I Would like someone to verify this proof, since I could not find it in my professors notes on properties of isomorphic groups.
Again, the statement: is $G\cong H$. $G$ is cyclic if and only if $H$ is cyclic. I will only demonstrate ($\Rightarrow$)
Proof. $G\cong H$. Suppose $G$ is cyclic. Then $G=\langle a\rangle $ for some $a\in G$, i.e. $c\in G\Rightarrow c=a^{s}$, for some $s\in \mathbb{Z}$. We want to show that $H=\langle b\rangle$ for some $b\in H$.
Pick any $h\in H$. Then $h=\phi (g)$ where $g\in G=\langle a\rangle$, i.e. $g=a^{k}$ for some $k\in \mathbb{Z}$.
Now, since $G\cong H$ we have that $h=\phi (g)=\phi (a^{k})=\phi (a\cdot a\cdots a)=\phi (a)\phi (a)\cdots\phi (a)=\bigl(\phi (a)\bigr)^{k}$. Furthermore $\phi (a)\in H$ since $a\in G$ and $\phi \colon G\rightarrow H$. Put $b=\phi (a)$ and we finally have that $h=\bigl(\phi (a)\bigr)^{k}=b^{k}$ for some $k\in \mathbb{Z}$. Thus $H=\langle b\rangle$, meaning $H$ is cyclic.