We say that a partially ordered group $(G,\cdot, \geq)$ is Archimedean if for any $g,h >1\in G$ there exists some n such that $g^n > h$.
All the non-archimedean groups I know of can be written as the product of archimedean groups. I'm wondering if this is generally true. I think I've found a proof, but I haven't heard this theorem anywhere, so I suspect my proof is flawed.
Proof: We know that convex subgroups can be ordered under inclusion. Say we have convex subgroups $H_1\subset H_2$; let $h_1,\dots$ be the generators of $H_1$, $a_1,\dots$ be the generators of $H_2$ not in $H_1$ and A the group generated by $a_1,\dots$. I claim that $H_2 \cong A\times H_1$ because there is a homomorphism $f(a_1+\dots+h_1+\dots)=(a_1+\dots,h_1+\dots)$. The kernel of $f$ is 0, so $f$ is an isomorphism. QED.
Is this correct?