Let $G=\mathbb{Z}\times\mathbb{Z}$ be a group under componentwise addition of integers and $H=3\mathbb{Z}\times 3\mathbb{Z}$. Except for $\mathbb{Z}\times 3\mathbb{Z}$ and $3\mathbb{Z}\times\mathbb{Z}$, determine all the proper subgroups of $G$ which properly contain $H.$ Explain why your list is complete.
I have to admit that for this one, I haven't had a nice working proof nor idea how to proceed. I hope someone understands me. Please, I need some help. At least make me understand how to proceed. Thank you.
Edit: I still haven't figured this out. Can someone please explain to me the proof so I would be able to study it for similar questions? I have prepared a bounty for this question. Thank you very much.
It is easy to check that the quotient group $G/H=\{(m,n)+H: m,n\in\{ 0,1,2\}\}$. That is $|G/H|=9$. Consider a quotient homomorphism $q: G\to G/H$, putting to each element $x\in G$ its coset $x+H$. Clearly, $q(G)=G/H$ and $q(H)=H+H=H$ is the identity of the group $G/H$.
Now let $F$ be a proper subgroup of $G$ properly containing $H$. It is well-known and easy to show that $q(F)$ is a subgroup of $q(G)$. If $q(F)=0+H$ then $F\subset H$, which is impossible, since $F$ contains $H$ properly. If $q(F)=G/H$ then $G=F+H=H$, which is impossible, since $G$ contains $F$ properly. Thus $q(F)$ is a proper non-identity subgroup of $G/H$. Since $|q(F)|$ divides $|G/H|=9$, we conclude $|q(F)|=3$. Since each group of order $3$ is cyclic, $q(F)$ is cyclic. Let $x+H$ be its generator. Then $x=(m,n)+H$ for some $m,n\in\{ 0,1,2\}$. The following cases are possible.
If $m=0$ and $n=0$ then $x+H=H$ and so $F=H$, a contradiction.
If $m=0$ and $n\ne 0$ then $F=\{(0,k)+H: k\in\{ 0,1,2\}\}=3\Bbb Z\times\Bbb Z$, but this is an excluded case.
If $m\ne 0$ and $n=0$ then $F=\{(k,0)+H: k\in\{ 0,1,2\}\}=\Bbb Z\times 3\Bbb Z$, but this is an excluded case.
If ($m=1$ and $n=1$) or ($m=2$ and $n=2$) then $F=\{(k,k)+H: k\in\{ 0,1,2\}\} $.
If ($m=1$ and $n=2$) or ($m=2$ and $n=1$) then $F=\{(k,2k)+H: k\in\{ 0,1,2\}\} $.