# 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 $a\leq b$, the poset interval $[a,b]\subseteq A$ also. In other words, $c\in A$ for any $c\in P$ such that $a\leq c$ and $c\leq b$. 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 $e\leq a\in H$. 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 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 $(a,b)\neq(0,0)=O$. We divide this into several cases:

1. 1.

$ab>0$. If $a>0$, then $b>0$ ($P$ in the first quadrant  ), so that $O\leq P$, which means $[O,P]\subseteq H$. If $a<0$, then $b<0$ ($P$ in the third quandrant), so that $O\leq-P$. In either case, $H$ contains a rectangle   ($[O,P]$ or $[O,-P]$) that generates $G$, so $H=G$.

2. 2.

One of $a$ or $b$ is $0$. Suppose $a=0$ for now. Then either $0 so that $[O,P]\subseteq H$ or $b<0$ 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. (a)

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

2. (b)

If $cd<0$, say $d<0$ (or $0), then for some positive integer $n$, $0, so that $O\leq Q+nP$, and $H=G$ as well. On the other hand, if $c<0$ (or $0), then $-Q$ returns us to the previous argument and $H=G$ again.

3. (c)

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

$ab<0$. It is enough to assume that $0 and $b<0$ (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 $(r,s)<(t,u)$ on $L$, then the slope of $L$ is positive

 $0<\frac{u-s}{t-r},$

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. (a)

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

2. (b)

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

3. (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 $0 and $d<0$. Since $L$ passes through $P$ and not $Q$, we have that

 $\frac{a}{c}\neq\frac{b}{d}.$

Let $0 and $0 and assume $r. Then there is a rational number $m/n$ (with $0) such that

 $r<\frac{m}{n}

This means that $na and $nb, or $nP. 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 $<0$, or $H=G$.

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 non-trivial, say $P=(a,b)\in H$, $P\neq O$. If $0, then $(c,d)\leq(a,b)$ for any $c regardless of $d$. Choose $Q=(c,d)$ to be in the first quadrant. Then $[O,Q]\subseteq H$, so that $H=G$. If $a<0$, then $-P$ takes us back to the previous argument. If $a=0$, then either $[O,P]$ (when $0), or $[O,-P]$ (when $b<0$) 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 $\mathbb{R}^{2}$ with lexicographic order  has the $y$-axis as the only non-trivial proper convex subgroup.

## References

Title convex subgroup ConvexSubgroup 2013-03-22 17:04:04 2013-03-22 17:04:04 CWoo (3771) CWoo (3771) 4 CWoo (3771) Definition msc 06A99 msc 06F15 msc 06F20 msc 20F60 convex subset