Jtdylktuy

The problem goes as follows:

Let $(a,b)$ and $(c,d)$ be two linearly independent vectors in $mathbb{Z}^2$. Show that the abelian group
$$ G = mathbb{Z}^2/langle(a,b); (c,d)rangle$$
is finite, and determine the cardinality of $G$ in terms of $a,b,c,d$.

It's related to the fundamental theorem of finitely generated modules over PID. Since $(a,b)$ and $(c,d)$ are linearly independent, form the vectors in a $2times2$ matrix
$begin{pmatrix}
a& b \
c & d
end{pmatrix}=:A$. Applying the theorem, we can find two invertible matrices, say, $S$ and $T$ over $mathbb{Z}$ such that $Sbegin{pmatrix}
a& b \
c & d
end{pmatrix}T = begin{pmatrix}
d_1& 0 \
0 & d_2
end{pmatrix}=:D$, where $d_1,d_2inmathbb{Z}$ and $d_1|d_2$. We see that $det A = det D = pm d_1d_2$. So $det S = pm 1 = det T$. Then we have that
$$G = mathbb{Z}^2/langle(a,b); (c,d)ranglecong mathbb{Z}_{d_1}times mathbb{Z}_{d_2}.$$
It seems to me that $G$ is finite and $|G| = |det A|$. But I am not sure how to prove it. Although I was told to consider for a particular case of the group, I still am having a hard time to come up a complete solution. Anything would be appreciated! Thanks in advance.

Since $Gequiv Bbb{Z}_{d_1}times Bbb{Z}_{d_2}$, the order of $G$ is $d_1d_2$. Now, obviously, $d_1d_2=det D$. Now, $SAT=D$, so $det Scdot det Adet T=det D$. Take the absolute value, and we get $|det S||det A||det T|=|det D|$. However, $det S,det T=pm 1$, so $|det S|=|det T|=1$. Therefore, they cancel out and we get $|det A|=|det D|=|d_1d_2|$. Thus, in short, the order of $G$ is $d_1d_2$ and $|det A|=d_1d_2$, so $|G|=|det A|$.

Also, I tried to figure this out in an alternate way to validate your proof, but ultimately failed. However, I still think it's interesting. First, let $N$ be the normal subgroup generated by $(a,b)$ and $(c,d)$. Then, let's consider the order of $(x,y)+N$, just to kind of explore the group. Eventually, we should get to some $(nx,ny)+N$ which is a linear combination of $(a,b)$ and $(c,d)$. That gives us the following system:

$$c_1a+c_2c=nx$$
$$c_1b+c_2d=ny$$

By Cramer's Rule, the solution to this system is:
$$c_1=frac{n(dx-cy)}{ad-bc}, c_2=frac{n(ay-bx)}{ad-bc}$$

Now, $n$ is the smallest integer where these two constants are integers, so at the very least, we know $n$ is a factor of $ad-bc$. From that and the fact that this is a finitely generated abelian group, we know that the group is finite and thus it's going to be in the form of $Bbb{Z}_{a_1}timesBbb{Z}_{a_2}times...$. Furthermore, we know that $(1,0)+N$ and $(0,1)+N$ generates the group, so it's going to be in the form of $Bbb{Z}_{a_1}timesBbb{Z}_{a_2}$ for some $a_1$ and $a_2$. (This has been proven in another StackExchange question.)

Now the proof linked above also implies that $|G/N|$ is the product of the order of $(1,0)+N$ and the order of $(0,1)+N$ divided by the order of $(G/N)/C$, where $C$ is the cyclic subgroup generated by $(1,0)+N$. Now, I don't know if it helps, but maybe it does. Anyway, that's as far as I got, so good luck with the rest.