Doubts about a Quotient group of $\mathbb{Z}\oplus\mathbb{Z}$

Question

Consider the subgroup $H$ of $\mathbb{Z}\oplus\mathbb{Z}$ generated by $(9,1)$ and $(1,4)$.

1. Is $H=\{(9r + t,r + 4t) \in \mathbb{Z} \times \mathbb{Z} \ |\ r,t \in \mathbb{Z}\}$?

2. I think that the order of $(\mathbb{Z} \oplus \mathbb{Z}) / H$ is infinite, is it so?

3. Is $(\mathbb{Z} \oplus \mathbb{Z}) / H$ isomorphic to a well-know group?

Answer 1

Consider the matrix: \begin{bmatrix}9&1\\1&4\end{bmatrix} whose image as a map $\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ is the group $H$. We can perform automorphisms (given by invertible matrices over $\mathbb Z$, realized by row and column operations) to obtain the matrix: \begin{bmatrix}1&0\\0&35\end{bmatrix} Thus, the group $\mathbb Z\oplus \mathbb Z/H$ is isomorphic to $\mathbb Z/35$.

Edit: (more detail) Note that we can view elements of $\mathbb Z\oplus \mathbb Z$ as column matrices \begin{bmatrix}r\\t\end{bmatrix} and we can view a $2\times 2$ matrix over $\mathbb Z$ as a homomorphism $\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ by matrix multiplication: $$\begin{bmatrix}9&1\\1&4\end{bmatrix} \cdot \begin{bmatrix}r\\t\end{bmatrix}= \begin{bmatrix}9r + t\\r+4t\end{bmatrix}$$ This explains how to represent $H$ as the image of this matrix. Now, note that, by easy algebra, if I apply an automorphism of $\mathbb Z\oplus \mathbb Z\to \mathbb Z\oplus \mathbb Z$ before or after the given matrix, the resulting quotients will be isomorphic (this is basic group theory): Given an isomorphism $G\to G'$, and a subgroup $H$ with image $H'$, there is an induced isomorphism $G/H \to G'/H'$. Now, composition of homomorphisms corresponds to matrix multiplication (you can check this easily). So, we can multiply on the left and the right of the above matrix by any matrices that are invertible over $\mathbb Z$, and thus represent automorphisms of $\mathbb Z\oplus \mathbb Z$. Now, observe that: $$\begin{bmatrix}1&0\\-1&1\end{bmatrix}\cdot \begin{bmatrix}1&-8\\0&1\end{bmatrix}\cdot \begin{bmatrix}9&1\\1&4\end{bmatrix} \cdot \begin{bmatrix}1&31\\0&1\end{bmatrix} =$$ $$\begin{bmatrix}1&0\\-1&1\end{bmatrix}\cdot \begin{bmatrix}1&-31\\1&4\end{bmatrix} \cdot \begin{bmatrix}1&31\\0&1\end{bmatrix} =$$ $$\begin{bmatrix}1&-31\\0&35\end{bmatrix}\cdot \begin{bmatrix}1&31\\0&1\end{bmatrix} =$$ $$\begin{bmatrix}1&0\\0&35\end{bmatrix}$$ Using the interpretation above, the image of the matrix \begin{bmatrix}1&0\\0&35\end{bmatrix} is all column matrices of the form: \begin{bmatrix}r\\35t\end{bmatrix} and the quotient of this subgroup is $0\oplus \mathbb Z/35 = \mathbb Z/35$.

Further edit (yet more detail): So, the idea is that the subgroup $H$ is the image of a homomorphism, so say we have a homomorphism: $f:G \to G$, and then we apply an isomorphism $i: G\to G$ before $f$, then we get the composition $fi: G\to G$, since $i$ is onto, the image of this map is equal to the image of $f$, and so we have an isomorphism (equality in this case) $G/fi(G) = G/f(G)$. On the other side, if we apply $i$ after $f$, then we observe that $if(G)$ is the image of $f(G)$ under $i$, and therefore we get an induced isomorphism $G/f(G) \to G/if(G)$. By iterating this, we can pre and post-compose the homomorphism $f$ by any number of isomorphisms and we will get an induced isomorphism on the quotient of the image. In this case, the matrix:\begin{bmatrix}9&1\\1&4\end{bmatrix} is the homomorphism $f$, and the isomorphisms are given by the invertible matrices I multiplied by on the left and right. You should verify that matrix multiplication corresponds to composition of homomorphisms.

Answer 2

1. Yeah, that's one way to describe $H$
2. Note that $H$ contains $(35,0)$ (set $r=4$ and $t=-1$) and $(0,35)$ (set $r=-1$ and $t=9$). So any element in $\Bbb Z\oplus\Bbb Z/H$ has at least one representative $(a,b)$ with $0\leq a,b<35$.