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Suppose I define a group action with group $G$ over some set $X$. The set is totally ordered in an arbitrary way $X = [x_1, \dots, x_N]$. Then, the group action defines permutations of the indices $[g(1), \dots, g(N)]$ for each element $g\in G$. Call the set of such permutations of the indices $S = \{[i_1, \dots, i_N]\mid \exists g \in G : \forall j : g \cdot x_j = x_{i_j}\}$.

For example, if $X$ contains 4 elements and $G = C_4$, I can choose an ordering of $X$ such that $S = \{[1,2,3,4],[2,3,4,1],[3,4,1,2],[4,1,2,3]\}$.

Suppose that I choose to interpret the different permutations as valid orderings of the indices.

Questions : For a given group $G$ is there a way to define an order relation on $X$ for which the elements of $S$ are the only valid orderings? Are there properties that these order relations will respect?

For the above example, the order relation I would be looking for is a cyclic order generated by the ternary relations $[1,2,3], [2,3,4], [3,4,1]$.

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    $\begingroup$ I don’t get the ending, why do the cycles suddenly have $3$ elements, and there are only $3$ of them? $\endgroup$ Sep 22, 2021 at 19:52
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    $\begingroup$ Also, although you don’t say it, I assume you want $X$ finite. Is $G$ finite? $\endgroup$ Sep 22, 2021 at 19:56
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    $\begingroup$ In particular, if $G=\Sigma_{X}$ and the map $G\to \Sigma_X$ is the identity, then you get all possible orderings of the cycles. $\endgroup$ Sep 22, 2021 at 20:02
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    $\begingroup$ OK corrected, there was a missing question mark. For the ending, I'm saying that cyclic order is defined by these ternary relations $\endgroup$
    – Undead
    Sep 22, 2021 at 20:03
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    $\begingroup$ Given an action of $G$ on an ordered $X$ it seems you are defining: $$S=\{[i_1,i_2,\dots,i_N]\mid \exists g\in G: \forall j: g\cdot x_j=x_{i_j}\}$$ We can definitely define $S,$ and it is unique, given the action and the ordering of $X.$ $\endgroup$ Sep 22, 2021 at 20:12

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The only way for a single cyclic order relation to apply to all elements of $S$ is if $G$ acts cyclically on $X$ in a particular way.

That can be phrased as,

  1. For some $g\in G,$ and $d\mid N,$ $g^d\cdot x=x$ for all $x\in X,$ and if $g^i\cdot x\neq x$ for any $x\in X$ and $0<i<d,$ and
  2. For any $h\in G,$ there is an $i=0,1,\dots,d-1$ so that $h\cdot x=g^{i}\cdot x$ for all $x\in X.$

A more advanced way of saying this is if the image of $G$ in $\Sigma_X$ is cyclic of order $d$ and the orbits of the action are all size $d.$

Let $g\in G$ be a generator of the cyclic group in the image.

Let $X_1,\dots, X_{N/d}$ be the orbits.

Pick any $x_i\in X_i.$

Then we can pick the total order:

$$ \begin{align} [&x_{1},\dots,x_{N/d},\\ &g\cdot x_1,\dots,g\circ x_{N/d},\\ &g^2\cdot x_1,\dots,g^2\cdot x_{N/d},\\ &\vdots\\ &g^{d-1}\cdot x_1,\dots,g^{d-1}\cdot x_{N/d}] \end{align} $$

Then every element of $S$ will have the same cyclic order as $[1,2,\dots,N].$

But it won’t be true that every tuple with that order will be in $S.$ There will only be $d$ elements of $S,$ and there are $N$ different tuples with the same cyclic order. So $S$ is only totally defined by a cyclic order if $d=N.$


If $G=C_2=\{e,g\}$ and $X=\{a,b,c,d\}$ with $$g\cdot a=d, g\cdot b=c, g\cdot c=b, g\cdot d =a,$$ then the orbits are $X_1=\{a,d\}$ and $X_2=\{b,c\}$ and we can pick $x_1=d,x_2=b$ and you get $$x_1=d<X x_2=b<x_3=g\cdot d=a<x_4=g\cdot b=c$$ and $$S=\{[1,2,3,4],[3,4,1,2]\}$$

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