Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to find all linear fractional transformations that fix the points 1 and -1. In particular i'd like to give this set a group structure and see if it is some familiar group or not. I wrote conditions for such $f$:


and i found

$$a+b-c-d=0$$ $$a-b+c-d=0$$ which give $a=d$ and $b=c$, thus the determinant of associated matrix is $a^2-b^2$. And now, how can i go on?

share|cite|improve this question
An alternative way to solve your question is to consider the subgroup of linear fractional transformations fixing $0$ and $\infty$, which is isomorphic to your group. – 23rd Dec 8 '12 at 17:14
up vote 0 down vote accepted

The function $f$ does not change if you scale the coefficients by a constant, so you can just as well assume that $a^2-b^2=1$.

But this gives (after scaling by $-1$ if necessary) the hyperbolic rotations $$\begin{pmatrix} \cosh x & \sinh x \\ \sinh x & \cosh x \end{pmatrix}.$$

Now, you can just have a look at the change of $x$ when you multiply two matrices to understand the group structure.

share|cite|improve this answer
the additive group of complex numbers? – Federica Maggioni Dec 8 '12 at 17:34

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.