Suppose that a function $f: G \to H$ is an isomorphism. Must $f$ inverse be an isomorphism as well? Prove it if it is so, else, provide a counterexample.
I think that it is, the following is my proof:
Since f is isomorphic, f is both bijective and homomorphic:
f is bijective implies that if s, t belongs to H, there exists a, b belongs to G such that $f(a) = s$ and $f(b) = t$. On top of that, $f^{-1} (s) = a$ and $f^{-1} (t) = b$.
f is homomorphic implies that $f(a \star b) = f(a) \circ f(b)$ which implies that $f(f^{-1}(s) \star f^{-1}(t)) = s \circ t$. Applying $f^{-1}$ to both sides, we have $f^{-1}(s) \star f^{-1}(t) = f^{-1} (s \circ t)$.
Hence, if s, t belongs to H, $f^{-1}(s) \star f^{-1}(t) = f^{-1} (s \circ t)$, which means $f^{-1}$ is homomorphic. Since $f^{-1}$ is also bijective, $f^{-1}$ is isomorphic.
Can someone verify if this is right. Thanks.