If every finitely generated subgroup of a group $G$ is totally ordered, what approach should I take to prove that $G$ is also totally ordered?
Should I define a map that takes elements from the subgroups and make an ordering (I have no idea if this can even work, it was just an idea) or should I build a set that satisfies the four properties from this page in its second paragraph?
Additionally, for such properties, what does it mean when it says $PP\subset P$?
We need to prove that for any two elements $a, b$, we have either $a\le b$ or $b\le a$.
But this already holds in the subgroup generated by $a, b$ by hypothesis.
Having $P=\{x:x\ge e\}$, with $x, y\in P$ we get $$xy\ \ge\ xe=x\ \ge\ e$$ so $xy\in P$, proving $PP\subseteq P$.
Similarly, if $x\in P\cap P^{-1}$, then both $x$ and $x^{-1}$ are $\ge e$, thus $e=xx^{-1}\ge xe=x\ge e$ yielding $e=x$.
If $x\ge e$, then $g^{-1}xg\ge g^{-1}eg=e$.
Finally, $P\cup P^{-1}=G$ expresses that the ordering is total.