[GAP Forum] Table of Marks
tendai shumba
tendshumba at yahoo.com
Wed Aug 19 13:53:34 BST 2015
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? If the answer to my question is in the affirmative
then there is something wrong with the representative of the conjugacy class of groups given above since the cycle (3,9)(4,7)(5,12)(10,11) because the third transposition (5,12) has a 12.
Can someone explain? Assistance offered will be greatly appreciated.
Regards,
Tendai
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