Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

As is well known, the group $PSL(2,\mathbb Z)$ is isomorphic to the free product $C_2 \ast C_3$ of cyclic groups of order $2$ and $3$. Call the generators of the cyclic groups $S$ and $T$.

Problem: Given a prime number $p$ and a natural number $n$, write a presentation of the quotient $PSL(2, \mathbb Z/p^n\mathbb Z)$ with the images of $S$ and $T$ as generators.

share|improve this question

3 Answers 3

up vote 5 down vote accepted

I usually use Sunday's presentation: see MR0311782. His T has order 2 but your S will be what he denotes ST.

share|improve this answer
A direct link with no proxy: math.ca/10.4153/CJM-1972-118-x –  Steve D Oct 27 '11 at 19:32
@SteveD: Oops, thanks. –  Eric Rowell Oct 27 '11 at 23:29

A method to do this for the group $\textrm{PSL}_2 (\mathbb{F}_{p^n})$ can be found in the papers by Glover and Sjerve:

Representing $PSl_2(p)$ on a Riemann surface of least genus, L'Enseignement Mathématique 31 (1985)

The genus of $PSl_2(q)$, Journal für die reine und angewandte Mathematik 380 (1987).

share|improve this answer
For some reason I cannot put the correct link to the first paper. Anyway, it is available for free download: just go to unige.ch/math/EnsMath/EM_en/welcome.html and follow the link to the left at the bottom of the page –  Francesco Polizzi Oct 26 '11 at 23:46

The group $PSL_2(\mathbb{Z}/p^n)$ is the automorphisms group of the $(p+1)$ regular tree of depth $n$, where at level $m$ of the tree you have the points of $\mathbb{P}(\mathbb{Z}/p^m)$. The main benefit of this view, is that you can understand the relations at each level, and then move inductively to the next one.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.