# 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 $F_2$.

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As Jonas Meyer points out, every quotient of $F_2$ is necessarily $2$-generated, and if $S$ is a generating set for $G$, then the image of $S$ is a generating set for $G/N$ for any $N$. Since $F_3$ has quotients which are $3$- but not $2$-generated, the answer is "no". In fact, you cannot have $F_k$ as a quotient of $F_{\ell}$ for any $\ell\lt k$ for precisely that reason (even going all the way to $k=\omega$). –  Arturo Magidin Dec 13 '10 at 21:34

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.

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Of course! I suppose it is trivial to see, since cosets are given by representatives "upstairs," which are spelled out with 2 generators in this case, so $F_2/H$ is generated by $\{aH,bH\}$ –  Andrew Marshall Dec 13 '10 at 21:46
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.