First a definition: a Fibonacci sequence in a group is a sequence in which the first two elements may be arbitrary, and from there on each element is a product (using the group operation) of the previous two elements. So, if the first two elements are x,y then the sequence starts x,y,xy,yxy,xyyxy, ...The question is: for which groups G and starting values x,y, do we have that all the elements in the sequence remain in the same conjugacy class? (I am particularly interested in the case where the group is Alt(n), the group of even permutations on n elements, and the pair x,y generate the whole group.) Clearly a necessary condition is that the first three elements x,y,xy all belong to the same conjugacy class. (There is some literature on when the product of elements in specific conjugacy classes give an element in a specified conjugacy class, the problem is addressed using character theory, see for example exercise 7.67 in Stanley's Enumerative combinatorics.) This necessary condition is not sufficient though. A sufficient condition, probably not necessary, is that there exist some element t such that x^t=y, y^t=xy. In this case conjugation by t takes xy to yxy, takes yxy to xyyxy, etc, in other words conjugation by t acts as a shift on the Fibonacci sequence starting with x,y, so all the elements are conjugate. If we write a=xt^(-1), b=t^(-1) we find that aab=bba, so that the group generated by x,t is a quotient of the Gieseking group.
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2$\begingroup$ In case $G={\rm Alt}(n)$, for $n \leq 7$ I found no pair of elements which generates the whole group and which satisfies your Fibonacci condition. Do you know whether for larger $n$ there are such elements at all? $\endgroup$– Stefan Kohl ♦Oct 24, 2014 at 10:00
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$\begingroup$ Moshe Noiman has emailed me that $n$ needs to be somewhat bigger, and explained his construction. There are examples e.g. for $n=17$ ($x = (1,15,17,14,8,13,11,9,7,3,4)(2,16,10,6,12)$, $y = (1,14,7,3,8)(2,15,11,17,10,12,9,13,6,5,16)$). He also wrote that he cannot comment here for some reason -- hopefully this problem will be resolved soon. $\endgroup$– Stefan Kohl ♦Oct 26, 2014 at 21:48
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$\begingroup$ just to set the record straight - i learned about these examples from derek holt $\endgroup$– moshe noimanOct 27, 2014 at 6:49
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