For what values of $q$ is the action of $SL_2(\mathbb{F}_q)$ on $\mathbb{P}^1(\mathbb{F}_q)$ alternating?

Question

The group $SL_2(\mathbb{F}_q)$ acts on the projective line $\mathbb{P}^1(\mathbb{F}_q)$ (faithfully if $q$ is a power of 2, otherwise with kernel $\{\pm I\}$). We say this action is alternating if the corresponding map to $S_{q+1}$ actually lands in $A_{q+1}$.

For what values of $q$ is the action alternating? (It's easy to show by hand that it's not alternating for $q=2$ and that it is for $q=3, 4, 5$. I've checked it for a smattering of other small powers of $2$ and small primes by computer, and the answer appears to be for all $q \neq 2$, although it's possible I've made a coding error as ChatGPT gives a different answer (but with a bogus proof).)

Answer 1

You are right, the action is alternating for all $q>2$.

For $q>2$, $SL_2(\mathbb{F}_q)$ is generated by an element of order $3$ (an odd order permutation must be even) and elements of the form $\mathrm{diag}(a,a^{-1})$. The latter fix the element $1:0$ and act on the elements $b:1$ as $b:1\mapsto a^2b:1$, which is the permutation induced by the mapping $b:1\mapsto ab:1$ applied twice, and so is an even permutation too.

Answer 2

You have already checked $q\leq3$, so let $q>3$ be a prime power and $g\in\Bbb{F}_q^{\times}$ a generator. Then the group $\operatorname{SL}_2(q)$ is generated by the two matrices (see this preprint on arxiv for example): $$ S=\begin{pmatrix}g&0\\0&g^{-1}\end{pmatrix} \qquad\text{ and }\qquad T=\begin{pmatrix}-1&1\\-1&0\end{pmatrix}. $$ The matrix $S$ induces the map $$\Bbb{P}^1(\Bbb{F}_q)\ \longrightarrow\ \Bbb{P}^1(\Bbb{F}_q):\ (a:b)\ \longmapsto\ (ga:g^{-1}b),$$ where of course $(ga:g^{-1}b)=(g^2a:b)$. This shows that $S=R^2$ where $R$ is the permutation $$\Bbb{P}^1(\Bbb{F}_q)\ \longrightarrow\ \Bbb{P}^1(\Bbb{F}_q):\ (a:b)\ \longmapsto\ (ga:b),$$ so in particular $S\in A_{q+1}$. Similarly, the matrix $T$ induces the map $$\Bbb{P}^1(\Bbb{F}_q)\ \longrightarrow\ \Bbb{P}^1(\Bbb{F}_q):\ (a:b)\ \longmapsto\ (b-a:-a),$$ where of course $(b-a:-a)=(a-b:a)$. Then $$T^3(a:b)=T^2(a-b:a)=T(-b:a-b)=(-a:-b)=(a:b),$$ which shows that $T^3=I$. This means that $T$ maps to a product of disjoint $3$-cycles in $S_{q+1}$, and so $T\in A_{q+1}$. As the generators of $\operatorname{SL}_2(q)$ map into $A_{q+1}$ it follows that the entire group $\operatorname{SL}_2(q)$ maps into $A_{q+1}$, so indeed $\operatorname{SL}_2(q)$ is alternating if and only if $q>2$.