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

[johnathon simons]

Dear Thomas,


Thank you for your helpful comments on calculating the symmetrised structure constant.


Perhaps my understanding of rational classes is incorrect, but I have taken it from this equivalence: https://math.stackexchange.com/questions/218302/a-conjugacy-class-c-is-rational-iff-cn-in-c-whenever-c-in-c-and-n-is-co

>From my understanding of that, we would say a conjugacy class is rational if every entry of it in the character table is rational under the respective column. However, using my reasoning, if we look at M11, under the columns 8a and 8b it has entries of A = -sqrt(-2) and also underneath 11a and 11b it has entries B = (-1 - sqrt(-11)/2) both which are not rational and so that would mean M11 has only 6 rational classes (it has a total of 10 conjugacy classes)


But, using the "RationalClasses" function on GAP we know that M11 has 8 rational classes.  I'm certain I am mistaken in my reasoning/interpretation of the character table and would very much be appreciative of an explanation.

Essentially, all I am trying to do is find a triple of conjugacy classes (that are rational) such that a triple (g_1, g_2, g_3) of elements  satisfies the rigidity condition of Thompson to realize the group M11 as Galois over Q.

I am very much appreciative for all your help,


John


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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>
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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 ...




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