Explanation:

To show that the correct answer is option 'C', we need to prove that a finitely generated subgroup G of the additive group Q of rational numbers is cyclic.

Definition: A group G is cyclic if there exists an element g in G such that every element of G can be written as a power of g.

Proof:

Step 1: Let's assume that G is generated by the elements {a1, a2, ..., an}, where ai are rational numbers.

Step 2: We can write any element g in G as a linear combination of the generators {a1, a2, ..., an} with integer coefficients. This is because the rational numbers are closed under addition and subtraction.

Step 3: Let's consider the element g = a1/k, where k is the least common multiple of the denominators of a1, a2, ..., an. Since k is a positive integer, g is a rational number.

Step 4: We need to show that every element in G can be written as a power of g.

Step 5: Let's consider an arbitrary element h in G. Since h is in G, it can be written as a linear combination of the generators {a1, a2, ..., an} with integer coefficients: h = c1*a1 + c2*a2 + ... + cn*an, where ci are integers.

Step 6: Now, let's consider the element h' = (c1*a1 + c2*a2 + ... + cn*an)/k. Since k is the least common multiple of the denominators, h' is a rational number.

Step 7: We can rewrite h' as h' = (c1*a1/k) + (c2*a2/k) + ... + (cn*an/k), which is a linear combination of g.

Step 8: Therefore, h' is an element of the subgroup generated by g.

Step 9: Since h' is an arbitrary element of G, this means that every element in G can be written as a power of g.

Conclusion: Therefore, the finitely generated subgroup G of the additive group Q of rational numbers is cyclic. Hence, the correct answer is option 'C'.