Is this subgroup of an $\ell$-group closed under this operation?

Question

I have an $\ell$-group $G$ and a subgroup $H$ such that $H\cap G^+=\{id\}$. Consider two elements $b,h\in H$. Is it necessarily true that $h^+-b-h^-\in H$? This is clearly true in the abelian case, as $h^+-b-h^-=h-b$, but I don't know if it holds in general.

My progress is that I have an example of a noncommutative $G$ where $h^+-b-h^-\neq h-b$, but membership in $H$ still holds. Note that it is tradition to use +/- in $\ell$-groups, but in an arbitrary group, the notation would be $h^+b^{-1}(h^-)^{-1}$.

Answer 1

First, a couple of simplifications: $(h ^ -) ^ {-1} =(h ^ {-1}\vee 1)^{-1} = (h\wedge 1)$, and if $b^ {-1}\in H $ then $b\in H$. So your above operation is equivalent to $h ^ + b (h\wedge 1)$.

Now, let $G= Aut(\mathbb R)$, the group of automorphisms of the real numbers under composition, where $f\leq g\equiv\forall x\in\mathbb R: x f\leq x g$. Let $h = x ^ 3, b = x ^ 5, H =\langle h, b\rangle$. It is clear that no elements of $H$ are strictly positive.

I claim that $h ^ + b (h\wedge 1)\not\in H $. Below is some Wolfram code verifying this, as the algebra is kind of annoying.

(There are probably easier examples, but I couldn't find one immediately. The key thing is that, not only can the group not be commmutative, but $b, h$ can't either.)

hV = Function[x, Max[x,x ^ 3]]
hWedge = Function[x, Min[x ^ 3, x]]
b = Function[x, x ^ 5]
product = Composition[hV, b, hWedge][x]
Plot[product, {x, -2, 2}]