An $order$ on a set $S$ is an (anti-symmetric) relation $<$ on $G$ so that for each $a,b\in G$ exactly one of the following is true: $a<b, b<a$ or $a=b$. A group $G$ is called left orderable if there is an order $<$ on $G$ so that whenever $a,b,g\in G$ with $a<b$, $ga<gb$. Show that every element of a left orderable group has infinite order.
Solution Attempt I think that because whenever $a < b$ then for whatever $g$ I choose than $ga$ is never the identity $e$ so does this mean because it never is equal to $e$ that the order is infinite?
Why is $ga\neq e$? You state this without proof.
But it's much easier to prove this via a contradiction. What happens if $e<a$ and $a^n=e$?