Proving the Kernel is a free group My goal is to show the kernel of the following is a free group with infinite generators, meaning there are no relations on the generators. 
If my group is the fundamental group with genus $g$ of surfaces
$S_g=⟨a_1,b_1,…,a_g,b_g\mid[a_1,b_1]...[a_g,b_g]=1⟩ $
and I have a map $H$ from my group to the integers:
$H: S_g\rightarrow\mathbb{Z}$
such that $a_1$ goes to $1$ and all other elements go to zero.
For example, 
$a_1b_1a_1^{-1}$ would be $1+0-1=0$ so that would be in the kernel along with any combination of elements which has the sum of the $a_1$ exponents equal to zero.
 A: To apply the Reidemeister-Schreier method, the first step is to find a set of coset representatives of the subgroup. For that we can take $T := \{ a_1^i : i \in {\mathbb Z} \}$.
If $X$ denotes the generating set of the group, then the subgroup generators in the presentation  are the nontrivial elements in the set $\{ hx\overline{hx}^{-1} : h \in T, x \in X \}$, where, for $g \in G$, $\bar{g}$ denotes the coset representative of $g$.
In this example, the subgroup generators are $a_{ij} := a_1^ja_ia_1^{-j}$ $(2 \le i \le g, j \in {\mathbb Z}$) and $b_{ij} := a_1^jb_ia_1^{-j}$ $(1 \le i \le g, j \in {\mathbb Z}$).
To get the relators of the subgroup presentation, we multiply each $a^i \in T$ on the right (I use right actions)  by the relators in the presentation of $G$   and rewrite the result as a word in the subgroup generators.
In this example, we have a single relator $[a_1,b_1][a_2,b_2]\cdots[a_g,b_g]$,
so we have to rewrite $a_1^j[a_1,b_1][a_2,b_2]\cdots [a_g,b_g]$ for each $j \in {\mathbb Z}$. 
I get (using the convention $[g,h] = g^{-1}h^{-1}gh$) the subgroup relators:
$$R_j := b_{1,j-1}^{-1}b_{1j}[a_{2j},b_{2j}] \cdots [a_{gj},b_{gj}].$$
So at the moment we have infinitely many relators, one for each $j \in {\mathbb Z}$, but we can use the $R_j$ with $j>0$ to eliminate the $b_{1j}$ with $j>0$, and the $R_j$ with $j \le 0$ to eliminate the $b_{1j}$ with $j<0$. So we end up with the free group on the generators $\{ b_{10},a_{ij},b_{ij} : 2 \le i \le g, j \in {\mathbb Z}\}$.
