# Order relations and group actions

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 $$[1,2,3], [2,3,4], [3,4,1]$$.

• I don’t get the ending, why do the cycles suddenly have $3$ elements, and there are only $3$ of them? Sep 22, 2021 at 19:52
• Also, although you don’t say it, I assume you want $X$ finite. Is $G$ finite? Sep 22, 2021 at 19:56
• 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. Sep 22, 2021 at 20:02
• OK corrected, there was a missing question mark. For the ending, I'm saying that cyclic order is defined by these ternary relations Sep 22, 2021 at 20:03
• 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 $X.$ Sep 22, 2021 at 20:12

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 $$0 and
2. For any $$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 $$x_1=d and $$S=\{[1,2,3,4],[3,4,1,2]\}$$