Inverse of Group Isomorphism is Isomorphism
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.
$\blacksquare$