The linearly ordered group $(\mathbb{Z},+)$$(\mathbb{Z},+,\le)$ is a counterexample, of course, but that is probably not what the OP had in mind. In factTo give a detailed description of the situation, let us use the following theorem holdsnotation:
By a linearly bi-ordered group we mean a tuple $(G,\cdot,\le)$ where $(G,\cdot)$ is a group and $\le$ is a linear order on $G$ such that $ac \le bc$ and $ca \le cb$ whenever $a,b,c \in G$ such that $a \le b$. We use the notion linearly ordered group as shorthand or linearly bi-ordered group.
An isomorphism between two linearly ordered groups $(G,\cdot,\le)$ and $(H,\cdot,\le)$ is a group isomorphism $\varphi: (G,\cdot) \to (H,\cdot)$ such that both $\varphi$ and $\varphi^{-1}$ are increasing.
A linearly ordered group $(G,\cdot,\le)$ (whose neutral element we denote by $e$) is called Archimedean if, for all $a,b > e$ there exists an integer $n \in \mathbb{N}$ such that $a^n \ge b$.
We call a linear order an a set $S$ complete if every non-empty subset of $S$ that is bounded above has a supremum in $S$ (equivalently, every non-empty subset of $S$ that is bounded below has an infimum in $S$). Note that this property is sometimes called conditionally complete (instead of complete) in the literature.
Theorem 1. Let $(G,\cdot)$$(G,\cdot,\le)$ be a totally orderedan Archimedean linearly ordered group. Then $(G,\cdot)$ is isomorphic to aan ordered subgroup of $(\mathbb{R},+)$$(\mathbb{R},+,\le)$ (whichi.e. a subgroup of $(\mathbb{R},+)$ which carries the order inducedinherited from $\mathbb{R}$). In particular, $(G,\cdot)$ is commutative.
This result can, for instance, be found in Theorem 1 in Section IV.1 of
Now, let $(H, +)$ be a subgroup of $(\mathbb{R},+)$. If the induced order on $H$ is dense (i.e. for all $a,c \in H$ which fulfil $a < c$ there exists $b \in H$ such that $a < b < c$), then it is not difficult to show that $H$ is also denseAs kindly pointed out by user Alec Rhea in $\mathbb{R}$ (note, however, that this observation relies on the fact that $(H,+)$comments, there is a subgroup of $(\mathbb{R},+)$; there are non-dense subsetsrelated result by Hahn which gives a description of all $\mathbb{R}$ which are denselycommutative linearly ordered with respect to the induced order!)groups.
If the order on $H$ is dense and complete, thenNext we concludenote that $H = \mathbb{R}$. This proves the following result:
Corollary 2. Let $(G,\cdot)$ be a totallylinearly ordered Archimedean groupgroups whose order is dense and complete. Then $(G,\cdot)$ is isomorphic to $(\mathbb{R},+)$.
Remark 3. By isomorphic I mean that there exists a group isomorphism $\varphi$ such that both $\varphi$ and $\varphi^{-1}$ are increasing.
-- Edit. -- In Corollary 2, the assumption that the group beautomatically Archimedean is redundant at least if the group is a priori known to be commutative. More precisely we have:
Theorem 42. Let $(G,+)$$(G,\cdot,\le)$ be a commutative and totallylinearly ordered group. If and assume that the order $\le$ on $G$ is complete, then. Then $(G,+)$$(G,\cdot,\le)$ is Archimedean.
Proof. Let $0 < b \in G$ and define $S := \{a \in G: \, a > 0 \text{ and } ma < b \text{ for all } m \in \mathbb{N}\}$. We have to show that$e$ denote the neutral element of $S$ is empty$(G,\cdot)$, solet $a,b > e$ and assume to the contraryfor a contradiction that $a \in S$. Since $S$ is bounded above by$a^n < b$ for all $b$,$n \in \mathbb{N}$. Then the set $S$$S := \{a^n: \, n \in \mathbb{N}\}$ has a supremum $s$ in $G$. Note thatWe have $s > 0$$a^{-1}s < s$, so $2s > s$ and thus $2s \not\in S$, which in turn implies$a^{-1}s < a^n$ for some $s \not\in S$$n \in \mathbb{N}$. ThereforeConsequently, the set $S$ coincides with the interval $(0,s) := \{g \in G: 0 < g < s\}$$s < a^{n+1} \le s$, which is a contradiction.
