Inverse of Group Isomorphism is Isomorphism

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Theorem

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.

Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a mapping.


Then $\phi$ is an isomorphism if and only if $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is also an isomorphism.


Proof 1

A specific instance of Inverse of Algebraic Structure Isomorphism is Isomorphism.

$\blacksquare$


Proof 2

Necessary Condition

Let $\phi: G \to H$ be an isomorphism.

Then by definition $\phi$ is a bijection.

From Bijection iff Inverse is Bijection it follows that:

$\exists \phi^{-1}: \struct {H, *} \to \struct {G, \circ}$

such that $\phi^{-1}$ is also a bijection.

Thus:

\(\ds \forall g \in G, h \in H: \, \) \(\ds \map \phi g = h\) \(\iff\) \(\ds \map {\phi^{-1} } h = g\) Inverse Element of Bijection
\(\ds \leadsto \ \ \) \(\ds \map \phi {g_1 \circ g_2}\) \(=\) \(\ds h_1 * h_2\) Definition of Morphism Property
\(\ds \leadsto \ \ \) \(\ds \map {\phi^{-1} } {h_1 * h_2}\) \(=\) \(\ds g_1 \circ g_2\) Inverse Element of Bijection
\(\ds \) \(=\) \(\ds \map {\phi^{-1} } {h_1} \circ \map {\phi^{-1} } {h_2}\)


So $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is a homomorphism.

$\phi^{-1}$ is also (from above) a bijection.

Thus, by definition, $\phi^{-1}$ is an isomorphism.

$\Box$


Sufficient Condition

Let $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ be an isomorphism.

Applying the same result as above in reverse, we have that $\paren {\phi^{-1} }^{-1}: \struct {G, \circ} \to \struct {H, *}$ is also an isomorphism.

But by Inverse of Inverse of Bijection:

$\paren {\phi^{-1} }^{-1} = \phi$

and hence the result.

$\blacksquare$