Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 4 (Partially Ordered Groups)

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.

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.

The Ohnishi orderability criterion states that, for each finite list $g_1,\dots ,g_n\ne 1$, there are signs ${\epsilon }_i\in \{-1,1\}$ such that the semigroup of nonempty products of conjugates of $g_i^{{\epsilon }_i}$ does not contain $1$. The conjugates are essential for a two-sided invariant total order. Necessity follows by choosing all signed elements positive: their conjugates and nonempty products are still positive.

For sufficiency, consider the compact product $\{+,-{\}}^{G\setminus \{1\}}$. Impose the closed conditions that an inverse has the opposite sign, that conjugation preserves sign, and that a product of two positive elements is positive whenever it is nonidentity. A positive pair cannot multiply to the identity because of the inverse condition. Each condition uses finitely many coordinates. Given finitely many conditions, collect every nonidentity element appearing in them into a finite list and choose signs using the criterion. These signs satisfy the conditions: an incorrectly negative product would contribute its positive inverse to the generated semigroup, producing identity with the positive factors; an incorrect conjugate or inverse gives the same contradiction. Extend the signs arbitrarily to the remaining coordinates. Thus the finite intersection property and compactness give a global sign assignment. Its positive elements form a conjugation-invariant semigroup $S$ with $S\cap S^{-1}=\varnothing$ and $S\cup S^{-1}=G\setminus \{1\}$. Defining $a<b$ by $a^{-1}b\in S$ gives the required linearly ordered group.

Restriction shows that every finitely generated subgroup of an orderable group is orderable. Conversely, suppose every such subgroup is orderable. If the criterion failed on $g_1,\dots ,g_n$, each of the finitely many sign choices would have an identity witness made of finitely many conjugates. Collect all $g_i$ and all conjugating elements in these witnesses into one finitely generated subgroup $H$. An order on $H$ supplies one positive sign choice, but its corresponding witness is a nonempty product of positive elements in $H$, a contradiction. Hence the criterion holds in $G$.

Finally, a finitely generated subgroup of a torsion-free abelian group is ${\mathbb{Z}}^r$ by the Fundamental theorem of finitely generated abelian groups. Its lexicographic order is two-sided invariant. Applying the local-to-global conclusion proves that every torsion-free abelian group admits the required order.

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.

Here transitivity is order n-transitivity: increasing tuples are compared, rather than arbitrary reorderings. For increasing tuples ${\alpha }_1<\cdots <{\alpha }_n$ and ${\beta }_1<\cdots <{\beta }_n$, choose $h_{ij}$ sending both ${\alpha }_i$ to ${\beta }_i$ and ${\alpha }_j$ to ${\beta }_j$ when $i\ne j$, using order two-transitivity. For $i=j$, choose an element sending the single point to its target, using transitivity. Define

This belongs to the lattice-ordered permutation group. At ${\alpha }_i$, all terms in its $i$th inner meet equal ${\beta }_i$. At ${\alpha }_k$, every inner meet is at most ${\beta }_k$, because its $j=k$ term takes that value. The outer join is consequently exactly ${\beta }_k$ there. Thus $h$ maps the whole tuple to its target. This finite interpolation by lattice operations proves every finite order-transitivity without assuming the group permits arbitrary cut-and-paste restrictions of its elements.

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.

Write the two supports as $I_f=(l_f,r_f)$ and $I_g=(l_g,r_g)$. Every point in a support moves strictly right: a positive nonidentity order automorphism has no fixed point inside its one support interval. Choose $u\in I_f$, $v\in I_g$ and an increasing bijection $\theta :[u,uf]\to [v,vg]$ matching endpoints. The half-open intervals $[u{f}^n,u{f}^{n+1})$, for $n\in \mathbb{Z}$, partition $I_f$, because the iterates tend to its endpoints. DefineThe endpoint choices make adjacent pieces fit, giving an increasing bijection $b:I_f\to I_g$ with $fb=bg$. Extend over the two complementary closed rays by increasing maps taking $l_f,r_f$ to $l_g,r_g$. Outside the supports both automorphisms are identity, so $fb=bg$ still holds. Therefore $\boxed{b^{-1}fb=g}$. This gives the conjugacy of positive real-line bumps.

Construct the conjugacy of positive real-line bumps using $u=v=\alpha$. Since both $\alpha f$ and $\alpha g$ are greater than $\beta$, choose the fundamental-interval map to be identity on $[\alpha ,\beta ]$ and then increase from $\beta$ to the respective right endpoints. Equivariant extension along $f$ and $g$ gives a conjugator on their supports fixing $[\alpha ,\beta ]$. Extend it over the remaining portions of $J=[\gamma ,\gamma h]$, fixing the endpoints of $J$; both support intervals lie strictly inside $J$. Call the resulting interval bijection $b_0$. It satisfies $f{b}_0=b_{0}g$ on $J$.

