Suppose that $\phi : F_n \to F_m$ is a surjective homomorphism between free groups of rank $n$ and $m$, respectively. Is the kernel necessarily finitely normally generated? If so, can we say anything about the number of normal generators that are needed in terms of $n$ and $m$?
Normal generation of the kernel of a surjection of free groups
free-groupsgroup-theory
Best Answer
The answer is yes, and $n$ is an upper bound on the required number of normal generators. I don't know whether there is a better bound (such as $n-m$ perhaps)?
Let $x_i$ ($1 \le i \le n$) and $y_j$ $(1 \le j \le m)$ be free generators of $F_n$ and $F_m$, and choose elements $g_j \in F_n$ with $f(g_j) = y_j$ for $1 \le j \le m$.
Now for each $x_i$, we can write $f(x_i)$ as a word $w_i(y_j)$ in $y_j^{\pm 1}$, and $z_i := x_i^{-1} w_i(g_j) \in \ker f$ for $1 \le i \le n$.
I claim that $\ker f$ is normally generated by $\{ z_i : 1 \le i \le n \}$. Let $N = \langle z_i \rangle^{F_n}$, so $N \le K$, and we want to prove equality.
Now, since any word in the $x_i$ can be rewritten modulo $N$ as a word in the $w_i(g_j)$, we see that $F/N$ is generated by the elements $w_i(g_j)$, and hence also by $\{ g_j : 1 \le j \le m \}$. So $F/N$ is isomorphic to a quotient of $F_m$ and, since free groups of finite rank are known to be Hopfian (see here for example), we have $K=N$ as claimed.