Dear Sir/Madam,
Let x and y be a finite permutation representation,
x:=((6,10,12,14,9)(4,7,11,13,8));
y:=((1,3,6,9,5,2,4)(8,13,14,10,11,7,12));
I want to find the sequence C here from these permutations as,
C=[(1,3),(5,2)];
Here C are free vertices that means the points that are fixed under x and moved
for y and are consecutive to each other. Let say 1 is fixed for x and 1^y=3 and
then 3 is again fixed for x so we will add it in sequence c that i used in
program c:=[j] and again 3^y=6 but 6 is not fixed for x so the loop will
terminate. further I want to check other moved points of y that are free for x.
Here 5 is fixed for x and moved for y, i.e. 5^y=2 and 2^x=2 so C will be
C:=[5,2] because 2^y=4 and (4^x)<>4.
so here 5 and 2 are consective points and 1 and 3 are consecutive points.
this is what I have written in my previous email that 1 and 3 are fixed under x
permutation and they are in y cylce next to each other, similarly, 5 and 2 are
fixed under x and in y cycle they are next to each other moreover all of its
subsequences.
For this I tried this program in GAP that gives me only C:=[5,2] ,indeed i want
to calculate other value [1,3] by doing the same program.
The program I made as follows
FindSequenceOfFreeVertices:=function(x,y)
local j,c,next;
for j in MovedPoints(y) do
if j^x=j then
c:=[j];
next:=j^y;
while next^x=next do
Add(c,next);
next:=next^y;
od;
fi;
od;
Print("sequence of Free Vertices =",c,"\n");
end;
I hope this program will be easier for you to understand my question. Thanks
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