Complement of Relation Compatible with Group is Compatible

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Theorem

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

Let $\mathcal R$ be a relation on $G$.

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

Let $\mathcal Q = \complement_{G \times G} \mathcal R$, so that:

$\forall a, b \in G: a \mathop {\mathcal Q} b \iff \neg \left({a \mathop {\mathcal R} b}\right)$


Then $\mathcal Q$ is a relation compatible with $\circ$.


Proof

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

Suppose that $\neg \left({\left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)}\right)$.

Then by the definition of $\mathcal Q$:

$\left({x \circ z}\right) \mathop {\mathcal R} \left({y \circ z}\right)$

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

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


By Group Axiom $(G1)$: Associativity and the Group Axiom $(G3)$: Inverses:

$x \mathop {\mathcal R} y$

so by the definition of $\mathcal Q$:

$\neg \left({x \mathop {\mathcal Q} y}\right)$


By the Rule of Transposition:

$\forall x, y, z \in G: x \mathop {\mathcal Q} y \implies \left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)$

A similar argument shows that:

$\forall x, y, z \in G: x \mathop {\mathcal Q} y \implies \left({z \circ x}\right) \mathop {\mathcal Q} \left({z \circ y}\right)$

Thus, by definition, $\mathcal Q$ is a relation compatible with $\circ$.

$\blacksquare$