Here I prove every free group is free abelian. Where is the mistake?

Question

Let $F$ be free over $X$. Then for any group $G$ and any $\alpha: X \to G$ there is a homomorphism $\beta: F \to G$ such that $\alpha = \beta|X$. Alright. Now, in particular, when $G$ is abelian $F$ satisfies the definition of free abelian group. That is, if a group $F$ is free then $F$ is free abelian. What am I doing wrong?

Answer 1

Part of the definition of a free abelian group is that the group is abelian (not only are the target groups $G$ in the universal property restricted to abelian groups, but $F$ itself is required to be abelian). So your proof shows that if $F$ is a free group on a set $X$ and $F$ happens to be abelian, then $F$ is also a free abelian group on $X$. But $F$ may not be abelian (and in fact is not if $X$ has more than one element).