This is simple and may have been asked before, but I couldn't find it.

I have been asked to 'Define the action of $SL_2(p)$ (the group of 2 by 2 matrices of determinant 1 with entries in $\mathbb{F_p}$, the integers mod some prime p) on $\mathbb{F_p} \cup {\infty}$ by Möbius transformations.

It seems to me that if one does this in the natural way, with $\frac{ax+b}{cx+d}$ for some $x\in\mathbb{F_p} \cup {\infty}$, then a lot of the time your result will no longer be in this set. Is this wrong, and how would you define the action if this is incorrect?

  • 2
    $\begingroup$ The answer will all of the time be in this set. Please give even one counterexample. $\endgroup$ – KCd May 10 '15 at 7:41
  • $\begingroup$ I think I've sorted it now, I was taking the fraction to be literally divide rather than a multiplicative inverse. Brain fuzz $\endgroup$ – George Moore May 10 '15 at 21:34
  • $\begingroup$ I'd say in $\mathbf F_p$ that it literally is division too. In any field $a/b = ab^{-1}$ (when $b \not= 0$). But I understand what you mean. $\endgroup$ – KCd May 10 '15 at 23:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.