In Glass' Partially Ordered Groups Corollary 7.4.4 says:

If $G$ is an ordered group and $(G,G)$ is the right regular representation, then $(G,G)$ is primitive if and only if $G$ is order-isomorphic to a subgroup of $\mathbb R$.

I believe that by "right regular representation" he means that the action is multiplication from the right. A group acting on a set is said to be primitive if it preserves no non-trivial convex subsets.

Now consider $G=\mathbb Z^2$, lexically ordered [i.e. $(x_1,x_2)\leq (y_1,y_2)$ iff $x_1< y_1$ or $x_1 = y_1$ and $x_2\leq y_2$]. I believe this group is primitive when acting on itself: consider a block $\Delta$. Because $\Delta$ is non-trivial, convex and the group is totally ordered there must be some $s=\sup \Delta$ and $i = \inf \Delta$. Now take addition by $i^{-1}+s$ as the action - this will map $i$ to $s$ and $s$ to something outside of $\Delta$, so this block cannot be preserved. Since $\Delta$ was arbitrary, no block can be preserved, and therefore $(G,G)$ is primitive.

But $\mathbb Z^2$ does not seem to be order-isomorphic to (a subgroup of) $\mathbb R$. It's not archimedean for example, and $\mathbb R$ is. What am I not understanding?

  • $\begingroup$ First of all, $\mathbb R$ is archimedean. Second of all, the statement is that it's order-isomorphic to a subgroup, not all reals. Does this help a little? $\endgroup$ – Marek Oct 20 '13 at 15:11
  • $\begingroup$ @Marek: Thanks! I did mix up which one is archimedean. But it doesn't seem to me that $\mathbb Z^2$ can be isomorphic to even a supgroup of $\mathbb R$, due to the aforementioned archimedean issue. $\endgroup$ – Xodarap Oct 20 '13 at 15:17

The problem with the above "proof" was that $\sup \Delta$ might not be in $\Delta$ (or even exist at all).

Consider $\Delta = \{(0,x):x\in \mathbb Z\}$. If $y=0$ then $(y,z)+\Delta=\Delta$, otherwise $\left[(y,z)+\Delta\right]\cap \Delta = \emptyset$. So $\Delta$ is a convex block, meaning that $\left(\mathbb Z^2, \mathbb Z^2\right)$ is not primitive.

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