Relation Compatible with Group Operation is Reflexive or Antireflexive

Theorem

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

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

Then $\RR$ is reflexive or antireflexive.

Proof

Suppose that $\RR$ is not antireflexive.

Then there is some $x \in G$ such that $x \mathrel \RR x$.

Let $y \in G$.

Then by the definition of compatibility:

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

By Group Axiom $\text G 1$: Associativity and Group Axiom $\text G 3$: Existence of Inverse Element:

$y \mathrel \RR y$

Since this holds for arbitrary $y$, $\RR$ is reflexive.

Thus $\RR$ is either reflexive or antireflexive.

$\blacksquare$

This article, or a section of it, needs explaining. In particular: While the above demonstrates that if such a relation is not antireflexive it must be reflexive, it does not approach the task of demonstrating that such a relation can actually *be* antireflexive. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.