Describing the Kernel of a map from The Fundamental Group to the integers

Question

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, how do I write the kernel?

I know that everything except $a_1$ is in the kernel obviously but in this case

$a_1b_1a_1^{-1}$ would be $1+0-1=0$

so this is in the kernel as well, along with anything else that combines elements in a similar way. So I know the kernel is infinite, but how do I write it mathematically?

Answer 1

I guess, you mean all other generators, i.e. $b_1,a_2,\dots,b_q$, by 'all other elements' and 'everything except $a_1$'.

In your equation with $1-0+1$, the element $a_1$ can be replaced by any element of the group: $H(xb_1x^{-1})=0$. That is - such as every kernel - $\,\ker H$ will be a normal subgroup.
Specifically, it is the normal subgroup $N$ generated by $b_1,a_2,b_2,\dots,a_q,b_q$.

Assume that $w\in S_g$ is in the kernel, take a represantative word and group together anything that is not $a_1$, obtaining $w=x_0a_1^{t_1}x_1a_1^{t_2}\dots$ with $t_i\in\Bbb Z$ and $x_i\in N$. Then the assumption implies $\sum_it_i=0$. Finally, bring all $a_1^{t_i}$ to e.g. leftmost, by exchanging like $\ x_0a_1^{t_1}=a_1^{t_1}(\underbrace{a_1^{-t_1}x_0a_1^{t_1}}_{y_0})\ $ where $y_0$ is also in $N$.