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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 $p^{-1}(ab)=p^{-1}(a)p^{-1}(b)$?
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 ;-)
