Dear all,
Back again for another round of fun - ha!
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.
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;
if g3 in cls3 and M12 = Group(g1, g2) then
return [g1, g2, g3];
fi;
od;
return fail;
end;
Then for example:
gap> M12:=MathieuGroup(12);
Group([ (1,2,3,4,5,6,7,8,9,10,11), (3,7,11,8)(4,10,5,6),
(1,12)(2,11)(3,6)(4,8),(5,9)(7,10) ])
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)]
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 - I have simply taken the
conjugacy classes assuming they appear in an ordered list as shown above (hence
cc[6] corresponds to the sixth conjugacy class which is 4A).
2) Furthermore, if the above approach is hopeless in determining rigid triples,
could someone please inform me as why that is the 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?
Thank you as always,
John
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