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 po-group $G$ 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 $G={ℝ}^2$ be the po-group 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. $ab>0$. If $a>0$, then $b>0$ ($P$ in the first quadrant), so that $O\le P$, which means $[O,P]\subseteq H$. If , then ($P$ in the third quandrant), so that $O\le -P$. In either case, $H$ contains a rectangle ($[O,P]$ or $[O,-P]$) that generates $G$, so $H=G$.

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:

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

   2. 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.

   3. If $d=0$ (so $c\ne 0$), then either $O\le P+Q$ (when ) or $O\le P-Q$ (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.

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:

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

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

   3. $Q$ 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 $R=mQ-nP\in H$, 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 $H=G$.

The three main cases above exhaust all convex subgroups of ${ℝ}^2$ 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 non-trivial, 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 po-group ${ℝ}^2$ with lexicographic order has the $y$-axis as the only non-trivial proper convex subgroup.