Relation Compatible with Group Operation is Strongly Compatible

Theorem
Let $\left({S,\circ}\right)$ be a group.

Let $\mathcal R$ be an endorelation on $S$ compatible with $\circ$.

Let $x, y, z \in S$.

Then
 * $x \mathop{\mathcal R} y \iff x \circ z \mathop{\mathcal R} y \circ z$

and
 * $x \mathop{\mathcal R} y \iff z \circ x \mathop{\mathcal R} z \circ y$.

Proof
Since $\mathcal R$ is compatible with $\circ$,


 * $\forall a,b,c \in S: a \mathop{\mathcal R} b \implies a \circ c \mathop{\mathcal R} b \circ c$.

In particular, letting $a=x$, $b=y$, and $c=z$, we see that
 * $x \mathop{\mathcal R} y \implies x \circ z \mathop{\mathcal R} y \circ z$.

On the other hand, letting $a = x \circ z$, $b = y \circ z$, and $c = z^{-1}$, we see that


 * $x \circ z \mathop{\mathcal R} y \circ z \implies \left({x \circ z}\right) \circ z^{-1} \mathop{\mathcal R} \left({y \circ z}\right) \circ z^{-1}$.

By the associative property and the definition of inverse, this reduces to


 * $x \circ z \mathop{\mathcal R} y \circ z \implies x \mathop{\mathcal R} y$.

We have thus shown that
 * $x \mathop{\mathcal R} y \iff x \circ z \mathop{\mathcal R} y \circ z$.

A precisely similar argument shows that
 * $x \mathop{\mathcal R} y \iff z \circ x \mathop{\mathcal R} z \circ y$.