# What does this group look like?

Let $F$ be a field of characteristic $p>0$, and consider the field rational fractions of $t$, $F(t)$. Consider these two automorphisms: $$\sigma :t \mapsto -t$$ $$\tau: t \mapsto 1-t$$ I managed to prove that $\left<\sigma,\tau\right>$ is actually finite (via induction). Here are some results: $$\sigma^2 = I = \tau^2$$ $$\sigma\tau: t \mapsto t-1$$ $$\tau\sigma: t \mapsto t+1$$ $$\tau\sigma\tau: t \mapsto 2-t$$ $$\sigma\tau\sigma: t \mapsto -t-1$$ $$\sigma\tau\sigma\tau: t \mapsto 2+t$$ $$\tau\sigma\tau\sigma: t \mapsto -2+t$$ $$\sigma\tau\sigma\tau\sigma: t \mapsto -2-t$$ $$\tau\sigma\tau\sigma\tau: t \mapsto 3-t$$ And so on until eventualy, some sequence of $\tau\sigma...\sigma\tau$ will reach $\pm p \pm t$, returning to $\sigma$ or to $I$. My questions are: How many elements does this group have? It clearly depends on $p$. Also, what group is it isomorphic to? I also managed to find this element in $\text{Fix}(\left<\sigma,\tau\right>)$: $$q(t) = (t^p - t)^2$$ However it is not clear to me whether $\text{Fix}(\left<\sigma,\tau\right>) = F(q)$

• In general, any group generated by two elements -- both of which have order 2 -- is dihedral. Oct 29, 2017 at 7:01

For odd $p$, it is the dihedral group of order $2p$, consisting of $t\mapsto\pm t+a$ for $a\in\Bbb F_p$.
$F(t)$ therefore has degree $2p$ over the fixed field of the group.
• Why "odd"? Does it not work for $p=2$? Oct 29, 2017 at 13:31
• If $p = 2$ It is called the Klein 4 group. Or $\mathbb Z_2 \times \mathbb Z_2$ Oct 29, 2017 at 17:04