# partially ordered group, positive cone, quotient (exercise)

Definitions: A partially ordered group or po-group is a po-set $(G,\leq)$, such that $G$ is a group and $\forall x,y,a,b\!\in\!G\!:x\!\leq\!y\Rightarrow axb\!\leq\!ayb$, i.e. a po-set that is a group in which left and right translations are isotone (= order preserving). A directed group or do-group is a po-group that is an (up and down) directed set. A lattice ordered group or lo-group is a po-group that is a lattice. A totally ordered group or to-group is a po-group that is a totally ordered set. The (positive) cone of $G$ is $G_{\geq1}\!=\!G_+\!:=\!\{g\!\in\!G;\, g\!\geq\!1\}$.

For a group $G$ and subsets $A,B\!\subseteq\!G$, let $\langle A\rangle$ denote the subgroup generated by $A$, and let $AB\!=\!\{ab; a\!\in\!A,b\!\in\!B\}$ and $A^{-1}\!=\!\{a^{-1}; a\!\in\!A\}$.

Theorem (order/cone correspondence for groups) [Steinberg: Lattice-ordered Rings and Modules p.34-35; Blyth: Lattices and Ordered Algebraic Structures p.144-145]: Let $G$ be a group and $P\!\subseteq\!G$ a subset. Consider the following conditions on $P$: $$\text{(i) } PP\!\subseteq\!P; \hspace{3mm} \text{(ii) } \forall g\!\in\!G\!: gPg^{-1}\!\subseteq\!P; \hspace{3mm} \text{(iii) } P\!\cap\!P^{-1}\!=\!\{1\}; \hspace{3mm} \text{(iv) } PP^{-1}\!=\!G; \hspace{3mm} \text{(v) } P\!\cup\!P^{-1}\!=\!G.$$

a) Define the relation $x\leq_P\,y\Leftrightarrow x^{-1}y\!\in\!P$ on $G$. Call $P$ a po-cone / do-cone / lo-cone / to-cone on $G$ when (i)-(iii) hold / (i)-(iii),(iv) hold / (i)-(iii),(iv) hold and $(P,\leq_P\!)$ is a lattice / (i)-(iii),(v) hold. Then the maps $\begin{smallmatrix} \scriptstyle\leq &\!\scriptstyle \mapsto \! &\scriptstyle G_{\geq1}\\ \scriptstyle\leq_P &\!\scriptstyle \leftarrow\! &\scriptstyle P \end{smallmatrix}$ are mutually inverse bijections on the following pairs of sets: $$\begin{array}{r @{\hspace{1mm}} c @{\hspace{1mm}} l} \{\leq\subseteq\!G^2; \text{ the pair }(G,\leq)\text{ is a po-group}\} & \rightleftarrows & \{P\!\subseteq\!G; \text{ the subset }P\text{ is a po-cone on }G\}\\ \bigcup\!\mathbf{|}\hspace{2.6cm} & & \hspace{0.0cm}\bigcup\!\mathbf{|}\\ \{\leq\subseteq\!G^2; \text{ the pair }(G,\leq)\text{ is a do-group}\} & \rightleftarrows & \{P\!\subseteq\!G; \text{ the subset }P\text{ is a do-cone on }G\}\\ \bigcup\!\mathbf{|}\hspace{2.6cm} & & \hspace{0.0cm}\bigcup\!\mathbf{|}\\ \{\leq\subseteq\!G^2; \text{ the pair }(G,\leq)\text{ is a lo-group}\} & \rightleftarrows & \{P\!\subseteq\!G; \text{ the subset }P\text{ is a lo-cone on }G\}\\ \bigcup\!\mathbf{|}\hspace{2.6cm} & & \hspace{0.0cm}\bigcup\!\mathbf{|}\\ \{\leq\subseteq\!G^2; \text{ the pair }(G,\leq)\text{ is a to-group}\} & \rightleftarrows & \{P\!\subseteq\!G; \text{ the subset }P\text{ is a to-cone on }G\}. \end{array}$$

b) Let $(G,\leq)$ be a po-group with $P\!=\!G_{\geq1}$. T.f.a.e.: $(G,\leq)$ is a do-group; $(G,\leq)$ is directed up; $(G,\leq)$ is directed down; $\langle P\rangle \!=\!G$. T.f.a.e.: $(G,\leq)$ is an lo-group; $\forall g\!\in\!G\,\exists g\!\wedge\!1 \in G$; $\forall g\!\in\!G\,\exists g\!\vee\!1 \in G$.