The point $\gamma$ lies in a positive orbital of $h$, because $\gamma h>\gamma$. Its half-open cells $J{h}^n$, $n\in \mathbb{Z}$, partition that orbital. Extend byand set $b$ to identity outside the orbital. Since $b_0$ fixes the cell endpoints, this is an order automorphism and satisfies $bh=hb$. Each cell is preserved; $f$ and $g$ are identity on all cells other than the original one, so $fb=bg$ holds everywhere. Finally $b$ fixes $[\alpha ,\beta ]$ pointwise, and $a$ is supported inside that interval. Hence $ba=ab$. ThusThis is conjugating positive bumps while preserving a fundamental interval; merely choosing an arbitrary conjugator would not ensure either commutation relation.

The additive Abelian group $G$ is divisible because it is a real vector space. Each convex subgroup $C_n$ is divisible too: for $c\ge 0$ in $C_n$ and a positive integer $m$, its unique additive quotient $c/m$ satisfies $0\le c/m\le c$, hence belongs to $C_n$. Negative elements follow by taking inverses. Thus all $C_n$ are rational vector subspaces, independently of whether the given real scalar multiplication respects the order.

Choose an ordered-group isomorphism ${\theta }_n:C_{n+1}/C_n\to \mathbb{R}$. It is rational-linear, since additive homomorphisms between divisible torsion-free groups preserve division by integers. Extend a rational basis of $C_n$ to one of $C_{n+1}$; equivalently choose a rational-linear section $s_n:\mathbb{R}\to C_{n+1}$ of ${\theta }_n$ composed with the quotient map. ThenSince $C_1=0$ and the union is $G$, each $g\in G$ has a unique expression $g={\sum }_{j=1}^N{s}_j(r_j)$ with only finitely many nonzero coefficients. Define $\Phi (g)=(r_1,r_2,\dots )$. This is an injective additive group homomorphism. If $r_N\ne 0$ is the largest active coefficient, the image of $g$ in the ordered quotient $C_{N+1}/C_N$ has sign equal to that of $r_N$. Convexity ensures that this is also the sign of $g$.

The maximum-support convention for the Hahn group $V({\mathbb{Z}}_{>0},\mathbb{R})$ uses precisely finite supports and the sign of the largest active coefficient. Hence $\Phi$ preserves and reflects the total order, and therefore preserves join and meet. This proves the finite-support Hahn embedding along a countable convex chain. In fact every finite coefficient sequence is obtained, so $\Phi$ is an ordered-group isomorphism in this convention; the requested embedding follows.

No: there are at most three, including zero and the whole group. A two-generator ordered Abelian group is torsion-free and has rank of an abelian group at most two. A convex subgroup of an ordered group is pure: if $ng\in C$ with $n>0$, then $g$ lies between zero and $ng$ or their negatives, so $g\in C$. Therefore a proper inclusion $C<D$ of convex subgroups increases rational rank: an element of $D\setminus C$ cannot have a nonzero integer multiple in $C$. Convex subgroups form a chain, so a rank-two group permits at most two strict inclusions. This is the bound for convex subgroups of a finite-rank ordered Abelian group. It is sharp: lexicographically ordered group ${\mathbb{Z}}^2$ has exactly $0$, $\{0\}\times \mathbb{Z}$, and ${\mathbb{Z}}^2$.

A concrete model of the two-generated free Abelian lattice-ordered group is the lattice subgroup of real-valued functions on ${\mathbb{R}}^2$ generated by the coordinate projections $X(x,y)=x$, $Y(x,y)=y$. Its elements are finite joins of finite meets of integer linear forms $ax+by$. Thus they are continuous positively homogeneous piecewise linear functions, with pointwise addition and lattice operations. This is the piecewise integer-linear representation of a free Abelian lattice-ordered group.

For the construction, first form the term algebra on two symbols using addition, negation, zero, join and meet, then identify terms that agree in every Abelian lattice-ordered group. Evaluation under any assignment of the two generators factors through this quotient, proving its universal property. To obtain the concrete function model, use the identity-detection theorem that the real line generates the variety of Abelian lattice-ordered groups. Its proof passes a failed identity to a prime ordered quotient, chooses its finitely many active linear pieces, and uses the rational-inequality alternative underlying Farkas' lemma to find a real witness. This makes evaluation on ${\mathbb{R}}^2$ injective as well as surjective onto the generated function group.

Its underlying group is not ${\mathbb{Z}}^n$ for any finite $n$. The integer-linear hinge functions have infinite independent rank: $h_k(x,y)=(x-ky)\vee 0$ are integer-linearly independent for distinct integers $k$. On $y=1$, only $h_k$ has a slope jump at $x=k$, so each coefficient in a finite relation must vanish. The underlying group therefore has infinite rank of an abelian group despite having two lattice generators.

No. The free Abelian lattice-ordered group on two generators is countable, since every element is represented by a finite term over a countable alphabet of operation symbols and the two generators. The additive group of all real-valued functions on the given half-plane is uncountable: it already contains a distinct constant function for every real number. A lattice-ordered group homomorphism that is an isomorphism would be a bijection, contradicting these cardinalities. The concrete free-group model consists of specially constrained integer-linear term functions, not all real-valued functions on a domain.

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.