## 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.