Exercise: Let $G$ be a group with $H\!\unlhd\!G$, and suppose that $H$ and $G/H$ are po-groups. Prove that $$P:=H_+\cup\{g\!\in\!G;\, H\!\neq\!gH\!\in\!(G/H)_+\}$$ is a po-cone on $G$ iff $\forall g\!\in\!G\!: gH_+g^{-1}\!\subseteq\!H_+$. From now on, assume that $P$ is a po-cone on $G$, and $H,G/H$ have po-cones $H_+,(G/H)_+$. Prove: a) $H$ is convex in $G$ [and $P$ induces $H_+,(G/H)_+$]. b) $H\!<\!P\!\setminus\!H$. c) $G$ is a to-group iff $H$ and $G/H$ are to-groups. d) If $H\!\neq\!\{1\}$, then $G$ is an lo-group iff ($H$ is an lo-[sub]group and $G/H$ is a to-group). e) If $G$ is an lo-group, then the inclusion $H\!\rightarrow\!G$ is a complete lattice morphism.

Comment: I added the [...] parts (they were not originally part of the exercise).

Partial Solution: $(\Rightarrow)$: If $P$ is a po-cone, then $\forall g\!\in\!G\!: gPg^{-1}\!\subseteq\!P$, so for $h\!\in\!H_+$, we have $ghg^{-1}\!\in\!P$. From $H\!\unlhd\!G$ we get $ghg^{-1}\!\in\!H$, so we cannot have $H\!\neq\!ghg^{-1}H\!\in\!(G/H)_+$, hence $ghg^{-1}\!\in\!H_+$. $(\Leftarrow)$: (i): If $h,h'\!\in\!H_+$, then $hh'\!\in\!H_+$. If $h\!\in\!H_+$ and $H\!\neq\!gH\!\in\!(G/H)_+$, then $H\!\neq\!hgH\!\in\!(G/H)_+$ (if $hg\!\in\!H$, then $h^{-1}hg\!=\!g\!\in\!H$, $\rightarrow\leftarrow$; since $gH\!\in\!(G/H)_+$ and $hH\!=\!1H\!\in\!(G/H)_+$, we have $hHgH\!=\!hgH\!\in\!(G/H)_+$), and similarly, $H\!\neq\!ghH\!\in\!(G/H)_+$. If $H\!\neq\!gH\!\in\!(G/H)_+$ and $H\!\neq\!g'H\!\in\!(G/H)_+$, then $gg'H\!=\!gHg'H\!\in\!(G/H)_+$, and if $gg'\!\in\!H$, then ???. (ii): By assumption, $gH_+g^{-1}\!\subseteq\!H_+$. Furthermore, if $H\!\neq\!g'H\!\in\!(G/H)_+$, then $H\!\neq\!gg'g^{-1}H$ (since $H\!\unlhd\!G$) and $gg'g^{-1}H\!=\!gH(g'H)g^{-1}H\!\in\!(G/H)_+$. Thus $gPg^{-1}\!\subseteq\!P$. (iii): Assume that $g,g^{-1}\!\in\!P$. If $g,g^{-1}\!\in\!H_+$, then $g\!=\!1$. If $g\!\in\!H_+$ and $H\!\neq\!g^{-1}H\!\in\!(G/H)_+$, then $g^{-1}\!\in\!H$, $\rightarrow\leftarrow$. If $H\!\neq\!gH\!\in\!(G/H)_+$ and $H\!\neq\!g^{-1}H\!\in\!(G/H)_+$, then $gH\!\in\!(G/H)_+\!\cap\!(G/H)_+^{-1}$, so $gH\!=\!H$, $\rightarrow\leftarrow$.

a) If $h\!\leq\!g\!\leq\!h'$ and $h,h'\!\in\!H$, then $h^{-1}g, g^{-1}h'\!\in\!P$. If $h^{-1}g\!\in\!H_+$ or $g^{-1}h'\!\in\!H_+$, then $g\!\in\!H$. But if $h^{-1}gH\!\in\!(G/H)_+$ and $g^{-1}h'H\!\in\!(G/H)_+$, then $gH\!=\!hHh^{-1}gH\!\in\!(G/H)_+$ and $g^{-1}H\!=\!g^{-1}h'Hh'^{-1}H\!\in\!(G/H)_+$, i.e. $gH\!\in\!(G/H)_+\!\cap\!(G/H)_+^{-1}$, which implies $gH\!=\!H$, i.e. $g\!\in\!H$.

