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Given a pair of non-commuting linear transformations, $A$ and $B$, define the "next pair" in a sequence as $A*B$ and $B*A$. I am interested in finite cycles (i.e., the sequence eventually returns to the original $A$ and $B$). Is this studied somewhere? What is known about the maximum order of such cycles? (By the way, it is easy to see that my original "non-commuting" requirement can be relaxed to "non-equal" because I am dealing with a cycle.)

In particular, I've actually been looking at quaternions and one example cycle (with order 8) can be generated from: $$A=\space\space\frac{1}{\sqrt{2}}-\frac{j}{2\sqrt{2+\sqrt{2}}} + \frac{k}{2\sqrt{2-\sqrt{2}}}$$ $$B=\space\space\frac{1}{\sqrt{2}}+\frac{j}{2\sqrt{2+\sqrt{2}}} + \frac{k}{2\sqrt{2-\sqrt{2}}}$$ I can find longer cycles, but only numerically. So, in the space of just quaternions, what is the largest cycle order possible? Is it infinite?

EDIT (Aug 10 2021): Without loss of generality for the quaternion problem, I believe this is just a matter of finding finite cycles which can be generated from two degrees of freedom (real $\alpha$ and $\beta$): $$A=\space\space\sqrt{1-\alpha^2-\beta^2}-\alpha{j} + \beta{k}$$ $$B=\space\space\sqrt{1-\alpha^2-\beta^2}+\alpha{j} + \beta{k}$$ Seems easy, doesn't it?

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  • $\begingroup$ It seems that you're looking for relations of the form $A=w(A,B)$ or $B=w(A,B)$ where $w$ is a positive word. This is rather a (semi)group theory problem; I've added the group theory and dynamics tags. $\endgroup$
    – YCor
    Commented Aug 10, 2021 at 7:40
  • $\begingroup$ First, I'm unclear what you mean by a cycle. Are you asking whether the sequence contains a single term twice, or are you asking whether the entire sequence contains a finite number of terms? Secondly, Is the question whether there are arbitrarily long finite cycles? or whether there are $A$ and $B$ for which your sequence has no repeats? $\endgroup$ Commented Aug 10, 2021 at 9:45
  • $\begingroup$ @AnthonyQuas I am only interested in cycles where "the entire sequence contains a finite number of terms" and I want to know "whether there are arbitrarily long finite cycles". $\endgroup$
    – bobuhito
    Commented Aug 10, 2021 at 12:34
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    $\begingroup$ Since the quaternion group contains a non-Abelian free group, it is easy to construct a pair A and B for which this sequence will not contain repetitions. It is sufficient to take such A and B that generate a free group. $\endgroup$
    – kabenyuk
    Commented Aug 11, 2021 at 4:49

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