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.

  • $\begingroup$ 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? $\endgroup$ May 1, 2018 at 22:16
  • $\begingroup$ 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. $\endgroup$
    – Derek Holt
    May 1, 2018 at 22:45
  • $\begingroup$ @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! $\endgroup$ May 2, 2018 at 1:27

1 Answer 1


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

  • $\begingroup$ 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 $ker(H)$. $\endgroup$
    – JDZ
    May 30, 2018 at 5:28
  • $\begingroup$ $R_j$ 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. $\endgroup$
    – Derek Holt
    May 30, 2018 at 7:29
  • $\begingroup$ 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? $\endgroup$ Jun 2, 2018 at 21:30
  • $\begingroup$ It's the one left over because we eliminated $b_{1j}$ with $j>0$ and $j<0$. $\endgroup$
    – Derek Holt
    Jun 2, 2018 at 21:39

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