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If $V$ is a value in a lattice-ordered group of $t>1$, every convex lattice subgroup strictly containing $V$ contains $t$. Thus the convex lattice subgroup generated by $V$ and $t$ is the unique least subgroup properly containing $V$. It is called the cover of $V$. There is no intermediate convex lattice subgroup. A normal-valued lattice-ordered group requires $V$ to be a normal subgroup of this cover, not necessarily of the whole group. Values are prime convex lattice subgroups: maximality omitting $t$ makes them meet-irreducible in the lattice of convex lattice subgroups, so their ordered right cosets form a chain.

The displayed relation makes the right cosets into a lattice, with $Pg\vee Ph=P(g\vee h)$ and $Pg\wedge Ph=P(g\wedge h)$. Antisymmetry uses convexity: $g\le ph$ and $h\le qg$ give $1\le ph{g}^{-1}\le pq\in P$, hence $h{g}^{-1}\in P$. A prime convex lattice subgroup makes this order total even if the subgroup is not normal.

Put $e=1$ and write $G_{\ge e}$ for the nonnegative cone, including the identity. Every element of this cone is its own absolute value in a lattice-ordered group, so the desired condition is equivalent toWe prove this equivalence with normal-valued lattice-ordered group without assuming any characterization of normal values. Membership of $g$ in a convex lattice subgroup is equivalent to membership of $|g|$, so it suffices to discuss values of elements $t>e$. A value in a lattice-ordered group of $t>e$ means a convex lattice subgroup maximal among those omitting $t$. Its cover of a value in a lattice-ordered group will be denoted $V^{\ast }$: normal-valued means $V◃V^{\ast }$, not $V◃G$.

We first record facts about values and the coset order. Zorn's lemma supplies a value omitting any $t>e$, and supplies one containing any specified convex lattice subgroup that omits $t$. If $V$ is a value of $t$, every convex lattice subgroup strictly containing $V$ contains $t$. Therefore $V^{\ast }=⟨V,t{⟩}_c$ is the least such subgroup and there are no intermediate convex lattice subgroups. Moreover, if $C\cap D=V$ and both strictly contain $V$, then both contain $t$, a contradiction. The criterion proved in part (i) thus makes $V$ a prime convex lattice subgroup: its right cosets are totally ordered. Equivalently, for $u,v\ge e$ with $u\wedge v=e$, at least one belongs to $V$. The equivalence follows because the coset projection preserves finite meets; in proving this, inequalities with two elements of $V$ can be replaced by their join.

The right-coset order gives the useful testIndeed, $u\le pv$ gives $(u{v}^{-1})\vee e\le p\vee e\in P$, and convexity gives membership. Conversely the displayed positive part itself supplies such a $p$. We will also use the basic lattice-ordered group identity that $(r\vee e)\wedge (r^{-1}\vee e)=e$: the positive and negative parts of one element are disjoint.

Here is a squared bound from comparison at values. Suppose $p,q,x\ge e$ and $Vpq\ge Vx$ for every value $V$ of $x$. Then $p^2{q}^2\ge x$. To prove it, putand suppose $d>e$. Choose a value $P$ of $d$. If $x\in P$, then $e\le d\le x$, so $d\in P$, a contradiction. Otherwise extend $P$ to a value $V$ of $x$.

If $p\notin V$, set $c=(x{q}^{-2}{p}^{-1})\vee e$. Since $p{q}^2\ge pq$ and $Vpq\ge Vx$, test (2) gives $c\in V$. The element $(p{c}^{-1})\vee e$ is outside $V$: membership would imply $e\le p\le ((p{c}^{-1})\vee e)c\in V$, contradicting $p\notin V$. It is consequently outside $P$. Its inverse positive part isbecause $p^{-1}\le e$. These two positive parts have meet $e$, so primeness of $P$ forces $d\in P$, again a contradiction.

