Group Isomorphic to $(\mathbb{Z} \times \mathbb{Z})/ \langle(a,b),(c,d)\rangle$

Question

I have come across an interesting question regarding quotient groups of $\mathbb{Z} \times \mathbb{Z}$:

Suppose $(a,b)$ and $(c,d)$ are two independent elements of $\mathbb{Z} \times \mathbb{Z}$. Suppose also that $\gcd(a,b) =1$. Show that $(\mathbb{Z} \times \mathbb{Z})/ \langle(a,b),(c,d)\rangle \, \cong \, \mathbb{Z}_{|ad-bc|}$.

I have managed to find a little about the geometric interpretation of this result, but to provide a clear algebraic proof I believe it is necessary to use some linear algebra that I can't quite see. I would appreciate it if someone might be able to provide an elementary proof of this fact, perhaps using the First Isomorphism theory and a judicious choice of map.

Answer 1

There is a standard method of finding the invariants of abelian groups defined as quotients of ${\mathbb Z}^n$, in which you write down the associated matrix, and then perform unimodular row and column operations to transform it to Smith Normal Form. In this example the matrix is $\left( \begin{array}{cc}a&b\\c&d\end{array}\right)$.

These operations do not change the absolute value of the determinant of the matrix. Also, since theit effect on matrix entries $x$ is to replace $x$ by either $-x$ or by$x+ky$ for some other matrix entry $y$ and some $k \in {\mathbb Z}$, they do not change the $\gcd$ of the set of all matrix entries.

So the Smith Normal Form of the matrix in your case must be $\left( \begin{array}{cc}t&0\\0&u \end{array}\right)$, where $t = \gcd(a,b,c,d)$ and $tu=|ad-bc|$, and then $G \cong C_t \times C_u$.

So if $\gcd(a,b,c,d)=1$, then $t=1$ and $G \cong C_{|ad-bc|}$.

Answer 2

Let $G=\mathbb Z×\mathbb Z,$ $(a,b)\in G$ s.t. $gcd(a,b)=1$. Then there exists $a^*,b^*\in \mathbb Z$ such that $aa^*+bb^*=1$. Now consider the group $G/H$ where $H=\langle (a,b)\rangle$. We are going to show that $G/H\cong \mathbb Z$.

Let $(m,n)+H\in G/H$. Consider the system of equations, $$b^*x+ay=m$$$$-a^*x+by=n$$ Since the value of the determinant of the coefficient matrix is $det\begin{pmatrix}b^*&a\\ {-a^*}&b\end{pmatrix}=1$, we have integral solutions of $x$ and $y$. That is, there is $i_1,i_2\in \mathbb Z$ s.t. $b^*i_1+ai_2=m$ and $-a^*i_1+bi_2=n$

So, $(m,n)+H$ $$=(b^*i_1+ai_2,-a^*i_1+bi_2)+H$$ $$=i_1((b^*,-a^*)+H)\in \langle (b^*,-a^*)+H\rangle$$ i.e. $G/H=\langle (b^*,-a^*)+H\rangle.$

Note that, if $a=b$ then, the order of $(1,0)+H$ is infinite. If $a\ne b$ then, the order of $(1,1)+H$ is infinite. Hence, $G/H=\langle (b^*,-a^*)+H\rangle \cong \mathbb Z$

Now let's consider another subgroup of $G$. Let $K=\langle (a,b),(c,d)\rangle$, where $(a,b)$ and $(c,d)$ are independent(i.e. $(a,b)\notin \langle (c,d)\rangle$ and $(c,d)\notin \langle (a,b)\rangle$) Consider the group $K/H$. Note that $K/H\trianglelefteq G/H$. Consider the element $(ad-bc)((b^*,-a^*)+H)$ $$=(adb^*-bb^*c, a^*bc-aa^*d)+H$$ $$=(\alpha, \beta)+H,$$ where $\alpha =adb^*-bb^*c$ and $\beta=a^*bc-aa^*d$ of $G/H$.

Let us consider the equation $AX=B$, where $A=\begin{pmatrix}a&c\\ b&d\end{pmatrix}$, $X=\begin{pmatrix}x_1\\ x_2\end{pmatrix}$, $B=\begin{pmatrix}\alpha \\ \beta \end{pmatrix}$. Since, $det$ $A=ad-bc\ne 0$(as $(a,b)$ and $(c,d)$ are independent), $X=A^{-1}B$. By some further calculation we get that $X=\begin{pmatrix}u_1\\ u_2\end{pmatrix}$ for some $u_1, u_2\in \mathbb Z$(for a matter of interest, you can check that $u_1=b^*d+a^*c, u_2=-1$). So, $au_1+cu_2=\alpha$ and $bu_1+du_2=\beta$ Therefore, $(\alpha ,\beta )+H$$$=(au_1+cu_2,bu_1+du_2)+H$$$$=u_1(a,b)+u_2(c,d)+H\in K/H$$

Now we are going to show that $K/H=\langle (\alpha ,\beta )+H\rangle.$ Let, $\theta +H\in K/H$ where $\theta=t_1(a,b)+t_2(c,d)=(t_1a+t_2c, t_1b+t_2d)$ for some $t_1,t_2\in \mathbb Z$.

Now consider the equation, $A^*X^*=B^*$ where $A^*=\begin{pmatrix}\alpha &a\\\beta &b\end{pmatrix}$, $X^*=\begin{pmatrix}x\\ y\end{pmatrix}$, $B^*=\begin{pmatrix}t_1a+t_2c\\ t_1b+t_2d \end{pmatrix}$. Here $det$ $A^*=ad-bc\ne 0$. So, $X^*=(A^*)^{-1}B^*$. By further calculation we get $X^*=\begin{pmatrix}v_1\\ v_2\end{pmatrix}$ for some $v_1,v_2\in \mathbb Z$(for a matter of interest you can check that $v_1=-t_2, v_2=t_1+t_2(a^*c+b^*d)$) Hence, $$t_1a+t_2c=\alpha v_1+av_2$$$$t_1b+t_2d=\beta v_1+bv_2$$ $\Rightarrow \theta +H=(\alpha v_1+av_2, \beta v_1+bv_2)+H$$$=v_1(\alpha ,\beta )+H\in \langle (\alpha ,\beta )+H\rangle$$ So, $K/H=\langle (\alpha ,\beta )+H\rangle$

Now, $H\trianglelefteq K\trianglelefteq G$ where $$G/H=\langle (b^*,-a^*)+H\rangle \cong \mathbb Z$$ and $$K/H=\langle (ad-bc)(b^*,-a^*)+H\rangle \cong \mathbb (ad-bc)Z$$ So, from isomorphism theorem, we get $$(G/H)/(K/H)\cong G/K$$$$\Rightarrow \mathbb Z×\mathbb Z/\langle (a,b),(c,d)\rangle \cong \mathbb Z_{\mid ad-bc\mid}$$