Is it true that if $H_1$ and $H_2$ are isomorphic cyclic subgroups of $G$, then $G/H_1\cong G/H_2$? I have a question: it is true that if $H_1$ and $H_2$ are cyclic groups that are isomorphic, then $G/H_1$ is isomorphic to $G/H_2$? I know that if I remove the condition "cyclic groups", the given statement is false and there are numerous counterexamples that disprove it, but I don't know if my statement is true and I don't how to create a counterexample or to prove it. If it is true, can you give me a hint about how this can be proven?
For example, I just have shown that $Z_{12}/ \langle 2 \rangle$ is isomorphic to $Z_{12}/Z_6$ which is isomorphic to $Z_3$ (since both $\langle 2 \rangle$ and $Z_6$ are isomorphic). How this can be generalized? 

No, that's not true. Take $G = C_2 \times C_4$ and note that since $G$ is abelian, every subgroup of $G$ is normal in $G$. Let $H = C_2$ be the first direct factor of $G$ and $K$ be the unique subgroup of order $2$ of $C_4$. Obviously, $H \cong K$ since both have order $2$, but $G/H \cong C_4$ while $G/K \cong C_2 \times C_2$.
