Inverse of Algebraic Structure Isomorphism is Isomorphism/General Result

Theorem

Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ and $\struct {T, *_1, *_2, \ldots, *_n}$ be algebraic structures.

Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ be a mapping.

Then:

$\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ is an isomorphism

$\phi^{-1}: \struct {T, *_1, *_2, \ldots, *_n} \to \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ is also an isomorphism.

As $\paren {\phi^{-1} }^{-1} = \phi$, it suffices to show the sufficient condition.

Suppose that $\phi$ is an isomorphism.

We shall show that $\phi^{-1}$ is also an isomorphism.

We have a fortiori that $\phi$ is a bijection.

Hence from Inverse of Bijection is Bijection we have that $\phi^{-1}$ is also a bijection.

Then for all $x, y \in T$ and $i \in \set {1, 2, \ldots, n}$:

$\ds x \ast_i y$

$=$

$\ds \map \phi {\map {\phi^{-1} } x} \ast_i \map \phi {\map {\phi^{-1} } y}$

Definition 2 of Inverse Mapping

$\ds $

$=$

$\ds \map \phi {\map {\phi^{-1} } x \circ_i \map {\phi^{-1} } y}$

as $\phi$ is a homomorphism

$\ds \leadsto \ \ $

$\ds \map {\phi^{-1} } {x \ast_i y}$

$=$

$\ds \map {\phi^{-1} } {\map \phi {\map {\phi^{-1} } x \circ_i \map {\phi^{-1} } y} }$

applying $\phi^{-1}$ to both sides

$\ds $

$=$

$\ds \map {\phi^{-1} } x \circ_i \map {\phi^{-1} } y$

Definition 2 of Inverse Mapping

Therefore $\phi^{-1}$ is a homomorphism.

$\blacksquare$