[GAP Forum] Generating M11 via rational conjugacy classes.
John Simons
johnasimons at outlook.com
Thu Aug 3 19:01:52 BST 2017
Dear Max,
Thanks for your help - I very much appreciate it. The script does churn out some triples of elements so that's great. I was wondering however, when I try to load the script from a text editor it brings up a syntax error stating: "Syntax warning: unbound global variable in C:/gap4r8/M11.g.txt line 6
if g3 in cls3 and M11 = Group(g1, g2) then
I can execute the commands "FindNiceTriple(M11, rc[2]...), however, I was wondering whether the above may cause any problems to the actual performance? I am running GAP 4.8.7 on Windows 10 if it helps.
Lastly, I was curious as to why the triples that you obtained are different to the ones I've obtained? For example, when I try to FindNiceTriple(M11, rc[2], rc[5], rc[8]);, I repeatedly end up with the triple [ (1,2,9,3,4)(5,11,6,7,10), (2,10)(3,7,8,9,4,5,11,6), (1,4,2,7,9,8,6,5,3,11,10)]. Would it be possible to obtain other triples running the script -i.e to list further examples triples than simply ending at locating one or is there some ancient sorcery at play here?
Many thanks once again,
John
________________________________
From: Max Horn <max at quendi.de>
Sent: Wednesday, August 2, 2017 11:58:45 AM
To: John Simons
Cc: GAP Forum
Subject: Re: [GAP Forum] Generating M11 via rational conjugacy classes.
Dear John,
> On 02 Aug 2017, at 12:47, John Simons <johnasimons at outlook.com> wrote:
>
> 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).
[https://cdn.sstatic.net/Sites/math/img/apple-touch-icon@2.png?v=4ec1df2e49b1]<https://math.stackexchange.com/questions/218302/a-conjugacy-class-c-is-rational-iff-cn-in-c-whenever-c-in-c-and-n-is-co>
A conjugacy class $C$ is rational iff $c^n\\in C$ whenever ...<https://math.stackexchange.com/questions/218302/a-conjugacy-class-c-is-rational-iff-cn-in-c-whenever-c-in-c-and-n-is-co>
math.stackexchange.com
Let $C$ be a conjugacy class of the finite group $G$. Say that $C$ is rational if for each character $\chi: G \rightarrow \mathbb C$ of $G$, for each $c\in C$, we ...
>
> 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?
Absolutely. First, simplify this further: If a triple (g_1, g_2, g_3) has the require properties, then so does any of its conjugates. So it suffices to pick in the first one element g1. Then, let g2 run through all elements of the second class. This then determines g3 uniquely, and you just have to check if it lies in the third class. Finally, you have to check that M11 is generated by g1 and g2 (the element g3 is irrelevant for this).
One can improve this further, but for M11 the above is enough. Try this:
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 M11 = Group(g1, g2) then
return [g1, g2, g3];
fi;
od;
return fail;
end;
Then for example:
gap> M11:=MathieuGroup(11);
Group([ (1,2,3,4,5,6,7,8,9,10,11), (3,7,11,8)(4,10,5,6) ])
gap> rc:=RationalClasses(M11);;
gap> Length(rc);
8
gap> findNiceTriple(M11, rc[2], rc[5], rc[8]);
[ (1,8)(3,9)(5,7)(10,11), (1,7,6,3,4,2,11,9,5,8,10), (1,11,2,4,9,10)(3,6,5)(7,8) ]
gap> findNiceTriple(M11, rc[8], rc[8], rc[8]);
[ (1,3,4,9,5,11)(2,6,10)(7,8), (1,4,7,5,10,9)(2,6,11)(3,8), (1,4,11,2,5,8)(3,7)(6,10,9) ]
>
> 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".
Replacing rational by conjugacy classes above is of course trivial.
Most classes of M11 are uniquely determine by their size. leaves the two classes of size 8, resp. 11. I think the AtlasRep package can help you with identifying these, but I am not an expert on that.
Cheers,
Max
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