TITLE: Why is "A free group on more than one generator is not abelian." true?
QUESTION [9 upvotes]: Let $F$ be the free group on the generating set $A$. Note that $F$ is not abelian if $A$ contains more than one element.

How am I suppose to note this? Doesn't $Z\times Z$ has 2 elements in it's generating set (more than one element) but abelian?
How?

REPLY [11 votes]: Let $u,v$ be any two distinct elements of $A$. Suppose that $F$ is commutative. 
Let $G$ be any group, and $x,y$ be any two elements of $G$.
There is a group homomorphism $\varphi:F \to G$ such that $\varphi(u) = x$ and $\varphi(v) = y$. Therefore,
$$ xy = \varphi(uv) = \varphi(vu) = yx $$
Thus, we've proven that every group is abelian.
Since this is not true, our assumption that $F$ is commutative must be false.