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I'm trying to solve the following problem but am not sure of the best way to start or the best strategy to use. Let $F$ be a finite field with $q$ elements, and let $K = F(x)$ be the field of rational functions over $F$. Let $G$ be the group of all automorphisms $\sigma$ of $K$ such that $$\sigma(x) = \frac{ax + b}{cx + d}$$ where $a, b, c, d \in F$ and $ad - bc\neq 0$.

(1) Show that the order of $G$ is $q^3 - q$.

(2) Show that the fixed field $K^G$ of $G$ is $F(y)$, where $$y = \frac{(x^{q^2} - x)^{q+1}}{(x^q - x)^{q^2 + 1}}.$$

Any hints or strategies would be appreciated!

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Since $F$ has $q$ elements there's $q^4$ choices for values of $a$, $b$, $c$, and $d$. Therefore $|G| \leq q^4$.

The requirement that $ad-bc \neq 0$ along with counting the number of values of $a$, $b$, $c$, and $d$ that give equivalent $\sigma$'s reduces $|G|$ below $q^4$. Try to start counting those possibilities.

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  • $\begingroup$ Can you actually provide another hint as to how to count these possibilities? $\endgroup$ – User7819 Feb 15 '17 at 17:13

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