The free group $F_3$ being a quotient of $F_2$ Every finitely generated free group is a subgroup of $F_2$, the free group on two generators.  This is an elementary fact, as is the fact that $G$, finitely presented, is the quotient of $F(|S|)$ the free group on some set of generators $S$ for $G$.  
My question is whether $F_3$, and hence any finitely presented group, is a quotient of $F_2$.
 A: Here's a slightly different way, perhaps a little more sophisticated, to see this.
Free groups are Hopfian, meaning that every surjective endomorphism is an isomorphism.  There are a variety of ways to prove this.  It's proved in Lyndon & Schupp, using Nielsen transformations. Alternatively, you can appeal to an (easy) result of Malcev, which states that every finitely generated, residually finite group is Hopfian.
Now, there is an obvious epimorphism $F_3\to F_2$ with non-trivial kernel, given by killing a generator.  If $F_3$ were a quotient of $F_2$, the composition of these two maps would give an epimorphism $F_2\to F_2$ with non-trivial kernel.
A: If $F_3$ were a quotient of $F_2$, then $\mathbb{Z}^3$ would be, but $\mathbb{Z}^3$ cannot be generated by fewer than $3$ elements. To me it seems easier to see directly that $\mathbb{Z}^3$ needs at least $3$ generators than the corresponding statement for $F_3$, perhaps because it's easy to visualize.
The rank of a group is the smallest cardinality of a generating set.  Here's a list of some facts about ranks of groups (including that the rank of $F_3$ is $3$) on Wikipedia.
