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A faithful transitive order-primitive lattice permutation group is either a regular Archimedean translation action, an order two-transitive action, or a periodic action. In the periodic case a fixed-point-free order automorphism of the Dedekind completion commutes with the action and has cofinal and coinitial iterates; the point stabilizers act order two-transitively within period intervals. Only the regular case satisfies $|f||g|\le |g{|}^2|f{|}^2$ for every pair. In either other case, order two-transitivity on the whole chain or on a period interval supplies positive $a,b$ and $x<y<z<t$ with $xa=y$, $ya=z$, $xb=x$, $yb=t$, giving $xab=t>z=x{b}^2{a}^2$. The classification itself is a theorem, not a consequence of that witness calculation.

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.

The diamond theorem in the constructible universe gives $L\models ♢$, and $L$ satisfies ZFC. Apply the preceding construction inside $L$ with $S={\omega }_1^L$. It produces a normal splitting Suslin tree.

For completeness, such a tree yields a Suslin line. Order its nodes lexicographically using the two successors at every split, treating a node itself as a position between its two successor subtrees. Each node is thus a cut point between a left and a right subtree. This gives a dense linear order without endpoints. Every nonempty interval contains a whole cone above some node: for comparable endpoints use the successor cone of the descendant endpoint directed toward the other endpoint; for incomparable endpoints use the right-successor cone of the lower endpoint. Disjoint intervals therefore supply pairwise incomparable cone roots, so the order has the countable chain condition for a linear order. A countable collection of nodes has bounded heights; a cone based above that bound contains none of them, so it is not an order-dense subset. Passing to the Dedekind completion using proper cuts, so that no endpoints are added, preserves density, the countable chain condition for a linear order, and nonseparability. For nonseparability, a countable dense set in the completion would give a countable dense set of original nodes by choosing one original node between each distinct pair of its points. This contradicts the preceding height-bound argument. The result is a Suslin line.

Thus the Suslin hypothesis fails in $L$. The constructible universe theorem is a theorem of ZFC, so this is a relative-consistency argument, without an additional assumption that a transitive model exists:

An Aronszajn tree with no uncountable tree antichain. A normal splitting Suslin tree yields a Suslin line by a lexicographic ordering followed by Dedekind completion.