Partially ordered group

A group with a partial order preserved by multiplication on both sides: $a\le b$ implies $xay\le xby$ for all $x,y$. In additive notation this says $a\le b\Rightarrow x+a+y\le x+b+y$. Preserving only one side is a different orderability condition.

The nonnegative cone determines the partial order by $a\le b⟺a^{-1}b\in G_{\ge 1}$. It contains the identity element, is closed under products and conjugation, and intersects its inverse only in the identity. Conversely these conditions define a two-sided invariant partial order. Its strictly positive part omits the identity; conventions for $G^{+}$ differ, so specify which is intended.

A subgroup $C$ is convex if $c_1\le g\le c_2$ with $c_1,c_2\in C$ implies $g\in C$. In a linearly ordered group, convex subgroups form a chain under inclusion: positive elements witnessing incomparability would be comparable and force the smaller one into the other subgroup by convexity.

An ordered group whose partial order is a lattice. Each pair has a join $f\vee g$ and a meet $f\wedge g$. Translations preserve these operations, for example $a(f\vee g)b=afb\vee agb$. Pointwise maximum and minimum make the order automorphisms of a chain into a lattice-ordered group.

For $a,b\ge 1$, take $a^{\prime }=x\wedge a$ and $b^{\prime }=(a^{\prime }{)}^{-1}x$. Then $a^{\prime }\ge 1$ and $b^{\prime }=1\vee a^{-1}x$ is between one and $b$. Thus $x=a^{\prime }{b}^{\prime }$. Repetition splits a positive element bounded by a finite product into factors bounded by the corresponding factors. Every resulting positive factor is also bounded by $x$.

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.

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.

The two positive parts satisfy $g=g^{+}(g^{-}{)}^{-1}$, commute with one another, and have meet one. They isolate positive and negative displacement in the pointwise model of a lattice-ordered permutation group. These are group elements, unlike the scalar positive part of a real number.

A convex subgroup of an ordered group that is also closed under join and meet. A normal convex lattice subgroup is the kernel of the quotient lattice-ordered group homomorphism.

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.

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.

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 $E\cap (C\vee D)=(E\cap C)\vee (E\cap D)$.

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.

A proper convex lattice subgroup $P$ is prime when $a,b\ge 1$ and $a\wedge b=1$ imply $a\in P$ or $b\in P$. Equivalently the ordered right cosets of $P$ form a totally ordered set. Prime refers to this order property, and does not require a normal subgroup or prime index.

A value of $g\ne 1$ is a convex lattice subgroup maximal among those omitting $g$. Zorn's lemma supplies a value: the union of a chain of convex lattice subgroups omitting $g$ still omits $g$. Such subgroups are prime. Their unique cover is the smallest convex lattice subgroup properly containing them.

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.

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.

Under the inequality in one-sided bounds for adjoining a positive element, if $a\ge 1$ belongs to the cover of a value in a lattice-ordered group $V^{\ast }$ but not to $V$, then $⟨V,a{⟩}_c=V^{\ast }$. Thus every $b\ge 1$ in the cover is bounded above by both $v{a}^n$ and $a^{m}w$ for suitable nonnegative $v,w\in V$ and integers $m,n\ge 0$. The corresponding right and left coset bounds give the displayed cofinality, without first assuming that $V$ is normal. Left-coset order is obtained by inversion of right-coset order.

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.

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.

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.

An ordered group is Archimedean when $a\ge 1$ and $a^n\le b$ for all positive integers force $a=1$. In an additive Abelian group this excludes a positive element whose every integer multiple stays below one fixed bound. This generalizes the Archimedean property of the real numbers.

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 $n$.

For a torsion-free abelian group of finite rational rank $r$, each convex subgroup of an ordered group is pure: if $ng\in C$ for a positive integer $n$, then $g$ lies between zero and $ng$ or their negatives and hence belongs to $C$. A proper inclusion of convex subgroups therefore increases rational rank. They form a chain, so there are at most $r+1$, including zero and the whole group. A lexicographically ordered copy of ${\mathbb{Z}}^r$ attains the bound.

