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An Archimedean ordered group with a two-sided invariant total order is commutative. If there is a least positive element, its cofinal powers exhaust the group. Otherwise each $c>1$ has $d>1$ with $d^2<c$. To see this, choose $1<s<r<c$ and use $r{s}^{-1}$ if its square is below $r$; if not, the inequality $(r{s}^{-1}{)}^2\ge r$ gives $s^2\le r<c$. If positive $g,h$ satisfied $gh<hg$, put $c=(gh{)}^{-1}hg$ and choose such a $d$. Cofinality gives $d^m\le g<d^{m+1}$ and $d^n\le h<d^{n+1}$. Two-sided order invariance then gives both $hg<d^{m+n+2}$ and $hg=ghc>d^{m+n+2}$, a contradiction. Replacing negative elements by their inverses reduces arbitrary commutativity to the positive case.

For ordered Abelian groups, compare tuples at their first differing coordinate. This gives a two-sided invariant total order on their direct product. The additive group ${\mathbb{Z}}^2$ with its first coordinate dominant is not an Archimedean ordered group, since $n(0,1)<(1,0)$ for every positive integer $n$.

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.

No. Assign the two free generators to $(1,0)$ and $(0,1)$ in the lexicographically ordered group ${\mathbb{Z}}^2$. The universal property gives a lattice-ordered group homomorphism onto this group: its image contains both ordinary group generators. It is not an Archimedean ordered group, sinceThus a lattice-homomorphic image can contain a nonzero positive infinitesimal even when the original free object has an Archimedean pointwise function representation.