Dear Forum, Dear Tendai Shumba,
> On Aug 19, 2015, at 6:53 AM, tendai shumba <tendshu...@yahoo.com> wrote:
>
> Dear forum,
> I have done the following calculation on a machine running GAP version 4.7.7
> as well as on machines running older versions
> of the program:gap>to:=TableOfMarks("M11");
> TableOfMarks( "M11" )gap> RepresentativeTom(to,19);
> Group([ (3,9)(4,7)(5,12)(10,11), (3,12)(4,5)(6,11)(7,9), (3,5)(4,6)(9,10)
> (11,12) ]).
> My question is that since we know that the Mathiue group M_11 is a
> permutation group of degree11, would it not mean that all the subgroups are
> of degree 11?
What is happening is that the group used for this table of marks:
gap> g:=UnderlyingGroup(to);
Group([ (1,2)(3,5)(4,9)(6,10), (1,3,4,7)(2,12,10,11) ])
is an M11 that acts on the points
gap> MovedPoints(g);
[ 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12 ]
which is degree 11 but somewhat nonstandard. (I suspect this stems from getting
information consistent with M12.)
If you want your ``natural’’ M11, you could use for example:
gap> myM11:=MathieuGroup(11); # or whichever version you like
Group([ (1,2,3,4,5,6,7,8,9,10,11), (3,7,11,8)(4,10,5,6) ])
gap> iso:=IsomorphismGroups(UnderlyingGroup(to),myM11);
[ (4,11)(5,7)(6,9)(10,12), (1,12,3,11)(2,6,5,10) ] -> [ (2,7)(3,6)(4,10)(5,11),
(1,6,4,5)(7,9,11,8) ]
gap> Image(iso,RepresentativeTom(to,19));
Group([ (2,10)(3,8)(4,11)(5,7), (2,4)(3,10)(6,7)(8,11), (2,6)(3,5)(4,8)(7,11) ])
I hope this helps,
Alexander Hulpke
-- Colorado State University, Department of Mathematics,
Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA
email: hul...@math.colostate.edu, Phone: ++1-970-4914288
http://www.math.colostate.edu/~hulpke
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