Prove that if then is nonabelian.
First we show that the free group of rank 2 is nonabelian. Suppose to the contrary that is abelian where ; say . Define by , . By the universal property of free groups, there exists a unique group homomorphism . This homomorphism is clearly surjective, so that by the First Isomorphism Theorem, . However, every quotient of an abelian group is abelian, but if , is not abelian. Thus is nonabelian.
Clearly every free group of rank at least 2 contains a subgroup which is free of rank 2. (For instance, use the universal property.) Thus no free group of rank at least 2 can be abelian.