2
$\begingroup$

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 $GL(2,\mathbb{Z}/p^n \mathbb{Z})$?

$\endgroup$
4
  • $\begingroup$ 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, 2016 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, 2016 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, 2016 at 18:06
  • $\begingroup$ Oh right, I see. Thanks, I'll change that now. $\endgroup$
    – user48096
    Feb 20, 2016 at 12:48

1 Answer 1

2
$\begingroup$

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 $p^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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