If $h,h'\!\in\!H$, then $h\!\leq_{H_+}\!h' \Leftrightarrow h^{-1}h'\!\in\!H_+ \Leftrightarrow h^{-1}h'\!\in\!P \Leftrightarrow h\!\leq_{P}\!h'$. If $gH,g'H\!\in\!G/H$, then $gH\!\leq_{\!(G/H)_+}\!\!g'H \Leftrightarrow g^{-1}g'H\!\in\!(G/H)_+ \Leftrightarrow \big((\exists h\!\in\!H\!:g^{-1}g'h\!\in\!H_+)\text{ or }(H\!\neq\!g^{-1}g'H\!\in\!(G/H)_+)\big) \Leftrightarrow \exists h\!\in\!H\!:\big(g^{-1}g'h\!\in\!H_+\text{ or }H\!\neq\!g^{-1}g'hH\!\in\!(G/H)_+)\big) \Leftrightarrow \exists h\!\in\!H\!: g\!\leq_P\!g'h$. In the second equivalence, $\Leftarrow$ is clear, but for $\Rightarrow$, either $g^{-1}g'H\!\neq\!H$ or $g^{-1}g'\!=\!h^{-1}$ for some $h\!\in\!H$, and then $g^{-1}g'h\!=\!1\!\in\!H_+$.

b) Let $h\!\in\!H$ and $H\!\neq\!gH\!\in\!(G/H)_+$. To prove $h\!<\!g$, we must show that $h^{-1}g\!\in\!P$. We have $h^{-1}gH\!=\!gH\!\neq\!H$ and $h^{-1}gH\!=\!gH\!\in\!(G/H)_+$.

c) $(\Rightarrow)$: Assume $P\!\cup\!P^{-1}\!=\!G$. For any $h\!\in\!H$, either $h\!\in\!P$ or $h^{-1}\!\in\!P$, but since $H\!=\!hH\!=\!h^{-1}H$, either $h\!\in\!H_+$ or $h^{-1}\!\in\!H_+$, i.e. $h\!\in\!H_+\!\cup\!H_+^{-1}$. For any $gH\!\in\!G/H$, either $g\!\in\!P$ or $g^{-1}\!\in\!P$. If $g\!\in\!H_+$ or $g^{-1}\!\in\!H_+$, then $gH\!=\!H\!\in\!(G/H)_+$ or $g^{-1}H\!=\!H\!\in\!(G/H)_+$. Otherwise $gH\!\in\!(G/H)_+$ or $g^{-1}H\!\in\!(G/H)_+$. $(\Leftarrow)$: Assume $H_+\!\cup\!H_+^{-1}\!=\!H$ and $(G/H)_+\!\cup\!(G/H)_+^{-1}\!=\!G/H$. For any $g\!\in\!G$, either $g\!\in\!H$ (then $g\!\in\!H_+\!\cup\!H_+^{-1}\!\subseteq\!P\!\cup\!P^{-1}$) or $gH\!\neq\!H$ (then $g^{-1}H\!\neq\!H$, and either $gH\!\in\!(G/H)_+$ or $g^{-1}H\!\in\!(G/H)_+$), so either $g\!\in\!P$ or $g^{-1}\!\in\!P$.

d) $(\Rightarrow)$: Let $G$ be an lo-group. If $h\!\in\!H$, then $\exists h\!\vee_P\!1\!=:\!g\!\in\!P$, and ???, so $g\!\in\!H$, hence $g\!=\!h\!\vee_{H_+}\!1$. $(\Leftarrow)$: Let $H$ be an lo-group and $G/H$ a to-group. It suffices to prove that $\forall g\!\in\!G\!\setminus\!H\!: \exists g\!\vee_P\!1$. We have either $gH\!\in\!(G/H)_+$ or $g^{-1}H\!\in\!(G/H)_+$, i.e. $hg\!\in\!P$ or $g^{-1}h\!\in\!P$ for some $h\!\in\!H$. ???

e) We must prove that $\exists\inf_H\{h_i; i\!\in\!I\}\!=:\!h_0\Rightarrow \inf_G\{h_i; i\!\in\!I\}\!=\!h_0$ and $\exists\sup_H\{h_i; i\!\in\!I\}\!=:\!h_1\Rightarrow \sup_G\{h_i; i\!\in\!I\}\!=\!h_1$. If $\exists g\!\in\!G\!\setminus\!H\!: g\!\leq\!\{h_i; i\!\in\!I\}$, then ???, so $g\!\leq\!h_0$.

