Composite of Automorphisms is Automorphism

Theorem

Let $\struct {S, \odot_1, \odot_2, \ldots, \odot_n}$ be an algebraic structure.

Let:

$\phi: \struct {S, \odot_1, \odot_2, \ldots, \odot_n} \to \struct {S, \odot_1, \odot_2, \ldots, \odot_n}$

$\psi: \struct {S, \odot_1, \odot_2, \ldots, \odot_n} \to \struct {S, \odot_1, \odot_2, \ldots, \odot_n}$

be automorphisms.

Then the composite of $\phi$ and $\psi$ is also an automorphism.

Proof

From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism.

By the definition of a composite mapping, $\psi \circ \phi$ is a mapping from $S$ into $S$.

Hence $\psi \circ \phi$ is an automorphism.

$\blacksquare$

Sources

• 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\S 1.2$