Describe all the quotient groups of $\mathbb{Z} \oplus \mathbb{Z}$ I'm thinking about a problem in algebraic topology of how to determine all the $k$-fold covers of $\mathbb{T}^2$, where $k$ is a certain integer. In thinking about the problem, I came upon the following question: 
It's clear that every homomorphism $\phi:\mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$ can be represented by a $2\times 2$ matrix over the integers (i.e. $\phi = \begin{pmatrix} a & b \\ c& d \end{pmatrix}$ where $a,b,c,d \in \mathbb{Z}$). If $\phi$ has a kernel, can we necessarily say that the quotient group $\mathbb{Z} \oplus \mathbb{Z} / \ker(\phi)$ has the form $\mathbb{Z}_m \oplus \mathbb{Z}_n$, where $m$ and $n$ are some integers? If not, can we still nicely represent all the quotient groups of $\mathbb{Z} \oplus \mathbb{Z} / \ker(\phi)$, where $\phi: \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$? Moreover, how does the matrix representation of $\phi$ relate to the quotient groups? For example, does the determinant of the matrix give you all or most of the information you need to find the quotient groups?
My motivation in considering this is, as I mentioned, $k$-fold covers of $\mathbb{T}^2$. The fundamental group of $\mathbb{T}^2$ is of course $\mathbb{Z} \oplus \mathbb{Z}$. I want to relate the determinant of $\phi$ to how many times a covering map covers $\mathbb{T}^2$. For example, a $k$-fold cover $p: \mathbb{T}^2 \to \mathbb{T}^2$ would have $\det(p_\star) = k$. Where $p_\star: \pi_1(\mathbb{T}^2) \to \pi_1(\mathbb{T}^2)$ is the induced homomorphism . I'm not sure that this is true, but it motivated my question.
Thanks for your help!
 A: $\newcommand{\ZZ}{\mathbb{Z}}$


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*Any quotient of $\ZZ\times\ZZ$ has the form $\ZZ_m\times\ZZ_n$, where $m,n$ could be zero (in which case we set $\ZZ_0 := \ZZ$). However, in your case $(\ZZ\times\ZZ)/\ker\phi$ is just the image of $\phi$, which lying in $\ZZ\times\ZZ$, cannot have torsion. Thus, $(\ZZ\times\ZZ)/\ker\phi$ necessarily has the form either $\ZZ\times\ZZ$ or $\ZZ$ or $0$.

*The absolute value of the determinant of $\phi$ is the index of the image of $\phi$ in $\ZZ\times\ZZ$ (unless the determinant is zero, in which case the image is either $0$ or $\ZZ$, and the index is infinite) This because you can use elementary row and column operations to the matrix of $\phi$ to diagonalize it, which corresponds to applying automorphisms to the domain/range of $\phi$, which of course does not change the index of the image. Of course, if you have a diagonal matrix with diagonal entries $a,d$, then the determinant is just $ad$, and the image is just $a\ZZ\times d\ZZ$, easily seen to have index $|ad|$.

*Yes, the index of the image of the induced map on fundamental groups is exactly the degree of the corresponding cover. Note here that because covering maps satisfy the homotopy extension property, the induced map on fundamental groups coming from a cover is necessarily injective. Hence, the determinant is nonzero and your map has trivial kernel.

*Connected degree $k$ covers of a space $X$ are in bijection with the conjugacy classes of subgroups of index $k$. The set of all degree $k$ covers of $X$, not necessarily connected, are in bijection with (equivalence classes of) homomorphisms $\pi_1(X)\rightarrow S_k$, where the connected covers correspond to those homomorphisms whose image is a transitive subgroup of $S_k$. (Can you figure out what the equivalence relation must be?)
