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$\DeclareMathOperator\GF{GF}$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*}

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    $\begingroup$ 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 $\endgroup$
    – François Brunault
    Commented Oct 3, 2020 at 19:28
  • $\begingroup$ Can one not prove by just plugging in and doing the algebra that $f(g(f(x)))=g(f(g(x)))$? $\endgroup$
    – Gerry Myerson
    Commented Oct 4, 2020 at 10:32
  • $\begingroup$ Cross-ratio can be used in finite field as well $\endgroup$
    – Alexander
    Commented Oct 5, 2020 at 10:34
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    $\begingroup$ please do not vandalize the question, it has received an answer which would make no sense if the question is deleted. $\endgroup$
    – Carlo Beenakker
    Commented Oct 10, 2020 at 7:59
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    $\begingroup$ 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. $\endgroup$
    – Gerry Myerson
    Commented Oct 10, 2020 at 11:27

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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 $0,1,\infty$.

This group also reflects the different values of the cross-ratio when you permute the arguments: https://en.wikipedia.org/wiki/Cross-ratio#Six_cross-ratios

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  • $\begingroup$ Great answer !!! Cross ratios - was the abstraction I was looking for. Thanks. $\endgroup$
    – Alexander
    Commented Oct 5, 2020 at 14:15

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