Simplifying presentations of modular subgroups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T06:07:48Z http://mathoverflow.net/feeds/question/105557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105557/simplifying-presentations-of-modular-subgroups Simplifying presentations of modular subgroups Jimeree 2012-08-26T17:33:43Z 2012-08-26T18:34:23Z <p>I've been using the Reidemeister-Schreier process (detailed in e.g. Holt et al. - <em>Handbook of Computational Group Theory</em>) 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):</p> <p>$\left\langle \left[a..y\right]|a,d,e,f,g,ho,p,q,kl,s,t,vx,b,c,i,j,m,n,wr,yu\right\rangle$</p> <p>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:</p> <p>$\left\langle h,k,w,x,y|\right\rangle$</p> <p>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.</p> <p>Thanks!</p> http://mathoverflow.net/questions/105557/simplifying-presentations-of-modular-subgroups/105565#105565 Answer by Will Sawin for Simplifying presentations of modular subgroups Will Sawin 2012-08-26T18:34:23Z 2012-08-26T18:34:23Z <p>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.</p>