Inversion Mapping Reverses Ordering in Ordered Group/Corollary

Theorem
Let $\left({G, \circ, \le}\right)$ be an ordered group with identity $e$.

Let $x \in \left({G, \circ, \le}\right)$.

Then the following equivalences hold:


 * $x \le e \iff e \le x^{-1}$
 * $e \le x \iff x^{-1} \le e$
 * $x < e \iff e < x^{-1}$
 * $e < x \iff x^{-1} < e$

Proof
Applying User:Dfeuer/OG2 to $x$ and $e$ gives


 * $x \le e \iff e \le e \circ x^{-1}$
 * $e \le x \iff e \circ x^{-1} \le e$
 * $x < e \iff e < e \circ x^{-1}$
 * $e < x \iff e \circ x^{-1} < e$

Since $e \circ x^{-1} = x^{-1}\circ e = x^{-1}$ for all $x \in G$, the theorem holds.