Partially Ordered Groups and Their Cones
Rodica Ceterchi
Faculty of Mathematics and Computer Science
University of Bucharest
[email protected], [email protected]
Abstract
We present results concerning the relation between partially ordered
groups and their positive (or, dually, negative) cones. The theorem of
Birkhoff gives an intrinsic characterization of positive cones of po-groups.
The theorem of Nakada characterizes po-semigroups as positive cones of
po-groups, stressing the importance of the ”natural order” and its inverse.
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Partially ordered groups and their cones
Definition 1.1 A partially ordered group (po-group) is a structure (G, ·, e, ≤)
such that:
(i) (G, ·, e) is a group
(ii) (G, ≤) is a poset
(iii) a ≤ b implies ac ≤ bc and ca ≤ cb.
Property (iii) relates the group operation · to the order, and we say that multiplication (both left and right) is increasing.
The positive cone of a po-group is the set G+ := {a ∈ G | a ≥ e}. The
negative cone of a po-group is the set G− := {a ∈ G | a ≤ e}. Note that
(G+ , ·, e, ≤) is a po-monoid, and only a monoid, because a ∈ G+ \ {e} implies
a−1 ∈ G− \ {e}, and G+ ∩ G− = {e}. The same is true for G− . Because a ≤ b
implies b−1 ≤ a−1 , the inverse −1 is a monoid isomorphism −1 : (G+ , ·, e) →
(G− , ·, e) which reverses the order (and the multiplication too (ab)−1 = b−1 a−1 ).
Because of this, results proven for positive cones can be immediately transferred
to negative cones, and viceversa.
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A characterization of positive cones
Let (G, ·, e) be a group (no order), and let P ⊆ G be a subset of G (which
implies (P, ·) semigroup of G). The question is: what conditions must P fulfill
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in order to be the positive cone of some order on G? Under what conditions
does a sub-semigroup P of G determine an order ≤ on G, such that P = G+ ?
The answer is given by the next result.
Theorem 2.1 (Theorem 2, chapter II of [2])
A subset P of a group G is the positive cone of some partial order on G iff
it satisfies the following conditions:
(α) P ∩ P −1 = {e}
(β) P P ⊆ P
(γ) xP x−1 ⊆ P for any x ∈ G.
Proof: If (G, ≤) is a po-group, G+ has properties (α), (β) and (γ): the first
two are immediate, for (γ): a ≥ e implies xa ≥ x implies xax−1 ≥ xx−1 = e,
by using twice the monotony of ·.
For the reverse implication, let G be a group and P ⊆ G with properties
(α), (β) and (γ). Define on G the relation ≤: for a, b ∈ G,
a ≤ b ⇔ ba−1 ∈ P
Note that P is the positive cone of this relation is immediate from the definition:
e ≤ a ⇔ ae−1 = a ∈ P . We show that it is an order relation.
a ≤ a ⇔ aa−1 = e ∈ P .
a ≤ b ⇔ ba−1 ∈ P
b ≤ a ⇔ ab−1 ∈ P ⇔ (ba−1 )−1 ∈ P ⇔ ba−1 ∈ P −1
and from these two last relations, by (α), ba−1 ∈ P ∩ P −1 = {e}, and ba−1 =
e ⇔ a = b. For transitivity,
a ≤ b ⇔ ba−1 ∈ P
b ≤ c ⇔ cb−1 ∈ P
and from this and (β), (cb−1 )(ba−1 ) ∈ P P ⊆ P , i.e. ca−1 ∈ P ⇔ a ≤ c.
Next we show the monotony of · w.r.t. ≤. Let a ≤ b, i.e.ba−1 ∈ P . We
compute
(bc)(ac)−1 = (bc)(c−1 a−1 ) = ba−1 ∈ P , thus ac ≤ bc.
(cb)(ca)−1 = c(ba−1 )c−1 ∈ cP c−1 ⊆ P , so c(ba−1 )c−1 ∈ P by (γ), thus
ca ≤ cb.
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A second characterization of positive cones
(Birkhoff )
The previous result characterizes the positive cone as a subset of G. An intrinsic
characterization of positive cones is given by the next Theorem due to Birkhoff
[1] (which is Theorem 4 in chapter II of [2]).
Theorem 3.1 (Birkhoff ) An arbitrary semigroup (P, ·) is the positive cone of
some po-group G, iff the following hold:
(i) The cancellation laws hold in P ;
(ii) There is a neutral element e ∈ P ;
(iii) a, b ∈ P , ab = e implies a = b = e;
(iv) P a = aP for any a ∈ P .
Proof: The necessity is immediate. Let P = G+ for some po-group G.
