Subgroups of finite index of $\mathbb Z/2\mathbb Z\times \mathbb Z$ Let $H$ be a subgroup of index $n$ in $(\mathbb Z/2\mathbb Z)\times \mathbb Z.$    

Is there finitely many subgroups of finite index of  $(\mathbb Z/2\mathbb Z)\times \mathbb Z$ ?

If yes, can we find all the explicit forms possible  of the subgroup $H$  ?
thanks.
 A: Every subgroup where there is an element with a nonzero right coordinate will have finite index, since $\langle (0,n) \rangle$ and $\langle (0,2n) \rangle \subseteq \langle (1,n) \rangle$ both would have finite index in the group, so we have there are infinitely many subgroups of finite index. 
There are finitely many subgroups of a given index $n\in \mathbb N$, and I will show which subgroups further in the post, but I would like to point out all finitely generated groups have finitely many subgroups of a given index $n \in \mathbb{N}$. The sketch of how you would show that for finitely generated groups is: there are finitely many transitive actions on a finite set which can all be seen as $G$ acting on $G/H$ for some $H$ of finite index (and $G$ always has a transitive action on $G/H$), but all transitive finite actions have an associated group homomorphism $G \to S_n$, which determine $H$, and there are only finitely many since a group homomorphism is determined by where its generating set is mapped to and we are assuming there are finitely many generators. 
Lets describe the subgroups of $\mathbb{Z}/ 2\mathbb Z \times \mathbb Z$:


*

*Obviously if there is no nonzero element in the right coordinate of any element of $H \leq \mathbb{Z}/ 2\mathbb Z \times \mathbb Z$ we have the subgroup $H$ is either trivial or $\mathbb Z/2 \mathbb Z \times \{0\}$ which is infinite index. 

*Subgroups generated by one element are of the form $\{0\} \times n \mathbb{Z}$  or $\langle (1,n) \rangle$ have index $2n$. 

*Subgroups $\mathbb{Z}/2\mathbb{Z} \times n\mathbb{Z}$ are of index $n$ and are generated by two elements. Not that if $(1,0)$ is in a subgroup generated by two elements, then it is of the above form, since we can consider an element with the least positive right coordinate $n$, so $(1,n)$ or $(0,n)$, but in either case we have $(0,n)$ is in the group, and $(1,0), (0,n)$ generate $\mathbb{Z}/2\mathbb{Z} \times n\mathbb{Z}$.


Using the above "classification" we get for index $n$ odd, there is only only one subgroup, $\mathbb{Z}/2\mathbb{Z} \times n\mathbb{Z}$, and for $n$ even there are three, $\mathbb{Z}/2\mathbb{Z} \times 2n\mathbb{Z}$, $\{0\} \times n \mathbb{Z}$, and $\langle (1,n) \rangle$.
We will now prove there are no other subgroups:
Say that we have a subgroup $H$ which is rank greater than or equal to two and not $\mathbb{Z}/2\mathbb{Z} \times n \mathbb{Z}$, so $(1,0) \not \in H$. I will show such an $H$ does not exist. Let $(0,m)$ and $(1,n)$ have the least positive possible $m,n$ with the respective coordinates. We will show $(1,0) \in H$ or $H$ has rank 1 which will contradict our assumptions.
Note that the group generated by $(0,m)$ and $(1,n)$ will have $(0,\gcd(n,m))$ or $(1,\gcd(n,m))$, we will now consider these cases.


*

*If $(0,\gcd(n,m))$, then $\gcd(n,m)=m$, so $m|n$. Also $\gcd(n,m)=n$ since $n-m$ would have to be $0$ since $m|n$ but $(1,n)$ has the least such $n$. So we have $(1,m),(0,m)$ in the group, so $(1,0)$ is in the group, a contradiction.

*If $(1,\gcd(n,m) )=(1,n)$ then $n|m$. Consider $(0,m), (1,m-n), (0,m-2n)$: by assumption $m-2n \leq 0$, but then $(1,m-n)$ must be $(1,0)$ or $(1,n)$, since $n=\gcd(n,m) | m$, the first case is an immediate contradiction, so we must have $(0,m)=(0,2n)$, notice that $(0,m) \in \langle (1,n) \rangle$. Suppose $(0,p)$ or $(1,p)$ is in $H$ but not in $\langle (1,n) \rangle$ with $p>0$. Well then in the case of $(0,p)$ we must have $2kn \neq p>2n$. If $n \not \mid p$ then $(1,\gcd(n,p)) \in H$ or $(0,\gcd(n,p)) \in H$ and $\gcd(n,p) <n$ a contradiction. If $n | p$, then $p=(2k+1)n$, so $(-2kn,p-2kn)=(0,n)$ and $(1,0) \in H$ a contradiction. If it is $(1,p)$ then $2kn=p>n$
(as otherwise $q=\gcd(n,p)<n$ and we would have $(1,q)$ or $(0,q)$, either would be a contradiction, or $(1,p) \in \langle (1,n) \rangle$ which is also a contradiction) but that leads to $(1,2kn)-(2k,2kn)=(1,0) \in H$, a contradiction. So there is no such $p$, but then $H$ actually had rank 1, a contradiction.

