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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.

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up vote 4 down vote accepted

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

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

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Thanks for the answer. But, could't we consider the elements as they are shown in the set $G$ and so trying to find inverses? – Babak S. May 25 '12 at 19:21
@Babak: Sure, it'd be the exact same thing. But why go through extra work if we can recognize the group in terms of familiar objects and operations for which we already know how to do computations? If you have $f$ and explicitly try to solve for the coefficients of $f^{-1}$, it will be the same as working to explicitly solve the coefficients of the inverse matrix. – anon May 25 '12 at 19:23
@BabakSorouh I added some extra if you're interested. (And had to edit my edit because apparently I can't remember the letters of the alphabet correctly...) – anon May 25 '12 at 19:42
Thanks for your complete help. Indeed, solving above system without assuming some restrictions would be a bit hard. Thanks again. – Babak S. May 25 '12 at 20:52

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