0
$\begingroup$

Let's say that we have finite field $\mathbb F_q$ and we have a couple of involutions $g,f$ with exactly one fixed point (zero).

Let's take any element $\alpha \in \mathbb F_q$ Let's start applying involutions to this element in the following way: $$ \ldots f(g(\alpha)) \ldots g(\alpha) \ldots \alpha \ldots f(\alpha) \ldots g(f(\alpha)) \ldots $$ Let's imagine that on some step the element on the left side become equal to the element on the right side for any $\alpha$.

What is the name for such couple of involutions with this property ?

Are there any theorems regarding existence of such couple of involutions ?

The example of such couple of involutions is here: Chains of numbers generated by 2 involutions.

Improve this question
$\endgroup$
5
  • $\begingroup$ The most important question is: Let's $f=1/x$, how to find all corresponding $g$ $\endgroup$
    – Alexander
    Commented Dec 22, 2020 at 14:23
  • 1
    $\begingroup$ If I am reading your question correctly, isn't it the case that any pair of involutions have this property. There are only finitely many permutations of $\mathbb{F}_{q}$. Hence for some $n$, we have that $(fg)^{n}$ is the identity permutation. Then $(gf)^{n}$ is also the identity permutation, since $gf = g(fg)g^{-1}.$ $\endgroup$
    – Geoff Robinson
    Commented Dec 22, 2020 at 14:40
  • $\begingroup$ Correct, you are right. $\endgroup$
    – Alexander
    Commented Dec 22, 2020 at 14:43
  • $\begingroup$ I suggest to close question. $\endgroup$
    – Alexander
    Commented Dec 22, 2020 at 14:43
  • $\begingroup$ The interesting questions here is about $n$ and $q=2^k-1$. Found couple of involutions when $n | (q-1)/k$ $\endgroup$
    – Alexander
    Commented Dec 22, 2020 at 14:54

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .