Hölder's theorem on Archimedean groups

Remarks

If $G = \set{e}$ there is nothing to prove. Otherwise one fixes an element $u > e$ and sends $g$ to the real number $\sup \set{m/n : n \ge 1, \ u^m \le g^n}$. The Archimedean property makes this set bounded above and nonempty, and an argument with Dedekind cuts shows that the map is an injective order-preserving homomorphism. Commutativity of $G$ then follows from commutativity of $\R$. Example: the subgroup $\Z + \Z\sqrt{2}$ of $\R$ is Archimedean and is already a subgroup of the reals. The Archimedean hypothesis is needed: $\Z^2$ with the lexicographic order is linearly ordered, but $n \cdot (0, 1) < (1, 0)$ for all $n$. No order-preserving injective homomorphism to $\R$ exists for it, since the image of $(0, 1)$ would be positive with all its multiples below the image of $(1, 0)$.

Going deeper

For a student

The theorem is the base case for the structure theory of linearly ordered abelian groups. A general linearly ordered abelian group splits into Archimedean classes, and the quotient attached to each class is Archimedean, so Hölder's theorem embeds it into $\R$. The Hahn embedding theorem puts these real pieces together in a lexicographic product. The theorem also explains why the value group of a rank one valuation can be taken inside $\R$.