**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.15**

Prove that $\mathbb{Q} \times \mathbb{Q}$ is not cyclic.

Solution: Suppose $\mathbb{Q} \times \mathbb{Q}$ is cyclic. By Theorem 7 in the text, then, every subgroup of $\mathbb{Q} \times \mathbb{Q}$ is also cyclic. However, $\mathbb{Q} \times \mathbb{Q}$ has a subgroup isomorphic to $\mathbb{Z} \times \mathbb{Z}$, which we know to be not cyclic by Exercise 2.3.12; this is a contradiction.