Relation Compatible with Group Operation is Strongly Compatible

Theorem
Let $\struct {G, \circ}$ be a group.

Let $\RR$ be a relation on $G$ compatible with $\circ$.

$\RR$ is strongly compatible with $\circ$:


 * $\forall x, y, z \in G:$
 * $x \mathrel \RR y \iff x \circ z \mathrel \RR y \circ z$
 * $x \mathrel \RR y \iff z \circ x \mathrel \RR z \circ y$

Proof
Since $\RR$ is compatible with $\circ$:


 * $\forall a, b, c \in G: a \mathrel \RR b \implies a \circ c \mathrel \RR b \circ c$

In particular, letting $a = x$, $b = y$, and $c = z$, we see that:


 * $x \mathrel \RR y \implies x \circ z \mathrel \RR y \circ z$

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


 * $x \circ z \mathrel \RR y \circ z \implies \paren {x \circ z} \circ z^{-1} \mathrel \RR \paren {y \circ z} \circ z^{-1}$

By the associativity of $\circ$ and the definition of inverse, this reduces to:


 * $x \circ z \mathrel \RR y \circ z \implies x \mathrel \RR y$

We have thus shown that:


 * $x \mathrel \RR y \iff x \circ z \mathrel \RR y \circ z$

A similar argument shows that:


 * $x \mathrel \RR y \iff z \circ x \mathrel \RR z \circ y$

so $\RR$ is strongly compatible with $\circ$.