Ordered groups over the same underlying set and total order, can they be group isomorphic without being order isomorphic?

Question

Given ordered groups $H:=(A,+)$ and $H':=(A, \boxplus)$ as per the same total order $ \leqslant$ of A,

can $H$ and $H'$ be group isomorphic without being order isomorphic?

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I understand that there can be a group isomorphism between $H$ and $H'$ which does not preserve the order. For example, if $H:=H':=( \mathbb Z, +)$, $f(n):=-n$ is such a group isomorphism which does not preserve the order. Still, the two groups are order isomorphic, i.e. there exists some other group isomorphism which also preserves the order.

Answer 1

Here's an example of ordered groups which are isomorphic as groups, and which are isomorphic as linear orders, but which are not isomorphic as ordered groups. I think this answers the body of your question. In particular, after relabelling, you can assume they have the same underlying set and the ordering on the groups is the same.

Let $H$ be the subgroup $\Bbb Q(\sqrt 2) = \{a + b \sqrt 2 : a, b \in \Bbb Q\}$ of $\Bbb R$ with the inherited order and let $H'$ be the subgroup $\Bbb Q(\sqrt 3)$. As groups, we have $H \cong H' \cong \Bbb Q^2$. Both are dense countable linear orders, so as linear orders, we have $H \cong H' \cong \Bbb Q$. But they aren't ordered-group-isomorphic, which is morally because the orderings have some "information" about the algebraic properties of $\sqrt 2$ and $\sqrt 3$.

Indeed suppose $f: \Bbb Q(\sqrt 2) \to \Bbb Q(\sqrt 3)$ is an ordered-group isomorphism. Note that $f(1) > f(0) = 0$, so we can assume without loss of generality that $f(1) = 1$ (because $x \mapsto x/f(1)$ is an ordered-group isomorphism of $\Bbb Q(\sqrt 3)$, since $\Bbb Q(\sqrt 3)$ is actually an ordered field). Now it follows that $f$ fixes every rational. In particular if $y = f(\sqrt 2)$, then for all rationals $q$ with $q < \sqrt 2$ we have $q = f(q) < f(\sqrt 2) = y$ and for all $q > \sqrt 2$ we have $q = f(q) > f(\sqrt 2) = y$. But there is no such element $y$ in $\Bbb Q(\sqrt 3)$ (remember for example that it's an ordered subgroup of $\Bbb R$ and the unique such element of $\Bbb R$ is $\sqrt 2$), so there can be no such $f$.

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Addendum..

Suppose $(H_1, +_1, \le_1)$ and $(H_2, +_2, \le_2)$ are ordered groups, and there is some ordered set $(A, \le_A)$ and there are isomorphisms of ordered sets $f_1: (H_1, \le_1) \to (A, \le_A)$ and $f_2: (H_2, \le_2) \to (A, \le_A)$, and $g: (H_1, +_1) \to (H_2, +_2)$ is an isomorphism of groups, but we know that $(H_1, +_1, \le_1)$ and $(H_2, +_2, \le_2)$ are not isomorphic as ordered groups.

Define an operation $+_1^A$ on $A$ by $a +_1^A b = f_1(f_1^{-1}(a) +_1 f_1^{-1}(b))$ and similarly define an operation $+_2^A$ on $A$ by $a +_2^A b = f_2(f_2^{-1}(a) +_2 f_2^{-1}(b))$.

Then it's not hard to check that $(A, +_1^A, \le_A)$ and $(A, +_2^A, \le_A)$ are ordered groups, $f_1: (H_1, +_1, \le_1) \to (A, +_1^A, \le_A)$ and $f_2: (H_2, +_2, \le_2) \to (A, +_2^A, \le_A)$ are isomorphisms of ordered groups, $f_2 g f_1^{-1} : (A, +_1^A) \to (A, +_2^A)$ is an isomorphism of groups.

But since $(H_1, +_1, \le_1)$ and $(H_2, +_2, \le_2)$ are not isomorphic as ordered groups, it follows that $(A, +_1^A, \le_A)$ and $(A, +_2^A, \le_A)$ are not isomorphic as ordered groups.

Hence, your question is equivalent to "if $H$ and $H'$ are ordered groups, and are isomorphic as groups and as linear orders, must they be isomorphic as ordered groups?". This idea is a standard one when you're proving things about isomorphism or non-isomorphism, called "transport of structure". I don't think the above symbols are a terribly helpful way to think about it - the point is that given any bijection between $H$ and $A$, we can give $A$ the same structure as $H$ by "pulling the structure through the bijection". In the special case where $A$ already has an ordering and the bijection is an order isomorphism, the order this gives is the usual order on $A$, so the addition this gives must make $A$ into an ordered group, since we know the addition and ordering are compatible in $H$.