= One-sided bounds for adjoining a positive element
{title2=$\langle H,a\rangle_c=\{g:|g|\leq ha^n\}=\{g:|g|\leq a^nh\}$}
Assume $uv\leq v^2u^2$ for all $u,v\geq1$ in a <lattice-ordered group>. For a <convex lattice subgroup> $H$ and $a\geq1$, the displayed bounds allow $h\in H$ with $h\geq1$ and integers $n\geq0$. To prove the first set is a subgroup, use $(ha^n)(ka^m)\leq hk^2a^{2n+m}$. Products and inverse products are both bounded because $xy\leq|x||y|$ and $(xy)^{-1}\leq|y||x|$. Taking the larger exponent and a <join> of the two bounds in $H$ bounds their <absolute value in a lattice-ordered group>. Inversion preserves that absolute value; common bounds using $h\vee k$ prove lattice closure and convexity. The set contains $H$ and $a$, and any convex lattice subgroup containing them contains the entire set. The second equality follows by applying the proof to the opposite group, where the required inequality is the original one with the variables interchanged.
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