Then (i) and (ii) follow immediately. For (iii) let a, b ∈ G+ such that ab = e,
which implies a = b−1 . But a ∈ G+ , so b−1 ∈ G+ , so b ∈ G+ ∩ G− , so b = e
and a = e. For (iv): from x ≥ e follows by monotony axa−1 ≥ aa−1 = e,
so aG+ a−1 ⊆ G+ , from which aG+ ⊆ G+ a follows, and the reverse inclusion
follows similarly.
Sufficiency. We will construct a po-group G for which P will be the positive
cone.
For a ∈ P and x ∈ P , by (iv), there exists an element xa ∈ P such that
xa = axa . By (i), this element is unique. Then, for a fixed, we have a function
P → P , x → xa . Moreover, this function is one-to-one: if xa = ya , then
xa = axa = aya = ya, and by (i) x = y follows. Let us prove some properties
of this family of functions.
(A) aa = a.
(B) (xy)a = xa ya .
We have: a(xy)a = (xy)a = x(ya) = x(aya ) = (xa)ya = axa ya , and apply (i).
(C) (xa )b = xab .
We have: (ab)xab = x(ab) = (xa)b = (axa )b = b(axa )b = b(ab (xa )b ) =
(bab )(xa )b = (ab)(xa )b , from the definition and (B), and now apply (i).
(D) ea = e and xe = x.
Take now the direct product P × P . We define on it the following binary
relation:
(a, b) ∼ (c, d) ⇐⇒ adb = cb.
We prove it is an equivalence relation.
(a) ∼ is reflexive:
(a, b) ∼ (a, b) ⇐⇒ abb = ab, and this last relation is true, because, by (A),
bb = b.
(b) ∼ is symmetric:
Let (a, b) ∼ (c, d), i.e. adb = cb. From this, adb d = cbd. Now,
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db d = d(db )d = ddbd
by (C)
bd = (bd)bd = bbd dbd = bd dbd
by (A), (B) and (C)
Replacing, we obtain, addbd = cbd dbd which, by (i), implies ad = cbd , i.e.
(c, d) ∼ (a, b).
(c) ∼ is transitive:
Let (a, b) ∼ (c, d), i.e. adb = cb. Let also (a, b) ∼ (g, h), i.e. ahb = gb. We
multiply this last relation with db to the right, and obtain ahb db = gbdb . We
compute to transform this relation:
hb db = db (hb )db = db hbdb = db hdb
by (C)
ahb db = adb hdb = cbhdb = chd b
by previous, first hypothesis, and (C)
bdb = db
The relation becomes chd b = gdb, which by (i) becomes chd = gd, i.e. (c, d) ∼
(g, h).
We define now G to be G = (P × P )/ ∼.
We define a multiplication in G by:
(a, b) ? (c, d) = (acb , db)
? is associative:
[(a, b) ? (c, d)] ? (g, h) = (acb , db) ? (g, h) = (acb gdb , hdb)
(a, b) ? [(c, d) ? (g, h)] = (a, b) ? (cgd , hd) = (a(cgd )b , hdb)
cb gdb cb (gd )b = (cgd )b
by (C) and (B)
? has neutral element (a, a):
(a, a) ? (c, d) = (aca , da) = (ca, da)
(ca, da) ∼ (c, d) ⇔ cadda = cda ⇔ adda = da
by definition of ∼ and (i)
but adda = a(dd )a = ada = da
by (C) and (A)
(b, a) is inverse to (a, b):
(b, a) ? (a, b) = (baa , ba) = (ba, ba) ∼ (b, b) neutral
There is a natural embedding of P in G = (P × P )/ ∼, given by a → (a, e).
This is moreover a monoid isomorphism between P and {(a, e) | a ∈ P } ⊂ G,
because:
e → (e, e) ∼ (a, a) for any a ∈ P
ab → (ab, e) ∼ (a, e) ? (b, e)
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(a, e) ∼ (b, e) ⇔ aee = be ⇔ a = b, which gives injectivity.
We will identify P with its image in G, P = {(a, e) | a ∈ P }. Then P −1 =
{(e, a) | a ∈ P }. We prove that P has properties (α), (β) and (γ) of Theorem
2.1, and thus P will be the positive cone of a partial order on G.
(α) An element in P ∩ P −1 is of the form (a, e) and also (e, b). But,
(a, e) ∼ (e, b) ⇔ abe = e ⇔ ab = e, which implies by (iii) a = b = e,
thus P ∩ P −1 = {(e, e)}.
(β) (a, e) ? (b, e) = (ab, e) ∈ P , thus P P ⊆ P .
(γ) We will show that xP ⊆ P x for any x ∈ G. Take x = (c, d) ∈ G. We
have:
(c, d) ? (a, e) = (cad , d) = (bc, d) = (b, e) ? (c, d),
because cad ∈ cP = P c by (iv), thus there is a b ∈ P such that cad = bc.
We can apply now Theorem 2.1, and we conclude that G is a po-group, with
the partial order determined by its positive cone P .
