RE: [math-fun] a problem of permutation group(please help me)

26 Dec 2003


      Here is how one can do that in Maple - very similar to Fred W. Helenius'
solution using GAP.
...
with(group):
...
pg:=permgroup(8,{s1=[[5,7],[6,8]],s2=[[5,6],[7,8]],s3=[[3,4]],s4=[[3,7
],[4,8]],s5=[[2,6],[4,8]],s6=[[2,4],[6,8]],s7=[[7,8]],s8=[[6,8]],s9=[[
4,8]]}):
...
grouporder(pg);
                                 5040
...
groupmember([[2,3,4,5,6,7,8]],pg);
true

Maple's library doesn't include a procedure for a decomposition of a
group element into a product of generators. For this case, with
generators of order 2, the following procedure works,
...
dec := proc(e, g)
local i, x, v, w;
    w := op(2, g);
    while not member(e, map(rhs, w)) do
        v := {};
        for i in op(2, g) do v := v union map(x ->
            cat(lhs(i), lhs(x)) =
            group:-mulperms(rhs(i), rhs(x)), w)
        end do;
        w := w union v
    end do;
    select(x -> rhs(x) = e, w)
end proc:
...
dec([[2,3,4,5,6,7,8]],pg);
{s9s6s3s2 = [[2, 3, 4, 5, 6, 7, 8]],

        s9s6s2s3 = [[2, 3, 4, 5, 6, 7, 8]]}

The same as in GAP, the composition of permutations in Maple is defined
from left to right. Check the result,
...
`&*`:=mulperms:
...
assign(op(2,pg));
...
s9&*s6&*s2&*s3;
                       [[2, 3, 4, 5, 6, 7, 8]]
s9&*s6&*s3&*s2;
                       [[2, 3, 4, 5, 6, 7, 8]]
It is interesting that Maple found shorter decompositions than GAP,
...
s9&*s2&*s1&*s6&*s3;
                       [[2, 3, 4, 5, 6, 7, 8]]
Alec Mihailovs 
http://webpages.shepherd.edu/amihailo/