2
$\begingroup$

We know that if $\Omega$ be set of all 1-dimension subspaces of $V=V_{2}(q)$ which $V$ is a vector space on a finite field $GF(q)$ and so $|V|=q^{2}$ then, group $PGL_2(q)$ acts on $\Omega$. Also, it can be proved that the below set $G$ is a group acting on $\Delta$=$GF(q)\cup$ $\left\{\infty\right\}$ and it is isomorphic with group $PGL_2(q)$ as permutation groups:

$G= \left\{f_{abcd}|f_{abcd}:\Delta→\Delta, f_{abcd}(z)=\frac{az+b}{cz+d}, ad-bc≠0; a,b,c,d\in GF(q)\right\}$ and $\infty:→\frac{a}{c}$ , $\frac{-d}{c}$:→$\infty$ .

Honestly, I am probing the set for being a group and could't find the inverse element for an element in it. Thanks for any help.

1 Answers 1

4

Hint: Let $f$ and $g$ have two $2\times 2$ nonsingular matrices of coefficients associated to them. Compute the matrix associated to $f\circ g$, and compare with the matrix product of the original two matrices.

$f=\frac{az+b}{cz+d}~\leftrightarrow~ A=\begin{pmatrix}a&b\\c&d\end{pmatrix};\quad g=\frac{\alpha z+\beta}{\gamma z+\delta}\leftrightarrow B=\begin{pmatrix}\alpha &\beta \\\gamma&\delta\end{pmatrix}; $

$f\circ g\leftrightarrow \begin{pmatrix}?&?\\?&?\end{pmatrix};\qquad AB=?. $

If you know how to invert matrices, you should be able to explicitly invert these transformations.

(Alse see here.)


I suppose you also want to see more explicitly how this is linear-algebraic. Say we have

$f(z)=\frac{az+b}{cz+d},\quad f^{-1}(z)=\frac{pz+q}{rz+s}.$

Composing them together,

$\begin{array}{c l}f\circ f^{-1} &=\frac{a\frac{pz+q}{rz+s}+b}{c\frac{pz+q}{rz+s}+d} \\ & =\frac{a(pz+q)+b(rz+s)}{c(pz+q)+d(rz+s)} \\ & = \frac{(ap+br)z+(aq+bs)}{(cp+dr)z+(cq+ds)} \\ & = \frac{1z+0}{0z+1}.\end{array}$

Writing as a system,

$\begin{cases}ap+br=1 \\ aq+bs=0 \\ cp+dr=0 \\ cq+ds=1. \end{cases}$

In terms of matrices:

$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}p&q\\r&s\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}. $

Of course scalar multiplication of these matrices does not affect the underlying functions, so set

$\begin{pmatrix}p&q\\r&s\end{pmatrix}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}. $

  • 0
    Thanks for your complete help. Indeed, solving above system without assuming some restrictions would be a bit hard. Thanks again.2012-05-25