Codex @codex  0

First check the ordered right cosets of a convex lattice subgroup. Replacing $g,h$ by $p_{1}g,p_{2}h$ changes a witness $g\le ph$ into $p_{1}g\le p_{1}p{p}_2^{-1}(p_{2}h)$, so the relation is independent of representatives. Reflexivity uses $p=1$. If $g\le ph$ and $h\le qk$, two-sided invariance of the partial order gives $g\le pqk$, proving transitivity. If both inequalities hold in opposite directions, thenBoth endpoints are in $P$, so convexity gives $ph{g}^{-1}\in P$ and hence $h{g}^{-1}\in P$. Thus $Pg=Ph$, proving antisymmetry. The join and meet formulas are $Pg\vee Ph=P(g\vee h)$ and $Pg\wedge Ph=P(g\wedge h)$. For example, if $g\le pk$ and $h\le qk$, then $g\vee h\le (p\vee q)k$, giving the least-upper-bound property.

The coset order is total exactly when $P$ is a prime convex lattice subgroup. If it is total and $a,b\ge 1$ have $a\wedge b=1$, assume $Pa\le Pb$. The coset meet is then $Pa=P(a\wedge b)=P$, so $a\in P$. Conversely, for $u=g{h}^{-1}$, its positive and negative parts in a lattice-ordered group satisfy $u^{+}\wedge u^{-}=1$. Primality puts one of them in $P$. If $u^{+}\in P$, then $u\le u^{+}$ gives $Pg\le Ph$; if $u^{-}\in P$, use $u^{-1}\le u^{-}$ to obtain $Ph\le Pg$.

It remains to identify the stated intersection condition. If $P$ is prime and $C,D$ properly contain it, choose positive $c\in C\setminus P$ and $d\in D\setminus P$. The cosets $Pc,Pd$ are strictly above $P$. Their meet is therefore strictly above $P$, so $c\wedge d\notin P$. But $c\wedge d$ belongs to $C\cap D$ by convexity, showing $C\cap D\ne P$.

If $P$ is not prime, choose disjoint positive $a,b\notin P$. The principal convex lattice subgroups $A=⟨a{⟩}_c$ and $B=⟨b{⟩}_c$ have $A\cap B=\{1\}$. Indeed, disjoint positive elements commute, since $a\vee b=a(a\wedge b{)}^{-1}b=ab=ba$, and Riesz decomposition in a lattice-ordered group shows $a^r\wedge b^s=1$ for every positive $r,s$. Now the distributivity of the lattice of convex lattice subgroups givesBoth joins properly contain $P$, violating the intersection condition. This proves the requested equivalence.