Characterizing $\mathbf{R}$ as an ordered group A standard characterization of $\mathbf{R}$ uses the order and the field structure: any linearly ordered field that is archimedean and complete is isomorphic to $(\mathbf{R}, +, \times, <)$ as an ordered field.
Is there a similar characterization of $\mathbf{R}$ as an ordered group?
Is any linearly ordered group that is archimedean and complete isomorphic to $(\mathbf{R}, +, <)$ as an ordered group or is some other assumption needed? Any reference is welcome.
 A: The linearly ordered group $(\mathbb{Z},+,\le)$ is a counterexample, but that is probably not what the OP had in mind. To give a detailed description of the situation, let us use the following notation:


*

*By a linearly bi-ordered group we mean a tuple $(G,\cdot,\le)$ where $(G,\cdot)$ is a group and $\le$ is a linear order on $G$ such that $ac \le bc$ and $ca \le cb$ whenever $a,b,c \in G$ such that $a \le b$. We use the notion linearly ordered group as shorthand or linearly bi-ordered group.

*An isomorphism between two linearly ordered groups $(G,\cdot,\le)$ and $(H,\cdot,\le)$ is a group isomorphism $\varphi: (G,\cdot) \to (H,\cdot)$ such that both $\varphi$ and $\varphi^{-1}$ are increasing.

*A linearly ordered group $(G,\cdot,\le)$ (whose neutral element we denote by $e$) is called Archimedean if, for all $a,b > e$ there exists an integer $n \in \mathbb{N}$ such that $a^n \ge b$.

*We call a linear order an a set $S$ complete if every non-empty subset of $S$ that is bounded above has a supremum in $S$ (equivalently, every non-empty subset of $S$ that is bounded below has an infimum in $S$). Note that this property is sometimes called conditionally complete (instead of complete) in the literature.
Theorem 1. Let $(G,\cdot,\le)$ be an Archimedean linearly ordered group. Then $(G,\cdot)$ is isomorphic to an ordered subgroup of $(\mathbb{R},+,\le)$ (i.e. a subgroup of $(\mathbb{R},+)$ which carries the order inherited from $\mathbb{R}$). In particular, $(G,\cdot)$ is commutative.
This result can, for instance, be found in Theorem 1 in Section IV.1 of
Fuchs, L., Partially ordered algebraic systems, Oxford-London-New York-Paris: Pergamon Press. IX, 229 p. (1963). ZBL0137.02001.
There, the theorem is attributed to Hölder.
As kindly pointed out by user Alec Rhea in the comments, there is a related result by Hahn which gives a description of all commutative linearly ordered groups.
Next we note that linearly ordered groups whose order is complete are automatically Archimedean:
Theorem 2. Let $(G,\cdot,\le)$ be a linearly ordered group and assume that the order $\le$ on $G$ is complete. Then $(G,\cdot,\le)$ is Archimedean.
Proof. Let $e$ denote the neutral element of $(G,\cdot)$, let $a,b > e$ and assume for a contradiction that $a^n < b$ for all $n \in \mathbb{N}$. Then the set $S := \{a^n: \, n \in \mathbb{N}\}$ has a supremum $s$ in $G$. We have $a^{-1}s < s$, so $a^{-1}s < a^n$ for some $n \in \mathbb{N}$. Consequently, $s < a^{n+1} \le s$, which is a contradiction.
In an earlier version of this post, a more complicated proof of Theorem 2 was given. The above version of the proof was kindly pointed out by user Emil Jeřábek in the comments.
Remark 3. Note that the existence of inverse elements is esssential not only for the proof, but also for the validity of Theorem 2. Indeed, the set $[0,\infty) \times [0,\infty)$, endowed with componentwise addition and the lexicographical order, is an example of a linearly ordered semigroup (whose composition operation is strictly monotone in both components) which is order complete but not Archimedean.
By combining Theorems 1 and 2 we arrive at the following corollary which, I think, answers the question of the OP:
Corollary 4. Let $(G,\cdot,\le)$ be a linearly ordered group. If the order $\le$ on $G$ is complete, then $(G,\cdot,\le)$ is isomorphic to one of the three linearly ordered groups $(\{0\},+,\le)$, $(\mathbb{Z},+,\le)$ and $(\mathbb{R},+,\le)$.
Proof. According to Theorem 2 $(G,\cdot,\le)$ is Archimedean, so it is isomorphic to an ordered subgroup $(H,+,\le)$ of $(\mathbb{R},+,\le)$ due to Theorem 1. If $H$ has only one element, then obviously $H = \{0\}$, so assume that $H$ has at least two elements. Now we distinguish two cases:
First case: $h_0 := \inf \{h \in H: \, h > 0\} > 0$. Then it is easy to see that $H = h_0 \mathbb{Z}$, so $(H,+,\le)$ is isomorphic to $(\mathbb{Z},+,\le)$.
Second case: $\inf \{h \in H: \, h > 0\} = 0$. Then one readily checks that the set $H$ is dense in $\mathbb{R}$, and the completeness of the order on $H$
implies that $H = \mathbb{R}$. This proves the corollary.
-- Note on edits made 2018-06-10. -- I rewrote the answer and consolidated the various edits from the previous versions in order to make this post better readable for future visitors. I also incorporated various suggestions by users  Alec Rhea, Emil Jeřábek and YCor, so let me thank them for their comments!
