Left-orderable group

Remarks

Every orderable group is left-orderable. Example: the Klein bottle group $K = \langle a, b \mid b a b^{-1} = a^{-1} \rangle$ is left-orderable but not orderable. Every element of $K$ is $b^n a^m$ for unique integers $n, m$. Set $b^n a^m \le b^{n'} a^{m'}$ when $n < n'$, or $n = n'$ and $m \le m'$. Left multiplication by $b$ adds $1$ to $n$, and left multiplication by $a$ sends $b^n a^m$ to $b^n a^{m + (-1)^n}$. Both keep this order, so it is a left order. In a two-sided order, $a > e$ would give $a^{-1} = b a b^{-1} > e$, which contradicts $a > e$. The case $a < e$ is the same, so $K$ is not orderable. Non-example: a group with an element $g \ne e$ of finite order $n$ is not left-orderable, since $e < g$ gives $e < g < g^2 < \cdots < g^n = e$. If $g < e$, the same argument applies to $g^{-1}$. So a left-orderable group is a torsion-free group. Equivalently, $G$ is left-orderable if and only if it has a subsemigroup $P$ with $G = P \sqcup \set{e} \sqcup P^{-1}$.