[GAP Forum] Generating a group from a triple of elements.

[johnathon simons]

Dear forum,

Thank you ever so much for all your help. I've come to terms with the issues mentioned above. Hope you all have a wonderful day!

Best,

John


Sent from Outlook<http://aka.ms/weboutlook>
________________________________
From: Frank Lübeck <frank.luebeck at math.rwth-aachen.de>
Sent: Tuesday, August 22, 2017 3:32:50 AM
To: johnathon simons; forum at gap-system.org
Subject: Re: [GAP Forum] Generating a group from a triple of elements.


Dear John, dear Forum,

It seems that the mix up of two terms for finite groups leads
to some confusion:

"rational class":  this is a union of conjugacy classes of the group,
     the rational class of an element x in a group G is the union of
     the conjugacy classes containing x^k for all k in Z with gcd(k,|x|) = 1.
     Equivalently, x and y in G are in the same rational class if the
     cyclic groups <x> and <y> are conjugate in G.

"rational conjugacy class": this is a conjugacy class which is a full
     rational class. Rational conjugacy classes can be read off from the
     character table of G, these are the conjugacy classes on which all
     irreducible characters have integer values.

For the example M11: There a 10 conjugacy classes and 6 rational conjugacy
     classes, the unions of classes 8a and 8b and of classes 11a with 11b form
     two more rational classes (so altogether there are 8 rational classes,
     this is what you get by the command 'RationalClasses' in GAP).


On Sat, Aug 19, 2017 at 06:17:30PM +0000, johnathon simons wrote:
> I seem to find myself running in circles with
> this topic. Could someone help me understand why it is not sufficient for
> a sporadic group G that if:
>
> 1) G = <g_1,g_2,g_3>, with g_i an element of a rational conjugacy class.
> 2) g_1*g_2*g_3 = 1
> 3) The symmetrised structure constant (as defined in the response by Thomas) equals 1.
> 4)
> that simply that for a sporadic group G, finding a triple generator
> <g_1,g_2,g_3> = G with g_1*g_2*g_3 = 1, and the symmetrised structure
> constant (as being defined in the previous response of Thomas) being equal
> to 1 does not sufficiently determine rational rigidity (i.e that G is a
> Galois group over Q)?

Yes, if G is simple and you find elements g_1, g_2, g_3 as in 1), 2), 3)
then this is a sufficient criterion for G being a Galois group over Q.

But your example M11 does not have a rational rigid triple (the structure
constant computation discussed before involved elements of order 11 which
do not ly in a "rational conjugacy class". There also is a generalization of
the rigidity criterion for n-tuples. M11 does also not contain rational
rigid n-tuples for any n.

Nevertheless, it can be  shown that M11 is a Galois group over Q with a more
complicated twisted version of the rigidity criterion. I recommend the book
  Malle, Matzat, "Inverse Galois Theory", Springer, 1999
for more details. Theorem I.6.12 handles M11.

Best regards,
  Frank

--
///  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/