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A commutative lattice-ordered group, usually written additively. Examples include real-valued function groups with pointwise addition, maximum and minimum. Such a group need not have a total order.

The displayed join is nonnegative in a lattice-ordered group. In an additive Abelian group it becomes $|g|=g\vee (-g)$. This order-theoretic absolute value need not be a real scalar.

Every lattice-ordered group embeds in a lattice-ordered permutation group on one totally ordered set. Choose a value in a lattice-ordered group for each nonidentity element, act on its ordered right cosets, and combine these chains in an order sum. The original element moves its own value's identity coset, making the action faithful. The coset lattice formulas make the embedding preserve both join and meet.

The convex lattice subgroups of a lattice-ordered group form a distributive lattice, with intersection as meet and the convex lattice subgroup generated by a union as join. A positive element of a join is bounded by a finite product of positive elements from the generating subgroups. Riesz decomposition in a lattice-ordered group splits such an element into positive factors from those subgroups. If the original element belongs to another convex subgroup $E$, every factor does too. Consequently $E\cap (C\vee D)=(E\cap C)\vee (E\cap D)$.

A group homomorphism preserving both join and meet. A bijective such map is an isomorphism of lattice-ordered groups; an injective one is an embedding.

A group of order automorphisms of a totally ordered set that is closed under pointwise maximum and minimum. Products act on the right when writing $\alpha f$; then $\alpha (fg)=(\alpha f)g$. The pointwise lattice operations make it a lattice-ordered group.

A group with a two-sided invariant total order. Thus it is a lattice-ordered group whose join and meet choose the larger and smaller element. An order on only one side is not sufficient here. Every linearly ordered group is torsion-free: a positive element has strictly increasing positive powers, and a negative element has a positive inverse.

A lattice-ordered group is normal-valued when every value in a lattice-ordered group is a normal subgroup of its cover. Normality in the cover is weaker than normality in the entire group. Every linearly ordered group is normal-valued.

Assume $uv\le v^2{u}^2$ for all $u,v\ge 1$ in a lattice-ordered group. For a convex lattice subgroup $H$ and $a\ge 1$, the displayed bounds allow $h\in H$ with $h\ge 1$ and integers $n\ge 0$. To prove the first set is a subgroup, use $(h{a}^n)(k{a}^m)\le h{k}^2{a}^{2n+m}$. Products and inverse products are both bounded because $xy\le |x||y|$ and $(xy{)}^{-1}\le |y||x|$. Taking the larger exponent and a join of the two bounds in $H$ bounds their absolute value in a lattice-ordered group. Inversion preserves that absolute value; common bounds using $h\vee k$ prove lattice closure and convexity. The set contains $H$ and $a$, and any convex lattice subgroup containing them contains the entire set. The second equality follows by applying the proof to the opposite group, where the required inequality is the original one with the variables interchanged.

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.

Evaluate the generators as coordinate projections on ${\mathbb{R}}^n$. The lattice-ordered group they generate consists of continuous positively homogeneous piecewise linear functions with finitely many integer-linear pieces. Equivalently its elements are finite joins of finite meets of integer linear forms. The universal property follows because the real line detects every identity of Abelian lattice-ordered groups: if an identity fails, pass to a prime ordered quotient and select the finitely many active linear pieces. The resulting rational linear inequalities have a real solution by the alternative underlying Farkas' lemma, giving a real witness to failure. This is a statement about integer coefficients and positive homogeneity, not arbitrary real coefficients or affine constant terms.

For $u\ge 1$ in a lattice-ordered group, the displayed set is the smallest convex lattice subgroup containing $u$. Products of two such bounds are bounded by a larger power of $u$; inversion and the lattice operations preserve the bounds. This gives an explicit test for membership.

Let $p,q,x\ge 1$ in a lattice-ordered group. Under the displayed hypothesis, suppose $d=(x{q}^{-2}{p}^{-2})\vee 1>1$ and choose a value $P$ of $d$. If $x\in P$, then $1\le d\le x$ contradicts its omission. Otherwise extend $P$ to a value $V$ of $x$. The right-coset test is $Vu\le Vv⟺(u{v}^{-1})\vee 1\in V$. If $p\notin V$, put $c=(x{q}^{-2}{p}^{-1})\vee 1\in V$. Then $(p{c}^{-1})\vee 1\notin V$, while its inverse positive part is $(c{p}^{-1})\vee 1=d$. These parts have meet one, so primeness of $P$ forces $d\in P$. If $p\in V$, put $c=(x{q}^{-1})\vee 1\in V$ and $t=p^{2}q\notin V$; the disjoint pair $(t{c}^{-1})\vee 1$ and $(c{t}^{-1})\vee 1=d$ gives the same contradiction. No normality assumption is used in this separation argument.

A lattice-ordered group is normal-valued precisely when the displayed inequality holds for all its elements. The theorem is proved without using it as a definition. In the forward direction a normal value has a linearly ordered quotient of its cover with no proper nontrivial convex subgroup. That quotient is Archimedean, hence Abelian by Archimedean linearly ordered groups are Abelian. Apply squared bound from comparison at values to $p=b,q=a,x=ab$. For the reverse direction, power cofinality in the cover of a value excludes a conjugate $u=a^{-1}va$ outside the value: its right cosets $V{u}^n$ are all bounded by $Va$, contradicting cofinality above $V{a}^2$. Left-coset cofinality treats $av{a}^{-1}$. Every element is a quotient of nonnegative lattice parts, proving normality in the cover. In a linearly ordered group, comparison of $a,b\ge 1$ immediately gives $ab\le \max (a,b{)}^2\le b^2{a}^2$, so every such group is normal-valued.