Isomorphisms preserve cyclic

Suppose $G$ is cyclic and $\phi: G\to H$ is surjective. Let $G$ be generated by $a\in G$, i.e. $G = \langle a\rangle$. Let $b\in H$. Then there exists some $x\in G$ so that $b = \phi(x)$, as $\phi$ is surjective. Further, since $G$ is cyclic, we have $x = a^n$ for some $n$. Then $b = \phi(x) = \phi(a^n) =\phi(a)^n$. This can be done for any element of $H$, so we see that $H = \langle \phi(a) \rangle$.

To give a direct proof: Let $y\in H$, and $\phi:G\rightarrow H$ an isomorphism of the groups $G$ and $H$. Consider $\phi^{-1}(y)\in G$. Since $G$ is cyclic, we can express $\phi^{-1}(y)$ in terms of a generator for $G$ ($G$ can be generated by a single non-identity element because it is cyclic; why?). Hence we have a generator $g$ for the group $G$ (that is, $\langle g\rangle=G$), and it follows that $\phi^{-1}(y)=g^k$ for some integer exponent $k$. Now we claim that $\phi (g)$ must be a generator for $H$, for every element of $G$ is of the form $g^k$ for some integer $k$. Moreover, because $\phi$ is a bijection, it follows $y=\phi(g^k)$. Now because $\phi$ is an isomorphism, we have that $\phi (g^k)=[\phi (g)]^k$, and thus we have expressed $y$ in terms of a single generator and since $y$ was arbitrary, it follows that $H$ is generated by a single element, and hence must be cyclic.