Identity Mapping is Automorphism

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Theorem

The identity mapping $I_S: \struct {S, \circ} \to \struct {S, \circ}$ on the algebraic structure $\struct {S, \circ}$ is an automorphism.

Its image is $S$.


Semigroup Automorphism

Let $\struct {S, \circ}$ be a semigroup.

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


Ordered Semigroup Automorphism

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.


Group Automorphism

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Then $I_G: \struct {G, \circ} \to \struct {G, \circ}$ is a group automorphism.

Its kernel is $\set e$.


Ring Automorphism

Let $\struct {R, +, \circ}$ be a ring whose zero is $0$.

Then $I_R: \struct {R, +, \circ} \to \struct {R, +, \circ}$ is a ring automorphism.

Its kernel is $\set 0$.


Proof

By definition, an automorphism is an isomorphism from an algebraic structure onto itself.

An isomorphism, in turn, is a bijective homomorphism.


From Identity Mapping is Bijection, the identity mapping $I_S: S \to S$ on the set $S$ is a bijection from $S$ onto itself.


Now we need to show it is a homomorphism.

Let $x, y \in S$. Then:

\(\ds \map {I_S} {x \circ y}\) \(=\) \(\ds x \circ y\) Definition of Identity Mapping
\(\ds \) \(=\) \(\ds \map {I_S} x \circ \map {I_S} y\) Definition of Identity Mapping


Thus $\map {I_S} {x \circ y} = \map {I_S} x \circ \map {I_S} y$ and the morphism property holds/

This proves that $I_S: S \to S$ is a homomorphism.

As $I_S$ is a bijection, its image is $S$.

$\blacksquare$


Sources