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.



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 $|G/H|$.

  • $\begingroup$ for example, if $G$ has rank $2$ and $\Bbb Z$-basis $x, y$; and if $H$ has $\Bbb Z$-basis $$2x+y, x+3y$$ then what are the elements of $G/ H$ ? or how can one find these elements? $\endgroup$ – Consider Non-Trivial Cases Aug 28 '20 at 18:12
  • $\begingroup$ To have an explicit list of elements, you need to find the invariant factors and the adapted basis. This is obtained vy the algorithm to find the Smith normal form of this matrix. A priori, we know it will have 5elemenst (hence it will be cyclic and the invariant factors will be $(1,5)$). $\endgroup$ – Bernard Aug 28 '20 at 18:44

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.

  • $\begingroup$ in which book or notes have these examples/ also can you plz address the example I used? thanks. $\endgroup$ – Consider Non-Trivial Cases Aug 28 '20 at 17:52
  • $\begingroup$ $2\cdot 1$ is 2 times 1. In your case the quotient is the abelian group of order $142$ generated by $\le3$ elements. What this group is exactly you can get by reducing the matrix to the triangular form. About the book, there should be a reference in the book you are reading. $\endgroup$ – Mark Sapir Aug 28 '20 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.