Chains of numbers generated by 2 involutions

$$\DeclareMathOperator\GF{GF}$$Consider the finite field $$\GF(p)$$ for prime $$p$$.

Consider the pair of involutions $$f(x) = 1-x$$ , $$g(x) = 1/x$$, and the chain of numbers generated by these 2 involutions in the following way: $$\cdots f(g(f(x))) \leftarrow g(f(x)) \leftarrow f(x) \to x \to g(x) \to f(g(x) \to g(f(g(x)) \cdots$$

Apparently the maximal length of this chain for specific $$x$$ is equal to 6.

Could you please explain if this construction has some special name in mathematics, or was studied in the theory of finite fields?

For example for $$\GF(31)$$ we have:

$$\begin{gather*} 12 \leftarrow 20 \leftarrow 14 \to 18 \to 19 \to 13 \to 12 \\ 12, 13, 14, 18, 19, 20. \end{gather*}$$

• These two involutions generate a group of order 6 isomorphic to the symmetric group $S_3$ (this is true over any field $k$, viewing these two involutions acting on the projective line $\mathbb{P}^1(k)$). This corresponds to the different values of the cross-ratio when you permute the arguments: en.wikipedia.org/wiki/Cross-ratio#Six_cross-ratios – François Brunault Oct 3 '20 at 19:28
• Can one not prove by just plugging in and doing the algebra that $f(g(f(x)))=g(f(g(x)))$? – Gerry Myerson Oct 4 '20 at 10:32
• Cross-ratio can be used in finite field as well – Alexander Oct 5 '20 at 10:34
• please do not vandalize the question, it has received an answer which would make no sense if the question is deleted. – Carlo Beenakker Oct 10 '20 at 7:59
• This is a site for questions of math research, @AVT. The $-2$ probably came from users who feel your question had no research angle. There's a fair chance that at some point other users will vote to close it, and then to delete it. Or, maybe not; maybe users will feel the answer redeems the question. But in any event, vandalizing the question is a significant breach of this website's norms. Please don't do it. – Gerry Myerson Oct 10 '20 at 11:27

The involutions $$x \to 1-x$$ and $$x \to 1/x$$ generate a group of order 6 isomorphic to the symmetric group $$S_3$$. This is true over any field $$k$$, viewing these two involutions acting on the projective line $$\mathbb{P}^1(k)$$. One way to see this: these involutions leave stable $$\{0,1,\infty\}$$, and any linear fractional transformation of $$\mathbb{P}^1(k)$$ is determined uniquely by its action on $$0,1,\infty$$.