convex subgroup
We begin this article with something more general. Let $P$ be a poset. A subset $A\subseteq P$ is said to be convex if for any $a,b\in A$ with $a\le b$, the poset interval $[a,b]\subseteq A$ also. In other words, $c\in A$ for any $c\in P$ such that $a\le c$ and $c\le b$. Examples of convex subsets are intervals^{} themselves, antichains^{}, whose intervals are singletons, and the empty set^{}.
One encounters convex sets most often in the study of partially ordered groups. A convex subgroup $H$ of a pogroup $G$ is a subgroup^{} of $G$ that is a convex subset of the poset $G$ at the same time. Since $e\in H$, we have that $[e,a]\subseteq H$ for any $e\le a\in H$. Conversely, if a subgroup $H$ satisfies the property that $[e,a]\subseteq H$ whenever $a\in H$, then $H$ is a convex subgroup: if $a,b\in H$, then ${a}^{1}b\in H$, so that $[e,{a}^{1}b]\subseteq H$, which implies that $[a,b]=a[e,{a}^{1}b]\subseteq H$ as well.
For example, let $G={\mathbb{R}}^{2}$ be the pogroup under the usual Cartesian ordering. $G$ and $0$ are both convex, but these are trivial examples. Let us see what other convex subgroups $H$ there are. Suppose $P=(a,b)\in H$ with $(a,b)\ne (0,0)=O$. We divide this into several cases:
 1.

2.
One of $a$ or $b$ is $0$. Suppose $a=0$ for now. Then either $$ so that $[O,P]\subseteq H$ or $$ so that $[O,P]\subseteq H$. In either case, $H$ contains a line segment^{} on the $y$axis. But this line segment generates the $y$axis. So $y$axis $\subseteq H$. If $H$ is a subgroup of the $y$axis, then $H$=$y$axis.
Otherwise, another point $Q=(c,d)\in H$ not on the $y$axis. We have the following subcases:

(a)
If $cd>0$, then $H=G$ as in the previous case.

(b)
If $$, say $$ (or $$), then for some positive integer $n$, $$, so that $O\le Q+nP$, and $H=G$ as well. On the other hand, if $$ (or $$), then $Q$ returns us to the previous argument and $H=G$ again.

(c)
If $d=0$ (so $c\ne 0$), then either $O\le P+Q$ (when $$) or $O\le PQ$ (when $$), so that $H=G$ once more.
A similar set of arguments shows that if $H$ contains a segment of the $x$axis, then either $H$ is the $x$axis or $H=G$. In conclusion^{}, in the case when $ab=0$, $H$ is either one of the two axes, or the entire group.

(a)

3.
$$. It is enough to assume that $$ and $$ (that $P$ lies in the fourth quadrant), for if $P$ lies in the second quadrant, $P$ lies in the fourth.
Since $O,P\in H$, $H$ could be a subgroup of the line group $L$ containing $O$ and $P$. No two points on $L$ are comparable^{}, for if $$ on $L$, then the slope of $L$ is positive
$$ a contradiction^{}. So $L$, and hence $H$, is an antichaine. This means that $H$ is convex.
Suppose now $H$ contains a point $Q=(c,d)$ not on $L$. We again break this down into subcases:

(a)
$Q$ is in the first or third quandrant. Then $H=G$ as in the very first case above.

(b)
$Q$ is on either of the axes. Then $H=G$ also, as in case 2(b) above.

(c)
$Q$ is in the second or fourth quadrant. It is enough to assume that $Q$ is in the same quadrant as $P$ (fourth). So we have $$ and $$. Since $L$ passes through $P$ and not $Q$, we have that
$$\frac{a}{c}\ne \frac{b}{d}.$$ Let $$ and $$ and assume $$. Then there is a rational number $m/n$ (with $$) such that
$$ This means that $$ and $$, or $$. But $nP,mQ\in H$, so is $R=mQnP\in H$, which is in the first quadrant. This implies that $H=G$ too.
In summary, if $H$ contains a point in the second or fourth quadrant, then either $H$ is a subgroup of a line with slope $$, or $H=G$.

(a)
The three main cases above exhaust all convex subgroups of ${\mathbb{R}}^{2}$ under the Cartesian ordering.
If the Euclidean plane^{} is equipped with the lexicographic ordering, then the story is quite different, but simpler. If $H$ is nontrivial, say $P=(a,b)\in H$, $P\ne O$. If $$, then $(c,d)\le (a,b)$ for any $$ regardless of $d$. Choose $Q=(c,d)$ to be in the first quadrant. Then $[O,Q]\subseteq H$, so that $H=G$. If $$, then $P$ takes us back to the previous argument. If $a=0$, then either $[O,P]$ (when $$), or $[O,P]$ (when $$) is a positive interval on the $y$axis. This implies that $H$ is at least the $y$axis. If $H$ contains no other points, then $H=y$axis. In summary, the pogroup ${\mathbb{R}}^{2}$ with lexicographic order^{} has the $y$axis as the only nontrivial proper convex subgroup.
References
 1 G. Birkhoff Lattice Theory, 3rd Edition, AMS Volume XXV, (1967).
Title  convex subgroup 

Canonical name  ConvexSubgroup 
Date of creation  20130322 17:04:04 
Last modified on  20130322 17:04:04 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  4 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 06A99 
Classification  msc 06F15 
Classification  msc 06F20 
Classification  msc 20F60 
Defines  convex subset 