Inversion Mapping Reverses Ordering in Ordered Group/Corollary

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

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

Then the following equivalences hold:


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

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


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

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