My question: Let $(\Gamma,+,\leq)$ be an ordered abelian group. We know that archimedean property can be stated as: for all $a,b\in\Gamma$ with $a>0,b\geq0$, there exists $n\geq0$ such that $b\leq na$. However, if we consider the multiplicative case, namely $(\Gamma,\cdot,\leq)$ is the ordered abelian group. Is there exists Archimedean property written multiplicatively? I think there is. And I state that as follows: for all $a,b\in\Gamma$ with $b<1,a\leq1$, there exists $n\geq0$ such that $b^{n}\leq a$. Is it correct?

In fact, I failed to show that it is equivalent to $\Gamma$ having convex rank 1.

Answer: You have correctly stated the multiplicative version of the Archimedean property.

Let $\Gamma$ be an ordered multiplicative group with the Archimedean property.

Suppose $H$ is a convex subgroup of $\Gamma$ with $H\ne \{1\}$. Let $1\ne x\in H$. Then $\{x,x^{-1}\}\subset H$ and one member of $\{x,x^{-1}\}$ is $>1$. So WLOG let $1<x\in H$.

(i). If $1\le y\in\Gamma$ there exists $n\in \Bbb N_0$ with $y\le x^n\in H $. But $H$ is convex with $\{1,x^n\}\subset H$ and $1\le y\le x^n$ so $y\in H$.

(ii). If $1>z\in \Gamma$ then $1<z^{-1}$ so $z^{-1}\in H$ by (i), so $z\in H$.

So $H=\Gamma$.

So the only convex subgroups of $\Gamma$ are $\Gamma$ and $\{1\}$.

Appendix. It was unnecessary to assume $\Gamma$ is Abelian. Non-Abelian ordered groups do exist. But by elementary (but not brief) methods we can show that if $\Gamma$ is an ordered group with the Archimedean property then there is an ordered-group-isomorphism from $\Gamma$ to a subgroup of the additive Reals, implying $\Gamma$ is Abelian.