User:Dfeuer/Cone Condition Equivalent to Congruence

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Theorem

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

Let $C$ be a cone compatible with $\circ$ with the following properties:

$C = C^{-1}$

$e \in C$

Let $\cong$ be the relation induced by $C$.

Then $\cong$ is a Congruence Relation.

Proof

By User:Dfeuer/Cone Compatible with Group Induces Transitive Compatible Relation, $\cong$ is compatible with $\circ$ and is transitive.

By User:Dfeuer/Cone Condition Equivalent to Symmetry and User:Dfeuer/Cone Condition Equivalent to Reflexivity, $\cong$ is symmetric and reflexive.

Thus $\cong$ is an Equivalence Relation compatible with $\circ$, and hence a congruence relation.

$\blacksquare$

Theorem

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

Let $\cong$ be a Congruence Relation on $G$.

Then there is a unique cone $C$ compatible with $\circ$ such that $C$ induces $\cong$, and $C$ has the following properties:

$C = C^{-1}$

$e \in C$

Proof

Since a congruence relation on $G$ is an equivalence relation compatible with $\circ$, it is in particular transitive.

Thus by User:Dfeuer/Transitive Relation Compatible with Group Operation Induced by Unique Cone $\cong$ is induced by a unique cone

$C = \{x \in G: e \cong x\}$

That is, $C$ is the $\cong$ congruence class of $e$.

By User:Dfeuer/Cone Condition Equivalent to Symmetry:

$C = C^{-1}$

Finally, by User:Dfeuer/Cone Condition Equivalent to Reflexivity:

$e \in C$

$\blacksquare$