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).