Inverse of Group Isomorphism is Isomorphism/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:

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.

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.