Inverse function of isomorphism is also isomorphism

Simply

$$p(p^{-1}(a)p^{-1}( b))=p(p^{-1}(a))p(p^{-1}(b))=ab$$ so apply $p^{-1}$ on two sides.

It is a standard exercise to check that $p^{-1}$ is indeed another bijection. So, $p^{-1}(a)=x$ if $p(x)=a$.

Now if $p^{-1}(ab)=z$ then $p(z)=ab$.

Take $p^{-1}(a)=x$ and $p^{-1}(b)=y$, then $p^{-1}(a)p^{-1}(b)=xy=z$ also.This implies $$p^{-1}(ab)=p^{-1}(a)p^{-1}(b).$$

Cause the inverse is strictly related to the function itself, that especially gives $p^{-1}(y\tilde{y})=p^{-1}(p(x)p(\tilde{x}))=p^{-1}(p(x\tilde{x}))=x\tilde{x}=p^{-1}(y)p^{-1}(\tilde{y})$. But this relation is really just a "happy accident". Most properties won't be heridated e.g. the inverse of a continuous function is not necesarily continuous or e.g. differentiability.

Moreover, I'd like to stress that -despite the fact that most textbook define isomorphism of groups to be bijective homomorphism- what one really desires is a homomorphism which inverse is homomorphism as well, what luckily comes for free ;-)