In my textbook, the proof has been done quite in a mainstream fashion. Which goes like the following:
Let $\phi:G \rightarrow G'$ be an isomorphism. We are to show that if either one be cyclic, then the other is too .
Using the fact that the mapping is surjective, for an arbitrary element $b \in G' $, $\exists$ some $a^r \in G$ such that $\phi (a^r) =b=\{\phi(a)\}^r$. Now, it proceeds to state that every element in $G'$ can be written in such a fashion, implying the cyclic nature of it. Further, in a similar way, it shows the converse.
My question is, why they simply did not use the fact that $o(a)=o(\phi(a))$ [ $\phi$ being an isomorphism] , in conjunction with the fact that they have the same cardinality [ the mapping being a bijection] and $o(a)=$ The order of $G$ ? In fact, the whole thing seems just to be a corollary of above. Again, the converse can be proven simply using the fact the $\phi^{-1}$ is an isomorphism from $G'$ to $G$.
Or am I missing something?