[GAP Forum] Maximal subgroups.
Frank Lübeck
frank.luebeck at math.rwth-aachen.de
Thu Aug 31 22:34:58 BST 2017
On Thu, Aug 31, 2017 at 09:15:51AM +0000, johnathon simons wrote:
> According to a certain paper I've found online, it states that for
> the Mathieu group M12, the triple (2A,4A,8A) is not a rigid generator
> of M12: that is, G =/= <g1, g2, g3> where g1 is contained in 2A,
> g2 is contained in 4A, g3 is contained in 8A and g1*g2*g3 = 1. In
> particular, it says that (2A,4A,8A) generates a proper subgroup of M12
> as can be seen from the character tables of the maximal subgroups.
Dear John Simons, dear Forum,
As was explained recently on this list, you can compute the structure
constant of this class triple from the character table of M12 to see
that triples from these classes with product 1 exist, but that all of such
triples (actually, there are several conjugacy classes of them) generate
proper subgroups.
> However, when I run the below algorithm (which attempts to find the
> rigid generating triple) I end up "apparently" finding one.
>
> findNiceTriple := function(G, cls1, cls2, cls3)
> local g1, g2, g3;
> g1 := Representative(cls1);
> for g2 in cls2 do
> g3 := (g1*g2)^-1;
You probably wanted to write 'G' instead of 'M12' in the following line:
> if g3 in cls3 and M12 = Group(g1, g2) then
> return [g1, g2, g3];
> fi;
> od;
> return fail;
> end;
(Otherwise the function looks alright.)
> Then for example:
>
> gap> M12:=MathieuGroup(12);;
> gap> cc:=ConjugacyClasses(M12);;
> gap> Length(cc);
> 15
>
> So then to see if the triple (2A, 4A, 8A) can generate a rigid triple
> of elements (g1,g1,g3) with g1*g2*g3 = 1 we have:
>
> gap> findNiceTriple(M12, cc[2], cc[6], cc[11]);
> [ (1,8,10,12,11)(3,7,4,5,9), (2,10,7,5)(3,7,8,9), (1,11,12,10,2,4,7,6,8,9)(3,5)]
Several things look strange here:
- If you had chosen the right classes, the result should be a list of
three elements of order 2, 4 and 8, resp.
- The second element is no permutation (where is 7 mapped to?)
Use something like
List(cc, c-> Order(Representative(c)));
List(cc, Representative);
to find the correct classes (and then you will get 'fail' from
'findNiceTriple'). Note that on different attempts the classes you
want can be in different positions because GAP uses the Random
function to find the conjugacy classes.
> 1) If I'm not mistaken, this implies that (2A, 4A, 8A) is such a rigid
> triple that generates M12? If not, could someone please clarify as to
> why this is not true (is there some issue with the above algorithm
If you have three conjugacy classes for which your function returns a
generating triple, then you cannot conclude that there is a rigid triple
(which means that the triple you found is the unique one up to conjugacy).
> 2) Furthermore, if the above approach is hopeless in determining
> rigid triples, could someone please inform me as why that is the
As said, to prove rigidness you need to check that there is up to conjugacy
only one triple with product one and that it generates the group.
> case and how one can determine whether such a triple generates a
> proper subgroup of M12 by simply looking at the characer tables of the
> maximal subgroups?
Also, as mentioned above, you do not need character tables of maximal
subgroups in this case, the character table of M12 is enough.
Best regards,
Frank Lübeck
--
/// Dr. Frank Lübeck, Lehrstuhl D für Mathematik, Pontdriesch 14/16,
\\\ 52062 Aachen, Germany
/// E-mail: Frank.Luebeck at Math.RWTH-Aachen.De
\\\ WWW: http://www.math.rwth-aachen.de/~Frank.Luebeck/
More information about the Forum
mailing list