Identity Mapping is Automorphism/Ordered Semigroups

Theorem

Let $\struct {S, \circ, \preccurlyeq}$ be an ordered semigroup.

Then $I_S: \struct {S, \circ, \preccurlyeq} \to \struct {S, \circ, \preccurlyeq}$ is a ordered semigroup automorphism.

Proof

From Identity Mapping is Semigroup Automorphism:

$I_S: \struct {S, \circ} \to \struct {S, \circ}$ is a semigroup automorphism.

From Identity Mapping is Order Isomorphism:

$I_S: \struct {S, \preccurlyeq} \to \struct {S, \preccurlyeq}$ is an order isomorphism.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.1$