Suppose $H$ is a finitely generated normal subgroup of a finitely generated free group $F(X)$ with basis $X = \{ x_1 , \dots , x_n \}$. Fix $1 \leq i \leq n$.
Is it true that the set of right cosets $\{ Hx_i^k \ | \ k \in \mathbb{Z} \}$ is finite? In other words, does the sequence $Hx_i , Hx_i^2 , Hx_i^3 , \dots $ terminate?
I have reason to believe this is true, but I'm not sure where to begin. I think this has to do with finite generation, but I'm not quite sure where that comes into play.
My first idea was that, because $H$ is normal, a coset $Hx_i^m$ is equal to the left coset $x_i^{-m} H$, which is an element of the quotient group $F/H$. In other words, what I'm asking is equivalent to showing that the element $x_iH$ has finite order in the group $F/H$.
Does anyone have any idea how one could show such a thing?
This is true. An easy way to prove it is to use Stallings' core.
A subgroup is finitely generated iff its core is finite. A subgroup is of finite index iff its core is a finite cover of the bouquet of $n$ circles. A subgroup is normal iff the core is $2n$-regular and "looks the same at every vertex". So a subgroup is normal and finitely generated $\to$ it is of finite index. See, for example, this text (Theorems 8.3, 8.11).