Ordering Compatible with Group Operation is Strongly Compatible/Corollary

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

Let $<$ be the reflexive reduction of $\le$.

Let $x, y \in G$.

Then the following equivalences hold:
 * $(\operatorname{OG}2.1):\quad x \le y \iff e \le y \circ x^{-1}$
 * $(\operatorname{OG}2.2):\quad x \le y \iff e \le x^{-1} \circ y$


 * $(\operatorname{OG}2.3):\quad x \le y \iff x \circ y^{-1} \le e$
 * $(\operatorname{OG}2.4):\quad x \le y \iff y^{-1} \circ x \le e$


 * $(\operatorname{OG}2.1'):\quad x < y \iff e < y \circ x^{-1}$
 * $(\operatorname{OG}2.2'):\quad x < y \iff e < x^{-1} \circ y$


 * $(\operatorname{OG}2.3'):\quad x < y \iff x \circ y^{-1} < e$
 * $(\operatorname{OG}2.4'):\quad x < y \iff y^{-1} \circ x < e$

Proof 1
By the definition of an ordered group, $\le$ is a relation compatible with $\circ$.

Thus by User:Dfeuer/CRG2, we obtain the first four results.

By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $<$ is compatible with $\circ$.

By again User:Dfeuer/CRG2, we obtain the remaining results.

Proof 2
Each result follows from User:Dfeuer/OG1. For example, by User:Dfeuer/OG1,


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

Since $x \circ x^{-1} = e$:


 * $(\operatorname{OG}2.1):\quad x \le y \iff e \le y \circ x^{-1}$