Question: How can I finish the '???' parts? I'm stuck at $PP\!\subseteq\!P$, d), e).

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For the first gap: if $gg'\in H$, then $g'\in g^{-1}H$, so $g'H=(gH)^{-1}$, $g'H\in(G/H)_+\cap\big((G/H)_+\big)^{-1}$, and therefore $g'H=H$, $\to\leftarrow$.

(d) Let $\pi:G\to G/H:g\mapsto gH$. For $g,g'\in G$, $g\le_G g'$ iff $g^{-1}g'\in P$; that that this is the case iff either

• $\pi(g)<_{G/H}\pi(g')$, or
• $\pi(g)=\pi(g')$ and $1\le_H g^{-1}g'$.

Note in particular that if $g,g'\in H$, then $g\le_G g'$ iff $g\le_H g'$.

Suppose that $G/H$ is a to-group, and let $g\in G$. Then exactly one of the following holds: $\pi(g)=\pi(1)$, $\pi(g)<_{G/H}\pi(1)$, or $\pi(g)>_{G/H}\pi(1)$. If $\pi(g)>_{G/H}\pi(1)$, then $g>_G 1$, and $g\lor_G 1=g$. If $\pi(g)<_{G/H}\pi(1)$, then $g<_G 1$, and $g\lor_G 1=1$. And if $\pi(g)=\pi(1)$, then $g\in H$.

We’ll show now that $G$ is a lo-group iff $H$ is a lo-group. Assume first that $H$ is a lo-group, and let $g\in G$. We just saw that if $g\notin H$, then $g\lor_G 1$ exists and is either $g$ or $1$, so assume now that $g\in H$, and let $u=g\lor_H 1$. Then if $g'\in G$ is any upper bound for $g$ and $1$ in $G$, either $\pi(g')>_{G/H}\pi(1)$, in which case $u<_G g'$, or $g'\in H$, in which case $u\le_G g'$, so $u=g\lor_G 1$.

Now suppose that $H$ is not a lo-group, and let $h\in H$ be such that $h$ and $1$ have no supremum in $H$; I claim that $h$ and $1$ can have no supremum in $G$, either. This is clear if there is a $u\in H$ such that $h,1\le_H u$: if $g$ is any upper bound of $h$ and $1$ in $G\setminus H$, $u<_G g$, so $g\ne h\lor_G 1$. Thus, we may assume that $\{h,1\}$ has no upper bound in $H$. Suppose that $g$ is an upper bound for $\{h,1\}$ in $G$; clearly $\pi(g)>_{G/H}\pi(1)$. If $H_+\ne\{1\}$, choose $p\in H_+\setminus\{1\}$; then $h,1<_G gp^{-1}<_G g$, so $g\ne h\lor_G 1$. The only remaining possibility is that $H_+=\{1\}$. In that case $\pi(gh)=\pi(g)$, so $gh$ is an upper bound for $\{h,1\}$. However, it’s clear that $h\ne 1$, so $g^{-1}(gh)=h\notin P$, $g\not\le_G gh$, and $g\ne h\lor_G 1$. Thus, $h$ and $1$ have no supremum in $G$, and $G$ is not a lo-group.

To complete the proof of (d) we must show that if $G/H$ is not a to-group, then $G$ is not a lo-group. Suppose, then, that $G$ is not a to-group; then there is a $g\in G$ such that $\pi(g)\not\le_{G/H}\pi(1)\not\le_{G/H}\pi(g)$. Suppose that $u$ an upper bound for $\{g,1\}$ in $G$; then $\pi(u)>_{G/H}\pi(g),\pi(1)$. As before, if $H$ has a strictly positive element $p$, then $up^{-1}$ is an upper bound for $\{g,1\}$ strictly smaller than $u$. If $H_+=\{1\}$, let $h\in H\setminus\{1\}$, and observe that $uh$ is an upper bound for $\{g,1\}$ such that $u\not\le_G uh$. In either case $u\ne g\lor_G 1$, and $G$ is not a lo-group.

(e) Recall from (d) that if $g,g'\in H$, then $g\le_G g'$ iff $g\le_H g'$.

