I am trying to figure out how specific generating triples in finite simple groups are calculated. My understanding is that it uses Frobenius's formula and character theory. I'm not an expert on either.
If I were to take the Mathieu group $H=M_{11}$ for example, then I can calculate its order, not sure about its exponent, but then I have read that Frobenius's formula and the character table show that there are
$$2^{6}\times3^{3}\times5^{2}\times11 = 20|H| = 20|Aut H|$$
triples of type $(3, 8, 11)$ in $H$.
How is this result obtained and can I use GAP to help with the process?
Am I correct in thinking I can do the below?
gap> G := MathieuGroup(11);
Group([ (1,2,3,4,5,6,7,8,9,10,11), (3,7,11,8)(4,10,5,6) ])
gap> C := CharacterTable(G);
CharacterTable( Group([ (1,2,3,4,5,6,7,8,9,10,11), (3,7,11,8)(4,10,5,6) ]) )
gap> List(ConjugacyClasses(G),x -> Order(Representative(x)));
[ 1, 2, 3, 6, 4, 8, 8, 5, 11, 11 ]
So elements of order 3, 8 and 11 are respectively in positions 3; 6 and 7; and 9 and 10.
gap> for i2 in [6,7] do
> for i3 in [9,10] do
> Display([3,i2,i3]);
> Display(ClassStructureCharTable(C,[3,i2,i3]));
> od;
> od;
[ 3, 6, 9 ]
39600
[ 3, 6, 10 ]
39600
[ 3, 7, 9 ]
39600
[ 3, 7, 10 ]
39600
So in all, we get 158400 = 20*7920 = 20|Aut(M11)| triples, as claimed.
