User:Dfeuer/Cone Condition Equivalent to Antisymmetry

Theorem

Let $(G,\circ)$ be a group with identity $e$.

Let $C$ be a cone compatible with $\circ$.

Let $\mathcal R$ be the transitive, compatible relation induced by $C$.

Then $\mathcal R$ is antisymmetric iff $C \cap C^{-1} \subseteq \{e\}$, where $C^{-1} = \{ x^{-1}: x \in C \}$.

Proof

Suppose $\mathcal R$ is antisymmetric and suppose that $x \in C \cap C^{-1}$.

Then $x^{-1} \in C$.

Thus $e \mathrel{\mathcal R} x$ and $e \mathrel{\mathcal R} x^{-1}$

By CRG3, $x \mathrel {\mathcal R}e^{-1} = e$, so by antisymmetry $x=e$.

Suppose on the other hand that $C \cap C^{-1} \subseteq \{e\}$.

Let $x,y \in G$.

Let $x \mathrel{\mathcal R} y$ and $y \mathrel{\mathcal R} x$.

Then $y \circ x^{-1} \in C$ and $x \circ y^{-1} \in C$.

But $x \circ y^{-1}$ is the inverse of $y \circ x^{-1}$, so $y \circ x^{-1} = e$.

Thus $y = x$.

$\blacksquare$