If $p\in V$, then $Vpq=Vq$, and $c=(x{q}^{-1})\vee e\in V$. Necessarily $q\notin V$, since otherwise $Vx\le V$ would put $x$ in $V$. Hence $t=p^{2}q\notin V$. As before, $(t{c}^{-1})\vee e$ is outside $V$ and outside $P$, while its inverse positive part ishere $t^{-1}\le e$. Primeness again yields the contradiction $d\in P$. Thus $d=e$, proving the bound. This auxiliary argument uses no normality assumption.

Suppose first that all values are normal in their covers. Given $a,b\ge e$, put $x=ab$. If $x=e$, both factors are $e$ and (1) holds. Otherwise choose any value $V$ of $x$. Since $e\le a,b\le x$, both factors belong to $V^{\ast }$. The quotient group $V^{\ast }/V$ is a linearly ordered group, because $V$ is prime and is normal in $V^{\ast }$. It has no nontrivial proper convex subgroup of an ordered group, since such a subgroup would lift to an intermediate convex lattice subgroup between $V$ and $V^{\ast }$.

This quotient is an Archimedean ordered group. If $u>e$ and all powers $u^n$ were bounded by some $v$, the principal convex lattice subgroup generated by $u$ would be a nontrivial proper subgroup: it cannot contain $v$, since $v\le u^m$ would contradict $u^{m+1}\le v$. That contradicts the preceding paragraph.

For completeness, Archimedean linearly ordered groups are Abelian by the following elementary argument. If there is a least positive element $s$, its powers are cofinal and every element lies between consecutive powers; absence of any element between $e$ and $s$ makes the group cyclic. Otherwise, for every $c>e$ there is $d>e$ with $d^2<c$: choose $e<s<r<c$ and put $d_0=r{s}^{-1}$. If $d_0^2<r$, use $d=d_0$; if not, multiplication of $d_0^2\ge r$ gives $s^2\le r<c$, so use $d=s$. If positive $g,h$ did not commute, interchange them to have $gh<hg$ and put $c=(gh{)}^{-1}hg>e$. Choose $d^2<c$ and, by the Archimedean property, integers $m,n\ge 0$ with $d^m\le g<d^{m+1}$ and $d^n\le h<d^{n+1}$. Two-sided order invariance givesa contradiction. Arbitrary nonidentity elements can be replaced by their positive inverses without changing whether they commute. This proves commutativity without assuming the characterization being proved.

It follows that $Vba=Vab=Vx$. Apply the squared-bound lemma with $p=b$, $q=a$ and $x=ab$, for all values of this $x$. We obtainTaking $a=|f|$ and $b=|g|$ gives the required inequality.

Conversely, assume (1). We need one-sided bounds for adjoining a positive element. If $H$ is any convex lattice subgroup and $a\ge e$, thenFor the first equality, denote the indicated set by $S$. The inequality (1) allows a positive power of $a$ to move past a positive element of $H$, at the cost of squaring:Also $xy\le |x||y|$ and $(xy{)}^{-1}\le |y||x|$. Applying the bound in both orders, and taking the join of the two resulting elements of $H$ and the larger exponent, bounds $|xy|$ in the same form. Inversion does not change $|g|$. A common bound obtained from $h\vee k$ and the larger exponent proves closure under join, meet and order intervals. Thus $S$ is a convex lattice subgroup containing $H$ and $a$. Conversely every such subgroup contains each $h{a}^n$ and everything whose absolute value it bounds, proving equality. The same proof in the opposite group proves the second equality, since (1) with the variables interchanged reads $ba\le a^2{b}^2$.

Let $V$ now be any value and $W=V^{\ast }$. For each $a\in W_{\ge e}\setminus V$, the subgroup $⟨V,a{⟩}_c$ strictly contains $V$ and is contained in $W$, so the cover property makes it equal to $W$. Formula (3) therefore gives power cofinality in the cover of a value: for every $b\in W_{\ge e}$ there are powers withHere the left cosets carry the order obtained by inversion from the right-coset order; equivalently $bV\le a^{m}V$ means $b\le a^{m}v$ for some $v\in V$.

