Relation Compatible with Group Operation is Strongly Compatible

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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 $\mathcal R$ is strongly compatible with $\circ$.

That is, the following equivalences hold:

$x \mathrel{\mathcal R} y \iff x \circ z \mathrel{\mathcal R} y \circ z$
$x \mathrel{\mathcal R} y \iff z \circ x \mathrel{\mathcal R} z \circ y$


Proof

Since $\mathcal R$ is compatible with $\circ$:

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


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

$x \mathrel{\mathcal R} y \implies x \circ z \mathrel{\mathcal R} y \circ z$


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

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

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

$x \circ z \mathrel{\mathcal R} y \circ z \implies x \mathrel{\mathcal R} y$


We have thus shown that:

$x \mathrel{\mathcal R} y \iff x \circ z \mathrel{\mathcal R} y \circ z$

A precisely similar argument shows that:

$x \mathrel{\mathcal R} y \iff z \circ x \mathrel{\mathcal R} z \circ y$

so $\mathcal R$ is strongly compatible with $\circ$.

$\blacksquare$