Explanation:
To understand why an abelian group of order 24 has exactly one subgroup of order 3, we need to consider the structure of abelian groups and their subgroups.

Abelian Groups:
An abelian group is a group in which the group operation is commutative. In other words, for any elements a and b in the group, a * b = b * a. Examples of abelian groups include the group of integers under addition and the group of real numbers under addition.

Subgroups:
A subgroup of a group G is a subset of G that forms a group under the same operation. In other words, a subgroup of G is a smaller group that retains the same group structure as G. For example, the set of even integers is a subgroup of the group of integers under addition.

Order of a Group:
The order of a group is the number of elements in the group. For example, the order of the group of integers under addition is infinite, while the order of the group of real numbers under addition is also infinite.

Sylow's Theorem:
Sylow's theorem states that if a prime number p divides the order of a finite group G, then G contains a subgroup of order p. In other words, for any prime number p, there exists a subgroup of G whose order is a power of p.

Applying Sylow's Theorem:
Since we are given that the abelian group has order 24, we can consider the prime factorization of 24: 24 = 2^3 * 3. According to Sylow's theorem, there must exist subgroups of order 2^3 = 8 and order 3.

However, since the group is abelian, all subgroups are normal subgroups. This means that the subgroups of order 2^3 = 8 and order 3 must intersect trivially, i.e., their intersection contains only the identity element.

Now, let's consider the subgroup of order 3. Since it intersects trivially with the subgroup of order 8, it must contain all elements of order 3 in the group. Any element of order 3 generates a cyclic subgroup of order 3. Therefore, there can only be one subgroup of order 3 in the abelian group of order 24.

Hence, the correct answer is option A: exactly one subgroup of order 3.