Prove that is $\phi :S \to S'$ is an isomorphism of $(S,*)$ with $(S',*')$ then
$\phi^{-1}$ is an isomorphism of $(S',*')$ with $(S,*)$
aside:
if $\phi$ is an isomorphism then we know it is one-to-one, onto and homomorphic.
so now we know $\phi^{-1}$ exists because it is one-to-one. Further, because $\phi$ is onto we know that $\forall s' \in S'$ and $\ s \in S$ that $\phi(s)=s'$
And, $\phi$ is homomorphic so we know $\phi(s_1 * s_2) = \phi(s_1) *' \phi(s_2)$
Proof (rough draft):
(i) $\phi(s_1) = \phi(s_2) \implies s_1 = s_2$ and so $\phi^{-1}(s_1') = \phi^{-1}(s_2') \implies s_1'=s_2'$ we have one-to-one
(ii) $\forall s' \in S'$ and $\forall s \in S$, $\phi(s)=s'$ thus $\phi^{-1}(s') = s$ so we have onto
(iii) Because we have one-to-one giving us the inverse and it is onto, we have
$\phi(s_1*s_2)=\phi(s_1)*'\phi(s_2)$ $\implies$ $\phi^{-1}(s_1'*'s_2')=\phi^{-1}(s_1') * \phi^{-1}(s_2')$ thus homomorphic
This looks pretty hideous even to me, advice on cleaning it up.
Best Answer
Let $X$ and $Y$ be two non-empty set and let $f: X \to Y$ be function. Then $f$ is a bijection if and only if there exists a function $g: Y \to X$ such that $fg = 1_Y$ and $gf = 1_X.$ This $g$ is denoted by $f^{-1}.$
Using the above, $\phi^{-1}$ is a bijection between $G$ and $G'.$ The only thing that needs to check is that it a group homomorphism.