# 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.

• Why do you think that it will be free? My first random guess: it looks like $b_1, a_2, \dots b_g$ generate the kernel, and that since the kernel is a subgroup, this is $S_{g-1}*\mathbb Z$ for each $g>1$. Either way, why do you think that it might be free? May 1, 2018 at 22:16
• There is a general result that all subgroups of infinite index in surface groups are free. But you could do this one directly using the reidemeister-Schreier method. May 1, 2018 at 22:45
• @DerekHolt holt the reidemeister-schreier method is what I was pointed to but I can't wrap my head around how to use it. If you can give me some tips on how to work with it then that would be very helpful! May 2, 2018 at 1:27

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 $\{ b_{10},a_{ij},b_{ij} : 2 \le i \le g, j \in {\mathbb Z}\}$.

• How does the subgroup $ker(H)$ end up being free (i.e. having no relators)? I thought the $R_j$ were the relators in the presentation of $ker(H)$.
– JDZ
May 30, 2018 at 5:28
• $R_j$ is used to eliminate a generator and then is removed itself. For example if $R_j = ab^2cbc$ then you can use that to eliminate $a$ by substituting $(b^2cbc)^{-1}$ for $a$ in the other relators, and then remove the relator $R_j$ from the presentation. This is an application of Tietze transformations. In this example all of the relators are eliminated so you end up with a free group. May 30, 2018 at 7:29
• Is $B_{10}$ just a representative or is it the specific $B_{ij}$ that is left over? If it is specifically that one that remains then why is it so? Jun 2, 2018 at 21:30
• It's the one left over because we eliminated $b_{1j}$ with $j>0$ and $j<0$. Jun 2, 2018 at 21:39