Dear all,
I'm currently new to GAP and computer algebra systems and was wondering whether
anyone knew of the easiest method to achieve the following.
Suppose I have a finite simple group - let us take $M_{11}$ and let $C_1, C_2,
C_3$ be a triple of rational conjugacy classes ( if necessary, for the
definition of rationality, see
https://math.stackexchange.com/questions/218302/a-conjugacy-class-c-is-rational-iff-cn-in-c-whenever-c-in-c-and-n-is-co).
I would like to find explicit computations in GAP (or another system), such
that I can find a triple of elements $g_1, g_2, g_3$ where $g_i \in C_i$
satisfies $g_1g_2g_3 = 1$ and $M_{11}$ $\cong$ $<g_1,g_2,g_3>$.
I know GAP has the command which allows one to see the standard generators of
$M_{11}$ in cycle notation, however, I cannot seem to find an easy way to do
the above. There is a command to allow one to obtain the conjugacy classes of
$M_{11}$ and also rational conjugacy classes in cycle notation (though I'm
unaware of how to convert this into ATLAS notation such as $1A, 2A$ etc.)
Essentially, does anyone know how to run through rational conjugacy classes and
arbitrarily take an element from three individual classes under the restriction
places above; that the product of the three elements must be identity? Perhaps
the fact that we can rewrite $g_3 = (g_1g_2)^{-1}$ simplifies the computational
procedure?
What I eventually would like to obtain is that the commands eventually spit out:
"The conjugacy classes $2A, 4A$ and $11A$ are a triple of rational conjugacy
classes satisfying the above".
I know that this is feasible since it has been done in this paper:
http://www.maths.qmul.ac.uk/~raw/pubs_files/sgensweb.pdf
However, I cannot seem to find any actual method of implementation/commands.
Thank you for your help!
Regards,
John
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