Power Function Preserves Ordering in Ordered Group/Corollary

Theorem
Let $\left({G, \circ, \le}\right)$ be an ordered group with identity $e$.

Let $<$ be the reflexive reduction of $\le$.

Let $x \in G$.

Let $n \in \N_{>0}$ be a strictly positive integer.

Then the following hold:
 * $x \le e \implies x^n \le e$
 * $e \le x \implies e \le x^n$
 * $x < e \implies x^n < e$
 * $e < x \implies e < x^n$

Proof 1
By User:Dfeuer/Power Function Strictly Preserves Ordering in Ordered Group:
 * $x \le e \implies x^n \le e^n$
 * $e \le x \implies e^n \le x^n$
 * $x < e \implies x^n < e^n$
 * $e < x \implies e^n < x^n$

By Identities are Idempotent, $e$ is idempotent with respect to $\circ$.

Therefore by the definition of an idempotent element, $e^n = e$.

Thus the theorem holds.

Proof 2
By Identities are Idempotent, $e$ is idempotent with respect to $\circ$.

By the definition of an ordered group, $\le$ is a transitive relation compatible with $\circ$.

Thus by User:Dfeuer/CTR5, we obtain the first two results:


 * $x \le e \implies x^n \le e$
 * $e \le x \implies e \le x^n$

By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $<$ is compatible with $\circ$.

By Reflexive Reduction of Transitive Antisymmetric Relation is Transitive, $<$ is transitive.

Thus by User:Dfeuer/CTR5, we obtain the remaining results:


 * $x < e \implies x^n < e$
 * $e < x \implies e < x^n$