Order Type of Totally Ordered Abelian Groups

As a preface, I don't know much group theory, so this might be a trivial question. I looked at some other questions on stack exchange, but I didn't notice a result which answers my question immediately.

I'm taking my first group theory course right now, and one of our assignments was intended to familiarize us with group isomorphisms. We were asked to prove that $\mathbb{Z}$ and $\mathbb{Q}$ are not group isomorphic under addition. My solution relied a lot on their order types being different, so I was wondering if that was the "underlying reason" for why they aren't group isomorphic. Written in full, my question is the following:

Suppose $X$ and $Y$ are totally ordered sets, and $(X, +_X)$ and $(Y, +_Y)$ are Abelian groups. Additionally, suppose that $+_X$ and $+_Y$ satisfy:$$a < b \implies a + c < b + c$$

For "$+$" and "$<$" as defined for $X$ and $Y$. If there exists a group isomorphism between $X$ and $Y$, must there also exist an order isomorphism between them?

I feel like it's true, but I got stuck when I tried to prove it. My strategy was:

1. show that $X$ and $Y$ are vector spaces over $\mathbb{Z}$
2. show that if $B_Y$ is a basis for $Y$ and $f:X\to Y$ is a group isomorphism, then $f^{-1}(B_Y)$ is a basis for $X$
3. fix a basis $B_Y$ consisting of positive elements, then construct a group isomorphism $g:X \to Y$ such that $g^{-1}(y) > 0$ for all $y \in B_Y$
4. Use 3. to show that $\ g(x) > 0 \iff x > 0 \ \ $ when $x \in \{ \sum_{n = 1}^{N} k_n g^{-1}(y_n): \ \, y_n \in B_Y, \ \ N, k_n \in \mathbb{Z}\} = \langle g^{-1}(B_Y) \rangle$.

I'm having trouble proving that $g$ is well-behaved under infinite sums of linearly independent vectors, so I hit a snag when trying to prove 2. Also, 4. only helps me if $B_Y$ generates $Y$, and I'm struggling to extend 4. to infinite sums when $B_Y$ doesn't generate $Y$.

Any tips/help is appreciated!