In 'A presentation of $PGL(2,p)$ with three defining relations' by E.F.Robertson and P.D.Williams, we can find a presentation of $PGL(2,p)$:

$\langle a,b | a^2 = b^p = (a b^2 a b^r)^2 = (abab^r)^3 = 1 \rangle$, s.t. $r$ is a square-free primitive element of $\mathbb{Z}/p \mathbb{Z}$.

Where we can associate $a$ to $\left( \begin{array}{cc} 0 & -r \\ 1 & 0 \\ \end{array} \right)$ and $b$ to $\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)$

It seems it is easy to extend this to $GL(2,\mathbb{Z}/p \mathbb{Z})$ by introducing another generator, $c$, which would be associated to the matrix $\left( \begin{array}{cc} r & 0 \\ 0 & r \\ \end{array} \right)$:

$\langle a,b,c | a^2 = b^p = c^{p-1} = (a b^2 a b^r)^2 = (abab^r)^3 = 1, ac=ca, bc=cb \rangle$

My question is if we can extend this again to $GL(2,\mathbb{Z}/p^n \mathbb{Z})$?

  • $\begingroup$ There is something wrong with your presentation. Since $a,b$ have determinant $2$ and $1$, all their powers have determinant a power of $2$. If $2$ is a square modulo $p$, this implies that $a,b$ is not a system of generator. $\endgroup$
    – Joël
    Feb 19 '16 at 0:33
  • $\begingroup$ Does it matter if 2 is a square? I thought that the important part is that it generates the group $\mathbb{Z}/p \mathbb{Z}$? $\endgroup$
    – user48096
    Feb 19 '16 at 16:22
  • $\begingroup$ [The point is that diag(x,1) = (word in a,b)*const implies, on taking determinants, that x is a square if 2 is a square.] $\endgroup$
    – alpoge
    Feb 19 '16 at 18:06
  • $\begingroup$ Oh right, I see. Thanks, I'll change that now. $\endgroup$
    – user48096
    Feb 20 '16 at 12:48

Since $GL(2,Z/p^nZ)$ is an iterated extension of $C_p^{2^2}$ by $GL(2,Z/pZ)$, one could form a presentation by adding further generators and modifying the existing relations (e.g. $a^2=1$ now becomes $a^2=n$ for a suitable element of the normal subgroup) -- this is a general technique for building presentations.

Since the pattern of repeated powering for larger $n$ is very regular, it will be sufficient to spell out the rules once for $p^2$.


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