Ordering Compatible with Group Operation is Strongly Compatible/Corollary

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

Let $x, y \in G$.

Then the following equivalences hold:


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


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


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


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

Proof
Since a group is closed under inverses, we can apply User:Dfeuer/OG1 to $x$, $y$, and $x^{-1}$ to obtain the following equivalences:


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


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

Applying User:Dfeuer/OG1 to $x$, $y$, and $y^{-1}$, on the other hand, yields


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


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

By the definition of group inverse,
 * $x \circ x^{-1} = x^{-1} \circ x = y \circ y^{-1} = y^{-1} \circ y = e$.

Making these substitutions proves the theorem.