Free groups of rank at least 2 are nonabelian

Prove that if |S| > 1 then F(S) is nonabelian.


First we show that the free group of rank 2 is nonabelian. Suppose to the contrary that F(S) is abelian where |S| = 2; say S = \{a,b\}. Define \varphi : S \rightarrow D_{2n} by \varphi(a) = r, \varphi(b) = s. By the universal property of free groups, there exists a unique group homomorphism \Phi : F(S) \rightarrow D_{2n}. This homomorphism is clearly surjective, so that by the First Isomorphism Theorem, D_{2n} \cong F(S)/\mathsf{ker}\ \Phi. However, every quotient of an abelian group is abelian, but if n \geq 3, D_{2n} is not abelian. Thus F(S) 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.

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