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Thank you for your reply. I do not understand your 5th (longest) paragraph. Why do we even need that "$h,1$ has no upper bound in $H$"? However, using your ideas, I found an analogous solution to (d) and (e) which I post below, since it's too long for a comment. Thank you again. – Leon Jul 27 '12 at 0:23

This is just a solution to (d) and (e) which I understand easier, but due to Brian M. Scott. It's too long for a comment.

d) Let $\pi:G\!\rightarrow\!G/H$, $g\!\mapsto\!gH$. For $g,g'\!\in\!G$, we have $g\!\leq_G\!g'$ iff $g^{-1}g'\!\in\!P$, iff either $\pi(g)\!<_{G/H}\!\pi(g')$ or ($\pi(g)\!=\!\pi(g')$ and $1\!\leq_H\!g^{-1}g'$).

$(\Leftarrow)$: Suppose that $G/H$ is a to-group. For $g\!\in\!G$, exactly one of the following holds: $\pi(g)\!=\!\pi(1)$, $\pi(g)\!<_{G/H}\!\pi(1)$, $\pi(g)\!>_{G/H}\!\pi(1)$. In the first case, $g\!\in\!H$. In the second, $g\!<_G\!1$, hence $g\!\vee_G\!1\!=\!1$. In the third, $g\!>_G\!1$, hence $g\!\vee_G\!1\!=\!g$. Thus if $g\!\notin\!H$, then $\exists\,g\!\vee_G\!1$. But if $g\!\in\!H$, let $s\!:=\!g\!\vee_H\!1$, so $\pi(s)\!=\!\pi(1)$. For any $g'\!\in\!G$ with $g,1\!\leq\!g'$, either $\pi(1)\!<_{G/H}\!\pi(g')$ (then $s\!<_G\!g'$), or $g'\!\in\!H$ (then $s\!\leq\!g'$). Thus $s\!=\!g\!\vee_G\!1$. We have proved that $g\!\vee_G\!1$ always exists, so by the correspondence theorem, $G$ is an lo-group.

$(\Rightarrow)$: Suppose that $G$ is an lo-group, but $H$ is not, i.e. $\exists h\!\in\!H\!:\nexists\,h\!\vee_H\!1$. Take $s\!:=\!h\!\vee_G\!1$. We see that $s\!\in\!G\!\setminus\!H$. By a), we have $H\!<\!s$, so $H\!<\!Hs$. If $H_+\!\neq\!\{1\}$, choose $p\!\in\!H_+\!\setminus\!\{1\}$, and then $h,1\!<\!p^{-1}\!s\!<\!s$, a contradiction with the minimality of $s$. But if $H_+\!=\!\{1\}$, then $h,1\!\leq\!hs$ gives (by minimality of $s$) $s\!\leq\!hs$, i.e. $h\!\in\!H_+$, so $h\!=\!1$, a contradiction.

Now suppose that $G$ is an lo-group, but $G/H$ is not a to-group, i.e. $\exists g\!: \pi(g)\!\nleq_{G/H}\!\pi(1)\!\nleq_{G/H}\!\pi(g)$. Take $s\!:=\!g\!\vee_G\!1$. Then $\forall h\!\in\!H\!:\pi(g),\pi(1)<_{G/H}\pi(s)=\pi(hs)$. If $H_+\!\neq\!\{1\}$, choose $p\!\in\!H_+\!\setminus\!\{1\}$, and then $g,1 < p^{-1}u < u$, a contradiction. If $H_+\!=\!\{1\}$, let $h\!\in\!H\!\setminus\!\{1\}$, and then $g,1\!\leq\!hs$, so $s\!\leq\!hs$, i.e. $h\!\in\!H_+$, a contradiction.

e) We must prove that $\exists\sup_H\{h_i;i\!\in\!I\}$ implies $\sup_H\{h_i;i\!\in\!I\}\!=\!\sup_G\{h_i;i\!\in\!I\}$, and that $\exists\inf_H\{h_i;i\!\in\!I\}$ implies $\inf_H\{h_i;i\!\in\!I\}\!=\!\inf_G\{h_i;i\!\in\!I\}$. By d), $H$ is an lo-group and $G/H$ is a to-group. We use the criterium from d). Suppose that $s\!:=\!\sup_H\{h_i\}\!\in\!H$ exists. For any $g\!\in\!G$ with $\{h_i\}\!\leq\!g$, either $\pi(s)\!=\!\pi(h_i)\!=\!\pi(g)$ (then $g\!\in\!H$, so $s\!\leq\!g$), or $\pi(s)\!=\!\pi(h_i)\!<\!\pi(g)$ (then $s\!<\!g$). Thus $s\!=\!\sup_G\{h_i\}$. The equality for infimums is proved analogously. $\blacksquare$

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