Abelian, divisible and finite subgroup implies trivial subgroup?

Question

I'm reading a proof from my lecture and don't understand one step:

We're given a group $(R,\cdot)$ and a proper subgroup $H$. We showed that

1. $H$ must be finite
2. $R$ is abelian and divisible

The proof then argues that $H$ therefore must be trivial. Can anyone explain to me why this already follows from these?

Also, in fact, we have even a stronger condition on $R$, namely that is an ordered group. Why can we not argue that $H$ hence must be trivial, since ordered groups are infinite? Note that we do not make use of 2. in this case. However, we explicitly proved '$H$ is trivial' as a corollary from 2.

Answer 1

The fact that the group is ordered is essential. Take any element $x \in H$. If $x > 0$, then $2x = x + x > x$, right? How about $3x$? Can $H$ be finite and non-trivial?