# [GAP Forum] Generating M11 via rational conjugacy classes.

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.