A group admits a two-sided invariant total order exactly when every finite list of nonidentity elements can be signed so that the conjugation-invariant semigroup generated by the signed elements excludes the identity. A compactness theorem argument supplies a compatible global sign assignment. Omitting the conjugates changes the criterion to a different orderability problem.

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.

For a totally ordered set $\Gamma$, take real coefficient families whose support satisfies the maximum condition: every nonempty subset of the support has a largest index. Add coefficients pointwise and decide the sign from the largest nonzero coordinate. This defines an ordered Abelian group. For $\Gamma ={\mathbb{Z}}_{>0}$ with its usual order, the maximum condition forces finite support. Some presentations reverse the index order and use a minimum instead; specify the convention when applying an embedding theorem.

Let a divisible ordered Abelian group be the union of a strictly increasing sequence of convex subgroups $C_1=0,C_2,\dots$, with each ordered quotient $C_{n+1}/C_n$ isomorphic to the additive real line. Convexity makes every $C_n$ divisible, so these are rational vector subspaces. Choose a rational-linear section for each quotient. Every element then has a unique finite sum of section values. Its sign is the sign of the coefficient of the largest active index, because convexity makes the corresponding ordered quotient decide its sign. The coefficients give an embedding, indeed an ordered-group isomorphism, into the Hahn group with positive integer indices in the maximum-support convention. No compatibility of a separate real scalar multiplication is needed for this group embedding.

The free object on generators $x_1,\dots ,x_n$ has the universal property that an arbitrary assignment of these generators to an Abelian lattice-ordered group extends uniquely to a lattice-ordered group homomorphism. It can be constructed from terms using addition, negation, zero, join and meet, modulo identities valid in every Abelian lattice-ordered group. Its underlying group need not be finitely generated even when there are only two lattice generators.

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.

The piecewise linear functions $h_n$, for distinct integer indices, are linearly independent over the integers. Restrict a finite integer relation to $y=1$. At $x=n$, only $h_n$ has a jump of one in its slope; hence the coefficient of that function must be zero. This proves that the underlying group of the two-generated free Abelian lattice-ordered group has infinite rank of an abelian group.

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 transitive lattice-ordered permutation group is order-primitive when its only invariant equivalence relations with convex classes are equality and the universal relation. This concerns convex blocks in a chain, rather than arbitrary blocks of a primitive group action.

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.

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.

An ordered permutation group is order n-transitive if it can send any increasing $n$-tuple to any other increasing $n$-tuple. This compares tuples with the same relative order, unlike unrestricted n-transitive group actions. In the usual transitive chain setting, order two-transitivity implies every finite order-transitivity for a lattice-ordered permutation group.

Suppose $h_{ij}$ sends the $i$th and $j$th nodes of an increasing input tuple to the corresponding output nodes. At the $i$th input, every $h_{ij}$ agrees, so their pointwise minimum takes the required value. At every other input it is bounded above by the required value, using the corresponding $j$ term. The pointwise maximum over $i$ therefore interpolates the entire tuple. The construction uses closure under finite join and meet, without assuming arbitrary restrictions of group elements belong to the group.

A positive bump of the real line is an order automorphism whose nonempty support of a permutation is one open interval and which moves every point of that interval to the right. It fixes the complement. Iterating any interior point tends to the right endpoint in positive time and to the left endpoint in negative time.

Choose fundamental intervals $[x,xf)$ and $[y,yg)$ for two positive bumps of an order automorphism. Any increasing bijection matching their endpoints extends by $b(x{f}^n)=(xb)g^n$ to their support intervals. Extend it monotonically over the two complementary rays. Then $fb=bg$, so $b^{-1}fb=g$. Bounded supports impose no further invariant for positive bumps in the full real-line order-automorphism group.

If two positive bumps of an order automorphism move a point $c$ beyond an interval $[c,d]$, their conjugacy can be chosen to fix that interval: begin with the identity there, complete the increasing map between fundamental intervals, and extend equivariantly along bump iterates. If both bump supports lie in one fundamental interval of another positive motion $h$, fix that cell's endpoints and repeat the conjugator on all its $h$-translates. The result commutes with $h$, and with any automorphism supported in the interval fixed pointwise.