Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 4 / 3

For each $g\ne 1$, choose a value in a lattice-ordered group $P_g$, using Zorn's lemma. Such a value is a prime convex lattice subgroup. Otherwise the last construction in Question 1(i) would give two larger convex lattice subgroups whose intersection is $P_g$; maximality among subgroups omitting $g$ would put $g$ in both, hence in their intersection. Thus the right coset set ${\Omega }_g=P_g\backslash G$ is a totally ordered set.

Right multiplication defines an action ${\rho }_g(a):P_{g}h\mapsto P_{g}ha$. It is well-defined, preserves the total order, and has inverse ${\rho }_g(a^{-1})$. The coset formulas from Question 1(i) show pointwise preservation of both join and meet, becauseand likewise for the meet.

Take the disjoint union of these chains, arranged in any fixed order sum of the indices $g\ne 1$. Let every $a\in G$ act on each component by ${\rho }_g(a)$. This gives a lattice-ordered group homomorphism into the order automorphisms of one chain. It is injective: a nonidentity $a$ moves the identity coset in ${\Omega }_a$, since $P_{a}a\ne P_a$. Its image is closed under pointwise join and meet, giving exactly the required lattice-ordered permutation group. This is the Holland representation theorem.

Here transitivity is order n-transitivity: increasing tuples are compared, rather than arbitrary reorderings. For increasing tuples ${\alpha }_1<\cdots <{\alpha }_n$ and ${\beta }_1<\cdots <{\beta }_n$, choose $h_{ij}$ sending both ${\alpha }_i$ to ${\beta }_i$ and ${\alpha }_j$ to ${\beta }_j$ when $i\ne j$, using order two-transitivity. For $i=j$, choose an element sending the single point to its target, using transitivity. Define

This belongs to the lattice-ordered permutation group. At ${\alpha }_i$, all terms in its $i$th inner meet equal ${\beta }_i$. At ${\alpha }_k$, every inner meet is at most ${\beta }_k$, because its $j=k$ term takes that value. The outer join is consequently exactly ${\beta }_k$ there. Thus $h$ maps the whole tuple to its target. This finite interpolation by lattice operations proves every finite order-transitivity without assuming the group permits arbitrary cut-and-paste restrictions of its elements.

For a faithful transitive ordered action, primitivity here means that the only invariant equivalence relations with convex classes are equality and the universal relation. The trichotomy is: a regular Archimedean action, an order two-transitive action, or a periodic action. In the regular case the chain is an ordered additive subgroup of $\mathbb{R}$, acting by translations. In the periodic case there is a fixed-point-free period $z$ on the Dedekind completion, commuting with the action and having cofinal and coinitial integer iterates; inside a period interval the relevant stabilizer acts order two-transitively. These are the cases of the McCleary trichotomy theorem.

Only the regular Archimedean case satisfies the inequality. It is a linearly ordered group. For positive $a,b$, if $a\le b$, then $ab\le b^2\le b^2{a}^2$; if $b\le a$, then $ab\le a^2\le b^2{a}^2$. Taking $a=|f|$, $b=|g|$ proves the displayed inequality.

To exclude an order two-transitive action, choose $x<y<z<t$. Choose an automorphism taking $(x,y)$ to $(y,z)$ and replace it by its pointwise join with the identity, giving positive $a$ with $xa=y$, $ya=z$. Similarly obtain positive $b$ with $xb=x$, $yb=t$. ThenThus $ab\nleqq b^2{a}^2$, while $|a|=a$ and $|b|=b$. The same construction inside a period interval excludes the periodic case, using the order two-transitive stabilizer supplied by the classification.