We define a subgroup $H$ as being convex if $g\in H\implies h\in H$ for all $1\leq h\leq g$. A convex subgroup $P$ is prime if for any two convex subgroups $X,Y$: $X\cap Y \subseteq P$ implies that $X\subseteq P$ or $Y\subseteq P$.
Say that $G$ is a latttice ordered subgroup of $A(S)$, the automorphisms of a totally ordered set $S$. Let $G_\alpha$ be the stabilizer of some $\alpha \in S$, i.e. $\{g:\alpha g=\alpha\}$.
Glass' Partially Ordered Groups states that $G_\alpha$ is "obviously" a prime (convex) subgroup of $G$, but I don't understand why.
I don't even see why it needs to be convex - while I can't come up with a counter example, I can't see any reason why if $g$ stabilizes $\alpha$ and $h\leq g$ then $h$ must also stabilize $\alpha$.
If $1 \leq h \leq g$ then $x \leq h(x) \leq g(x)$ for all $x \in S$ by definition of the lattice order on $A(S)$. Taking $x=\alpha$ we get $\alpha \leq h(\alpha) \leq \alpha$, as was to be shown ($G_a$ is convex).
Now suppose $h \wedge k =1$ for $h,k \in G$. Then $\min(h(\alpha),k(\alpha))=1(\alpha)$, that is,either $h(\alpha)=\alpha$ or $k(\alpha)=\alpha$. Thus either $h$ or $k$ in $G_a$. Hence $G_a$ is prime by Glass's Lemma 3.3.3.ii on page 37. This completes the proof of the “clearly” claim on page 140 (next to a proof of Holland's theorem).
The lattice order of $A(S)$ is described on the first page of Holland (1963) and also as Example 1.3.19 in Glass's Partially Ordered Groups. The prime part might also be proven in that paper (check the "tower" stuff), but this is not my area.