Permutation of Set is Automorphism of Set under Right Operation

Theorem

Let $S$ be a set.

Let $\struct {S, \to}$ be the algebraic structure formed from $S$ under the right operation.

Let $f$ be a permutation on $S$.

Then $f$ is an automorphism of $f$.

Proof

We have by hypothesis that $f$ is a permutation and so a fortiori a bijection.

It remains to show that $f$ is a homomorphism.

So, let $a, b \in S$ be arbitrary.

We have:

\(\ds \map f {a \to b}\)

\(=\)

\(\ds \map f b\)

Definition of Right Operation

\(\ds \)

\(=\)

\(\ds \map f a \to \map f b\)

Definition of Right Operation

The result follows.

$\blacksquare$

Also see

• Permutation of Set is Automorphism of Set under Left Operation

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.11 \ \text {(a)}$