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I've been using the Reidemeister-Schreier process (detailed in e.g. Holt et al. - Handbook of Computational Group Theory) to find the presentations of various modular subgroups. For example, this process tells us that the presentation of the principal congruence subgroup $\Gamma\left(4\right)$ is (before simplification):

$\left\langle \left[a..y\right]|a,d,e,f,g,ho,p,q,kl,s,t,vx,b,c,i,j,m,n,wr,yu\right\rangle$

I want to simplify this expression using Tietze transformations, again as described in Holt. However, doing so seems to give some worrying results. The first simplification is to remove all the "trivial" generators with monadic relators (e.g. $a,d,e,\ldots$). The second is to remove one of the two generators in each of the "double" relators, since the one will simply be the inverse of the other (i.e. delete one of $h$ or $o$, delete one of $k$ or $l$, etc.). However, doing these simplifications seems to leave the following presentation:

$\left\langle h,k,w,x,y|\right\rangle$

And this doesn't look like any good kind of presentation at all! If someone's able to tell me where I'm going wrong here, that would be hugely appreciated.

Thanks!

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    $\begingroup$ So, this a free group on five generators. What's wrong with this? $\endgroup$
    – Igor Rivin
    Commented Aug 26, 2012 at 18:30
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    $\begingroup$ An easier approach to computing presentations of congruence subgroups is to analyze their action on $\mathbb{H}$: if this action is properly discontinuous (which happens most of the time) then the corresponding congruence subgroup is the fundamental group of the quotient, which is a compact Riemann surface minus finitely many points (as in Will Sawin's answer). $\endgroup$
    – Qiaochu Yuan
    Commented Aug 26, 2012 at 18:43
  • $\begingroup$ In particular in that case it's a free group on $N/6+1$ generators, where $N$ is the index of that subgroup in $PSL_2(\mathbb Z)$, by an Euler characteristic argument. $\endgroup$
    – Will Sawin
    Commented Aug 26, 2012 at 20:10

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If I understand your presentation correctly, you have determined that $\Gamma(4)$ is a free group on $5$ generators. This is not surprising. The modular curve $X(4)$ is $\mathbb P^1$ with six cusps and no elliptic points, so its fundamental group is the free group on $5$ generators. You appear to have found one set of $5$ generators. There is nothing wrong with this.

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  • $\begingroup$ Yes, you're right. I don't know why I thought the presentation couldn't be right. Anyway, thanks! $\endgroup$
    – Jimeree
    Commented Aug 26, 2012 at 20:36

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