In particular $s > a$, so $s-a \in (0,s) = S$. We now show that $s-a$ is actually not an elementearlier version of $S$this post, which is a contradiction. As $s \not\in S$ we can find an integer $m \in \mathbb{N}$ such that $ms \ge b$more complicated proof of Theorem 2 was given. Hence, \begin{align*} (m+1)(s-a) = ms + s - (m+1)a \ge ms \ge b; \end{align*} for the equality on the left we used that our group is commutative and forThe above version of the inequality onproof was kindly pointed out by user Emil Jeřábek in the right we used that $(m+1)a \in S$ (since $a \in S$) and hence, $(m+1)a < s$. We have shown that $s-a \in S$, which is a contradictioncomments.
Remarks 5Remark 3. (a) I'm not sure whether Theorem 4 remains true if we do not assume $(G,+)$ to be commutative.
(b) Note that the existence of inverse elements is esssential not only for the proof, but also for the validity of Theorem 42. Indeed, the set $[0,\infty) \times [0,\infty)$, endowed with componentwise addition and the lexicographical order, is an example of a totallylinearly ordered semigroup (whose composition operation is strictly monotone in both components) which is order complete but not Archimedean.
-- Edit 2 (a few hours later). -- Well, it seems that it is not too difficult to answer the question in Remark 5(a) above. So in order to complete By combining Theorems 1 and 2 we arrive at the picturefollowing corollary which, let us add that Theorem 4 is actually true for non-commutative groupsI think, too - and the proof is almostanswers the same. Here arequestion of the detailsOP:
Theorem 6Corollary 4. Let $(G,\cdot)$$(G,\cdot,\le)$ be a totallylinearly ordered group. If the order $\le$ on $G$ is complete, then $(G,\cdot)$$(G,\cdot,\le)$ is Archimedeanisomorphic to one of the three linearly ordered groups $(\{0\},+,\le)$, $(\mathbb{Z},+,\le)$ and $(\mathbb{R},+,\le)$.
Proof. Let $e \in G$ denote the neutral element of $G$. Let $e < b \in G$ and define $S := \{a \in G: \, a > e \text{ and } a^m < b \text{ for all } m \in \mathbb{N}\}$. We haveAccording to show thatTheorem 2 $S$$(G,\cdot,\le)$ is emptyArchimedean, so assume to the contrary that $a \in S$. Since $S$it is bounded above by $b$, $S$ has a supremumisomorphic to an ordered subgroup $s$ in$(H,+,\le)$ of $G$$(\mathbb{R},+,\le)$ due to Theorem 1. Note that $s > e$, so $s^2 > s$ and thus $s^2 \not\in S$, which in turn impliesIf $s \not\in S$. Therefore$H$ has only one element, the set $S$ coincides with the interval $(e,s) := \{g \in G: e < g < s\}$. In particularthen obviously $s > a$$H = \{0\}$, so $sa^{-1} \in (e,s) = S$. Moreover, since $s \not\in S$, we can find an integer $m \in \mathbb{N}$ suchassume that $s^m \ge b$.
Up to now, the proof was exactly the same as for Theorem 4$H$ has at least two elements. Now waswe distinguish two cases.:
First case: $h_0 := \inf \{h \in H: \, h > 0\} > 0$. $sa^{-1} \ge a^{-1}s$. In this case we have \begin{align*} (sa^{-1})^{m+1} \ge a^{-(m+1)} s s^m \ge s^m \ge b \end{align*} since $s > a^{m+1}$ (asThen it is easy to see that $a^{m+1} \in S$). Hence$H = h_0 \mathbb{Z}$, so $sa^{-1} \not\in S$, which$(H,+,\le)$ is a contradictionisomorphic to $(\mathbb{Z},+,\le)$.
Second case: $\inf \{h \in H: \, h > 0\} = 0$. Then one readily checks that the set $sa^{-1} < a^{-1}s$. In this case we have \begin{align*} (sa^{-1})^{m+1} \ge s^m s a^{-(m+1)} \ge s^m \ge b, \end{align*} again since$H$ is dense in $s > a^{m+1}$. Hence$\mathbb{R}$, we arrive atand the same contradiction $sa^{-1} \not\in S$ as incompleteness of the first caseorder on $H$ implies that $H = \mathbb{R}$. This proves the theoremcorollary.
As a consequence, we get the following stronger version of Corollary 2:
Corollary 7-- Note on edits made 2018-06-10. -- Let $(G,\cdot)$ be a totally ordered group whose order is denseI rewrote the answer and complete. Then $(G,\cdot)$ is isomorphicconsolidated the various edits from the previous versions in order to $(\mathbb{R},+)$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!
