Kernel of a homomorphism from a free group into $\mathbb{Z}$. Let $F$ be a non-abelian free group, that is, a free group of rank at least $2$, and let $\phi: F \rightarrow \mathbb{Z}$ be a nontrivial group homomorphism. How to prove that the kernel of $\phi$ is not finitely generated?
 A: There is a useful result you could use to prove this.
Let $ F $ be a free group and $ E \leq F $ with $ |F:E| = \infty $. Suppose that $ \exists \: \{1\} \neq N \unlhd G $ with $ N \leq E $. Then the rank of $E$ is infinite.
The proof is as follows and assumes familiarity with the theory of Schreier generators of subgroups of free groups.
Let $F$ be free on $X$, let $ U $ be a Schreier transversal of $ E $ in $ F $ and, for $g \in F$, denote the element in $U \cap Eg$ by $\overline{g}$.
Let $ 1 \neq w = a_1 \cdots a_l \in N \leq E $, with  $a_i \in X^{\pm 1}$. For $ u \in U $, $ Euw = Euwu^{-1}u = Eu $, since $ uwu^{-1} \in N \leq E $. So $ \overline{uw} = u $, and $ uw \not\in U $, so there is a least $ k $ such that $ ua_1 \cdots a_k \notin U $. Let $ u_k := ua_1 \cdots a_{k-1} $. Then $ u_k \in U $ and $ u_ka_k \notin U $, so $ u_ka_k\overline{u_ka_k}^{-1} $ is not trivial. Since $ U $ is infinite and $l$ is
fixed, there is an infinite subset $ V $ of $ U $ and a fixed $k$ with
$1 \le k \le l$, such that $k$ is minimal with $u_ka_k \notin U$ for
all $u \in V$.  Then $ \left\{ u_ka_k\overline{u_ka_k}^{-1} : u \in V \right\} $ is an infinite subset of the set of Schreier generators of $E$, and hence $E$ has infinite rank.
A: For a completely different proof, that uses only the universal property of free groups, you could factor your map $\phi:F \to {\mathbb Z}$ as $\phi:F \to H \to {\mathbb Z}$, where $H$ is the wreath product ${\mathbb Z} \wr  {\mathbb Z}$, and observe that the kernel of $H \to {\mathbb Z}$ is infinitely generated.
A: This follows since the rank of any infinite-index subgroup of a nonabelian free group is infinite. One way to prove this is by covering space theory. An infinite index subgroup corresponds to a cover of a wedge of circles where every point has infinitely many preimages. Then, if the cover is not a tree, there is at least one loop, $L$. Acting on $L$ by the infinite group of deck transformations gives infinitely many copies of $L$, but some of them may intersect or even coincide. However, since each loop has finitely many edges, only finitely many copies of $L$ can intersect $L$. So we can produce infinitely many disjoint loops. Thus the graph has infinite rank.
