TITLE: Inverse function of isomorphism is also isomorphism
QUESTION [17 upvotes]: Let $G$ be a group, and let $p:G\rightarrow G$ be an isomorphism. Why is $p^{-1}$ also an isomorphism?
We know that $p(a)p(b)=p(ab)$ for any elements $a,b\in G$. We also know $p(a^{-1})=p(a)^{-1}$ for any element $a\in G$ (follows from the first statement.) How would it show $p^{-1}(ab)=p^{-1}(a)p^{-1}(b)$?

REPLY [3 votes]: 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 ;-)