Relation Compatible with Group Operation is Strongly Compatible/Corollary

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

Let $\mathcal R$ be a relation compatible with $\circ$.

Let $x, y \in G$.

Then the following equivalences hold:
 * $(1): \quad x \mathrel {\mathcal R} y \iff e \mathrel {\mathcal R} y \circ x^{-1}$
 * $(2): \quad x \mathrel {\mathcal R} y \iff e \mathrel {\mathcal R} x^{-1} \circ y$


 * $(3): \quad x \mathrel {\mathcal R} y \iff x \circ y^{-1} \mathrel {\mathcal R} e$
 * $(4): \quad x \mathrel {\mathcal R} y \iff y^{-1} \circ x \mathrel {\mathcal R} e$

Proof
We can apply CRG1 to $x$, $y$, and $x^{-1}$ to obtain the following equivalences:


 * $x \mathrel {\mathcal R} y \iff x \circ x^{-1} \mathrel {\mathcal R} y \circ x^{-1}$
 * $x \mathrel {\mathcal R} y \iff x^{-1} \circ x \mathrel {\mathcal R} x^{-1} \circ y$

Applying CRG1 to $x$, $y$, and $y^{-1}$, on the other hand, yields:


 * $x \mathrel {\mathcal R} y \iff x \circ y^{-1} \mathrel {\mathcal R} y \circ y^{-1}$
 * $x \mathrel {\mathcal R} y \iff y^{-1} \circ x \mathrel {\mathcal R} y^{-1} \circ y$

By the definition of inverses:


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

Making these substitutions proves the theorem.