Why is "A free group on more than one generator is not abelian." true? 
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?
 A: $\mathbb Z\times \mathbb Z$ is the free abelian group of two generators, but it is not the free group of two generators. The free group of two generators $F(a,b)$ is the set of all (reduced) finite strings containing the letters $a, b, a^{-1},$ and $b^{-1}$, whose group operation is concatenation-and-reduction. It, then, should be clear that $ab$ and $ba$ are two different elements of $F(a,b)$, and so this group is not abelian. 
A: By definition of a free group, all words in the alphabet $a,b,\ldots ,$ of $A$ are different if they are not identical after cancelling all occurences of $xx^{-1}$, etc. In particualr, the words $w_1=ab$ and $w_2=ba$ are different. Hence the group is non-abelian for rank at least $2$.
A: 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.
