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Given a finite group $G < Sym(\Omega)$; $\Omega$ finite, and $X \subset \Omega$, I can define a by the function $H(g) = \{x^g \| x \in X\}$ for each $g \in G$. Of course, each $H$ has the same size as $X$. Two different elements of $G$, $g_1,g_2$ may generate the same set $H(g_1)=H(g_2)$. Taking the collection of all such sets, $\mathscr{H} = \bigcup_{g \in G} H(g)$

I need to find $\mathscr{H}$, so what is the fastest way to find $F \subset G$ such that $\mathscr{H}$ is the disjoint union over $F$?

(I'm more interested mathematically how to do so, but I also need to compute the result. I am using GAP currently, with the following script:

setToThe := function(set,g)
    local x, l, i, s;
    l := [];
    i := 1;
    for s in set do
        x := s^g;
        l[i] := x;
        i := i + 1;
    return l;

grp := ...;
set := ...;

unique := [];
unique[1] := set;

for g in grp do
    allGood := true;
    startUnique := Size(unique);

    for j in [1..startUnique] do
        newset := setToThe(set,g);
        if newset = unique[j] then
            allGood := false;

    if allGood then
        unique[Size(unique)+1] := newset;
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Forgive my ignorance: what's $x^g$? – Dylan Moreland Jul 14 '11 at 3:44
So if $g \in G$ is a permutation of $\Omega$ (a bijection of $\Omega$ to itself), and $x \in \Omega$ then $x^g=g(x)$, that is, what $g$ takes $x$ to – Zeophlite Jul 14 '11 at 3:47
Does X have some nice properties, or is it an arbitrary subset? Is G some nice group (abelian? cyclic? :p), or is it some arbitrary finite group? – ShreevatsaR Jul 14 '11 at 4:57
It may, but I'd have to find out. What would be "nice" properties, in regards to finding $F$? – Zeophlite Jul 14 '11 at 5:30
What makes you think you can find an $F$ such that the union defining $\mathcal{H}$ is a disjoint union? This seems hard to pull off if, for example, $X$ contains more than half of the elements of $\Omega$. Anyway, it seems like a more efficient way of calculating $\mathcal{H}$ is to union up the $G$-orbits of points in $X$. – user83827 Jul 14 '11 at 9:57
up vote 0 down vote accepted

It's simple, I feel silly:


So the trick is instead of thinking $Sym(\Omega)$, think $Sym(\alpha)$ where $\beta \subset \mathscr{P}(\Omega)$ such that for all $\alpha \in \beta$, $|\alpha| = |X|$

share|cite|improve this answer
How does this answer the question you asked? – user83827 Jul 14 '11 at 14:28

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