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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 $f$:

$$f(z)=\frac{az+b}{cz+d}$$

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?

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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
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1 Answer

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.

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the additive group of complex numbers? –  Federica Maggioni Dec 8 '12 at 17:34
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