If you have a finitely generated group given by a presentation, is there a good method to determine a presentation for a subgroup generated by some subset of the generators given in the presentation? Specifically: I have found a presentation for the group $GL_2(\mathbb{Z})$, which is $$\bigl\langle R,T,X\;\bigm|\; X^2=T^2=(XT)^4=(RT)^3=(RTX)^3=(RXT)^3=1\bigr\rangle,$$ with the matrices given by $$R=\left(\begin{array}{cc} 0 & 1\\1 & 1\end{array}\right),\space T=\left(\begin{array}{cc} 1 & 0\\0 & -1\end{array}\right),\space X=\left(\begin{array}{cc} 0 & 1\\1 & 0\end{array}\right)$$ Can one easily find a presentation for the subgroup generated by R and T?

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    $\begingroup$ Yes. Look up the Reidemeister-Schreier rewriting process, which can be found in Magnus, Karrass and Solitar "Combinatorial Group Theory". There is a geometric alternative, which is easier, but I forget its name. If your subgroup has finite index, then it too will be finitely presented (assuming the first group is). Otherwise, anything can happen (well, within reason...)! $\endgroup$ – user1729 Aug 23 '11 at 18:20
  • $\begingroup$ I like to think of Reidemeister-Schreier as obtained by finding a 2-dimensional CW-complex whose fundamental group is the "big" group, then you find the covering space corresponding to the subgroup you're interested in, lift the CW-structure and compute its fundamental group. $\endgroup$ – Ryan Budney Aug 24 '11 at 3:20

Sorry for the comment with little explanation yesterday - my dinner was about to be on the table!

I decided it would be useful to work through the Reidemeister-Schreier rewriting process. So here it is!

So, we know that your subgroup has index 4, because of what Derek Holt has said. However, this is not too difficult to do by hand - you just end up with your (right) cosets being,

$H$, $Hx$, $Hxt$ and $Hxtx$.

(I've un-capitalised the matrices and have called the subgroup we want to find the presentation for $H$). Because your group has finite index it will have a finite presentation.

Note: finding the cosets is the annoying bit. If your subgroup is normal, however, then it is dead easy!

So basically you now want to do what Ryan Budney said. That is, draw your four cosets in a square (0-cells) and draw the lines where each generator sends each coset (1-cells). Index the generators by the cosets they leave from (so, for example, $x_{Hxtx}$ takes $Hxtx$ to $Hxt$). You should have $4$ lines labelled by an $x_i$, $4$ by a $t_i$ and $4$ by an $r_i$. (I apologise for not drawing a picture - I have no idea how to on here!)

Then add in your 2-cells. Basically, start at each of the four cosets and trace out where your original defining relations take each coset. Write down these paths (they should all be loops).

Your new group is the fundamental group of this 2-complex. So, your generators are the 0-cells, relations the 2-cells, and then take a collapsing tree.

Writing $x_0$ for $x_{H}$, $x_1$ for $x_{Hx}$, $x_2$ for $x_{Hxt}$ and $x_3$ for $x_{Hxtx}$ I got the following 2-cells,

$x_0x_1$, $x_2x_3$, $t_0^2$, $t_1t_2$, $t_3^2$, $x_0t_1x_2t_3x_3t_2x_1t_0$, $x_1t_0x_0t_1x_2t_3x_3t_2$, $x_2t_3x_3t_2x_1t_0x_0t_1$, $x_3t_2x_1t_0x_0t_1x_2t_3$, $(r_0t_0)^3$, $(r_1t_2)^3$, $(r_2t_1)^3$, $(r_3t_3)^3$, $r_0t_0r_1t_2x_1$, $r_2t_1x_2r_3t_3x_3$, $r_0x_0t_1r_2x_1t_0$, $r_1x_2t_3r_3x_3t_2$, $r_2x_1t_0r_0x_0t_1$, $r_3x_3t_2r_1x_2t_3$

and $x_0$, $t_1$, $x_2$ looks like a nice candidate for a collapsing tree.

Thus, we have the presentation,

$\langle x_1, x_3, r_0, r_1, r_2, r_3, t_0, t_2, t_3;$

$x_1, x_3, t_0^2, t_2, t_3^2, t_3x_3t_2x_1t_0, x_1t_0t_3x_3t_2, t_3x_3t_2x_1t_0, x_3t_2x_1t_0t_3, (r_0t_0)^3, (r_1t_2)^3, r_2^3, (r_3t_3)^3,$

$r_0t_0r_1t_2x_1, r_2r_3t_3x_3, r_0r_2x_1t_0, r_1t_3r_3x_3t_2, r_2x_1t_0r_0, r_3x_3t_2r_1t_3\rangle$

Noting that $x_1$, $x_3$ and $t_2$ are all trivial, and eliminating the appropriate elements (which takes quite a few steps) we get

$\langle r_0, r_3; (r_3r_0r_3)^2, (r_0r_3)^3\rangle$

which, applying the appropriate Tietze transformations, becomes

$\langle s, t; t^2, (rt)^3\rangle$

which is (thankfully - you have no idea how long this post took!) the same as Derek Holt's answer.


It would help if you got the presentation right! The final two relations should presumably be $(RTX)^2=(RXT)^2=1$. Using the modified presentation, an easy computer calculation in GAP or Magma shows that $\langle R,T \rangle$ has index 4 in $G$, and it is a free product of cyclic groups of orders 2 and 3, with defining relations $T^2 = (RT)^3=1$.


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