Take $v\in V_{\ge e}$ and $a\in W_{\ge e}\setminus V$. If $u=a^{-1}va\notin V$, then $u\in W_{\ge e}\setminus V$ andThe inequality uses $a\ge e$. But (4), applied to $u$ and the target $a^2$, gives $V{a}^2\le V{u}^n$ for some $n$. This contradicts $Va<V{a}^2$; equality of those two cosets would itself imply $a\in V$. Hence $a^{-1}va\in V$.

To obtain the other inclusion, suppose $w=av{a}^{-1}\notin V$. ThenLeft-coset cofinality in (4) gives $a^{2}V\le w^{n}V$, contradicting $aV<a^{2}V$. Thus $av{a}^{-1}\in V$ too. Every element of $V$ is a quotient of its two nonnegative lattice parts, so both conjugation inclusions hold for all $v\in V$. Positive elements of $W$ lying in $V$ already normalize $V$, and every element of $W$ is a quotient of two nonnegative elements. ConsequentlyAll values have this property, so $G$ is normal-valued. At no stage was normality in the whole group assumed or required.

Finally, in a linearly ordered group, nonnegative $a,b$ are comparable. If $a\le b$, then $ab\le b^2\le b^2{a}^2$; if $b\le a$, then $ab\le a^2\le b^2{a}^2$. Thus (1) holds in every o-group. The proved converse gives every o-group is normal-valued.

First check the ordered right cosets of a convex lattice subgroup. Replacing $g,h$ by $p_{1}g,p_{2}h$ changes a witness $g\le ph$ into $p_{1}g\le p_{1}p{p}_2^{-1}(p_{2}h)$, so the relation is independent of representatives. Reflexivity uses $p=1$. If $g\le ph$ and $h\le qk$, two-sided invariance of the partial order gives $g\le pqk$, proving transitivity. If both inequalities hold in opposite directions, thenBoth endpoints are in $P$, so convexity gives $ph{g}^{-1}\in P$ and hence $h{g}^{-1}\in P$. Thus $Pg=Ph$, proving antisymmetry. The join and meet formulas are $Pg\vee Ph=P(g\vee h)$ and $Pg\wedge Ph=P(g\wedge h)$. For example, if $g\le pk$ and $h\le qk$, then $g\vee h\le (p\vee q)k$, giving the least-upper-bound property.

The coset order is total exactly when $P$ is a prime convex lattice subgroup. If it is total and $a,b\ge 1$ have $a\wedge b=1$, assume $Pa\le Pb$. The coset meet is then $Pa=P(a\wedge b)=P$, so $a\in P$. Conversely, for $u=g{h}^{-1}$, its positive and negative parts in a lattice-ordered group satisfy $u^{+}\wedge u^{-}=1$. Primality puts one of them in $P$. If $u^{+}\in P$, then $u\le u^{+}$ gives $Pg\le Ph$; if $u^{-}\in P$, use $u^{-1}\le u^{-}$ to obtain $Ph\le Pg$.

It remains to identify the stated intersection condition. If $P$ is prime and $C,D$ properly contain it, choose positive $c\in C\setminus P$ and $d\in D\setminus P$. The cosets $Pc,Pd$ are strictly above $P$. Their meet is therefore strictly above $P$, so $c\wedge d\notin P$. But $c\wedge d$ belongs to $C\cap D$ by convexity, showing $C\cap D\ne P$.

If $P$ is not prime, choose disjoint positive $a,b\notin P$. The principal convex lattice subgroups $A=⟨a{⟩}_c$ and $B=⟨b{⟩}_c$ have $A\cap B=\{1\}$. Indeed, disjoint positive elements commute, since $a\vee b=a(a\wedge b{)}^{-1}b=ab=ba$, and Riesz decomposition in a lattice-ordered group shows $a^r\wedge b^s=1$ for every positive $r,s$. Now the distributivity of the lattice of convex lattice subgroups givesBoth joins properly contain $P$, violating the intersection condition. This proves the requested equivalence.

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.