Example/Demonstration of Quotient group $G/ H$ of Free Abelian Group

Question

Let $G$ be a free abelian group of rank $r$, and $H$ a subgroup of $G$. Then $G/ H$ is finite if and only if the ranks of $G$ and $H$ are equal. If this is the case, and if $G$ and $H$ have $\Bbb Z$-bases $x_1, \cdots , x_r$ and $y_1,\cdots , y_r$ with $y_i=\sum_j a_{ij}x_j$, then $|G/H|=|\det(a_{ij})|$

Can anyone show an example or demonstration of above Theorem for free Abelian group $G$, subgroup $H$, quotient group $G/ H$?

In the book, it is written that ,

for example, if $G$ has rank $3$ and $\Bbb Z$-basis $x, y, z$; and if $H$ has $\Bbb Z$-basis $$3x+y-2z, 4x-5y+ z, x +7z,$$ then $|G/ H|$ is the absolute value of $\begin{bmatrix} 3 & 1 & -2\\ 4 & -5 & 1\\ 1 & 0 & 7 \end{bmatrix}$, namely 142.

but in this case what is quotient group $G/H$? What are the elements of quotient group $G/H$?

I am having trouble to visualize how would $G/H$ look when $G$ and $H$ have $\Bbb Z$-bases $x_1, \cdots , x_r$ and $y_1,\cdots , y_r$ with $y_i=\sum_j a_{ij}x_j$.

This is Theorem Theorem 1.1 7. in the book Algebraic-Number Theory by Ian Stewart and David Tall, on page 30.

Answer 1

Sketch:

The general proof uses the Fundamental structure theorem for finitely generated abelian groups (which has a generalisation for finitely generated modules over a P.I.D.).

This theorem asserts that a subgroup $H$ of a finitely generated free abelian group $G$ of rank $r$ is free of rank $s\le r$, and that furthermore, there exists a basis $(x_1,\dots, x_r)$ and integers $(d_1,\dots, d_s)$ such that

• $(d_1x_1, \dots, d_s x_s)$ is a basis of $H$;
• $d_1\mid d_2\mid\dots\mid d_s.\,$ The $d_i$s are the invariant factors of the quotient $G/H$.

If $G$ and $H$ have the same rank, i.e. if $s=r$, in the new basis, the matrix $A=\bigl(a_{ij}\bigr)$ becomes the diagonal matrix $D=\bigl(d_i\bigr)$, and the change of basis matrix is in $SL(\mathbf Z)$, so that $\det(a_{ij})=\det D(d_i)=d_1\dots d_r$.

Now from the new basis, we deduce instantly that $$G/H\simeq\Bbb Z^r\big/(d_1\mathbf Z\times\dots\times\mathbf Z/d_r\mathbf Z) \simeq (\mathbf Z/d_1\mathbf Z)\times\dots \times (\mathbf Z/d_r\mathbf Z),$$ whence the assertion for $|G/H|$.

Answer 2

This answers the original question, before the edit.

The easiest example is if $r=1$, $G=\Bbb{Z}, H=2\Bbb{Z}$. The index is $2$, $G=\langle 1\rangle, H=\langle 2\rangle$, $2=2\cdot 1$ so the matrix is $(2)$ and the det is $2=|G:H|$. Less easy example is $G=\Bbb{Z}\times \Bbb{Z}=\langle (1,0),(0,1)\rangle$, $H=\langle (a,b),(c,d)\rangle$. Then $H$ has rank 2 iff $(a,b), (c,d)$ are not parallel. In that case the index $|G:H|$ is the area of the parallelogram spanned by $(a,b), (c,d)$ which is the determinant of the matrix.