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Let us have a quick glimpse at the order on G which has P as its positive
cone:
(a, b) ≤ (c, d) ⇔ (c, d) ? (a, b)−1 ∈ P ⇔ (c, d) ? (b, a) ∈ P ⇔ (c, d) ? (b, a) ∼
(x, e) for some x ∈ P
⇔ (cbd , ad) ∼ (x, e) for some x ∈ P ⇔ cbd = x(ad) for some x ∈ P ,
and note that this last condition involves only elements of P .
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A consequence involving the order (Nakada)
In what follows we will be concerned with the following question: if the monoid
P has already a partial order on it, what conditions must this order fulfill,
such that the above construction holds, P is embedded in a po-group G, as the
positive cone of its order, and the order on G extends that on P ?
Let us note that in a po-group the order is strongly related to the divisibility
relation. We call a ∈ G a right divisor of b ∈ G if there exists an x ∈ G such
that xa = b, and a is a left divisor of b if there exists an y such that ay = b.
Now, if the group is partially ordered, then we have:
a ≤ b ⇐⇒ a is a left divisor of b ⇐⇒ there exists an x ∈ G+ such that
xa = b
⇐⇒ a is a right divisor of b ⇐⇒ there exists an y ∈ G+ such that
ay = b
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The proof of =⇒ is immediate by taking x = ba−1 and y = a−1 b. For the proof
of ⇐= it is essential that x, y ∈ G+ : only then we have ba−1 ≥ e, thus b ≥ a,
and a−1 b ≥ e, thus b ≥ a. We say that the group is naturally ordered.
Definition 4.1 We call a po-monoid (P, ·, e, ≤) naturally ordered iff
a ≤ b ⇐⇒ there exist elements x, y ∈ P such that ax = b = ya.
We call a po-monoid (P, ·, e, ≤0 ) inverse naturally ordered iff
a ≤0 b ⇐⇒ there exist elements x, y ∈ P such that bx = a = yb.
We call a po-monoid (P, ·, e, ≤) positive iff all elements of P are positive.
We call a po-monoid (P, ·, e, ≤) negative iff all elements of P are negative.
Note that that the order ≤0 is the inverse of the order ≤:
a ≤0 b iff b ≤ a.
If (P, ·, e, ≤) is naturally ordered and positive, then (P, ·, e, ≤0 ) is inverse naturally ordered and negative, and viceversa.
As we expected, the answer to our question is that P must be naturally
ordered. We have the following result, due to Nakada [3] (also, Proposition 1 of
chapter X of [2]) .
Theorem 4.2 (Nakada) The following conditions are necessary and sufficient
for a po-semigroup (P, ·, ≤) to be the positive cone of a po-group G:
(a) P is cancellative.
(b) There exists a neutral element e ∈ P .
(c) P is naturally ordered and positive.
Proof: Necessity is obvious in view of the above remarks.
Sufficiency. We will prove that the conditions of Birkhoff’s Theorem 3.1
are fulfilled. But (a) is precisely (i) and (b) is (ii). Let us note that P positive
implies ab ≥ a, b for all a, b ∈ P .
(iii) Let a, b ∈ P such that ab = e. Then
ab ≥ a = ae ≥ e ⇒ ab = a = e,
ab ≥ b = be ≥ e ⇒ ab = b = e,
thus a = b = e and (iii) is fulfilled.
(iv) Let us show aP ⊆ P a for any a ∈ P . For a ∈ P fixed, consider any
b ∈ P \{e}. From ab ≥ a, because P is naturally ordered, a will be a right divisor
for ab, i.e. there exists an x ∈ P such that ab = xa, from which ab = xa ∈ P a.
The reverse inclusion is proved similarly, using the naturality of order and left
divisibility.
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In view of the remarks which preceeded Theorem 4.2 we have a dual of the
previous result.
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Theorem 4.3 (dual Nakada) In order for a po-semigroup (P, ·, ≤) to be the
negative cone of a po-group G, the following conditions are necessary and sufficient:
(a) P is cancellative.
(b) There exists a neutral element e ∈ P .
(c) P is inverse naturally ordered and negative.
A negative cone completely characterizes an order, precisely like the positive
cone. In fact, the negative cone of an order ≤ is nothing but the positive cone
of its reverse order ≤0 . Thus Theorems 2.1 and 3.1 hold if we replace positive
cone by negative cone.
References
[1] G. Birkhoff, Lattice-ordered groups, Annals Math. 43 (1942), 298-331
[2] L. Fuchs, Partially ordered algebraic systems, Pergamon Press, 1963
[3] O. Nakada, Partially ordered Abelian semigroups, I-II, Journ. Fac. Sci.
Hokkaido Univ. 11 (1951), 181-189, 12 (1952), 73-86
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