.

Answer 1

The Correct Option is B

We are asked to determine which of the following statements about the mathematical structure \( (F, \circ) \) is correct, given the operations defined on the set \( F \) of functions from \( A \) to itself.

1. Monoid:
A monoid is a set equipped with an associative operation and an identity element. Let's check the properties for \( (F, \circ) \):
Associativity:
The operation \( \circ \) is defined as \( (f_1 \circ f_2)(n) = f_1(n) + f_2(n) \), which is the addition of two functions' values at each point. Addition of integers is associative. Therefore, the operation is associative.

Identity Element:
The identity element for this operation would be a function \( e(n) \) such that \( (f_1 \circ e)(n) = f_1(n) \) for any function \( f_1 \). The function \( e(n) = 0 \) (the zero function) works as the identity because \( f_1(n) + 0 = f_1(n) \).

Therefore, \( (F, \circ) \) is a monoid.

2. Abelian Property:
An operation is Abelian if it is commutative, i.e., \( (f_1 \circ f_2)(n) = (f_2 \circ f_1)(n) \). Since \( (f_1 \circ f_2)(n) = f_1(n) + f_2(n) \) and addition of integers is commutative, the operation \( \circ \) is commutative.

Hence, \( (F, \circ) \) is an Abelian monoid.

3. Group:
A group requires the existence of an inverse element for every element in the set. In this case, for every function \( f_1 \in F \), we would need a function \( f_2 \) such that \( (f_1 \circ f_2)(n) = 0 \) for all \( n \). This would mean that \( f_2(n) = -f_1(n) \). Since the set \( F \) consists of functions mapping to non-negative integers, there is no function that can serve as the inverse of another. Therefore, \( (F, \circ) \) is not a group.

Conclusion:
Thus, the correct answer is \( \boxed{B} \): \( (F, \circ) \) is an Abelian monoid.