# Extending a partial order while preserving an automorphism

It is well known that if $(P, \leq)$ is a partial order then $\leq$ can always be extended to a linear order. This is sometimes called Szpilrajn´s theorem although it had been previously proved by Banach, Kuratowski and Tarski.

Now suppose that $f$ is an automorphism of $(P, \leq)$ and we want to extend $\leq$ to a linear order in such a way that $f$ remains an automorphism. Of course, this is not always possible since $f$ could have a finite orbit and automorphisms of linear orders can´t have finite orbits; but I wonder if this is the only obstruction. So let $A$ be the collection of all those $f$´s for which it is possible (for instance $Id_P \in A$); here are my questions:

1) Is it true that If $f$ has no finite orbits then $f \in A$?

2) Is $A$ a subgroup of $Aut(P, \leq)$?

and the vaguer

3) If the answer to 1 is negative. Can we somehow characterize the elements of $A$?

Perhaps this is all well known and studied but I couldn´t find anything at all in the literature, so references are also appreciated.

-

I couldn't find anything in the literature either, but the answer to the first question is positive. Let $G$ be a group acting on a space $X$. Say that $G$'s action on $X$ has the invariant order-extension property provided that every $G$-invariant partial order on $X$ (i.e., partial order $\le$ such that $x\le y$ iff $gx\le gy$, for each $g\in G$) extends to a $G$-invariant linear order on $X$.

I managed to show:

Theorem. The following are equivalent for an abelian group $G$:

1. $G$'s action on $X$ has the invariant order extension property.
2. $X$ has a linear $G$-invariant order.
3. No element of $G$ has an orbit of finite size greater than one.

Here's the proof.

The answer to Question 1 follows by letting $G$ be the subgroup of permutations generated by $f$.

(By the way, I assume that your no-finite-orbits condition really means: no finite orbits of size greater than one. Fixed points aren't a problem.)

-
Thank you Alexander. The MO question that you mention should be easy to answer now by taking first an appropiate quotient. –  Ramiro de la Vega Sep 14 '13 at 20:45
That's the plan. I'm working on it now. –  Alexander Pruss Sep 14 '13 at 20:58

The answer to question $2$ is negative. Let $P$ be any infinite set and let $\leq$ be the trivial partial order on $P$. In other words, $x\leq y\Rightarrow x=y$. Then $A$ is the set of all bijections from $P$ to $P$ with no finite cycles. If $f:P\rightarrow P$ has no finite cycles, then let $C$ be the partition of $P$ into its cycles. Give $C$ a total order $\preceq$. Finally, linearly order $P$ by letting $x\preceq y$ if i$x\in R\in C,y\in S\in C$ and $R\preceq S$ or where $f^{n}(x)=y$ for some $n>0$. Then $f$ is an automorphism of $(P,\preceq)$.

On the other hand, $A$ is generally not a subgroup of $Aut(P,\leq)=Sym(A)$. If $P=\mathbb{Z}$ and $f,g:\mathbb{Z}\rightarrow\mathbb{Z}$ are the maps where $f(x)=x+2$ for all $x$ and $g(x)=x+1$ whenever $x$ is even and $g(x)=x-1$ whenever $x$ is odd, then $f^{-1}g,f\in A$, but $ff^{-1}g=g\not\in A$.

-
Thanks Joseph for a nice example. –  Ramiro de la Vega Sep 12 '13 at 17:23