I've been given the finitely generated abelian group: $$\langle x_1, x_2 \mid 6x_1-6x_2, -6x_1-12x_2, 4x_1-8x_2\rangle$$ and written the corresponding matrix: $$A=\begin{pmatrix} 6 & -6 \\ -6 & -12 \\ 4 & -8 \end{pmatrix}$$ I now need to reduce this to Smith Normal form using the unimodular elementary row and column operations.

I keep running into difficulties that are usually because I'm not allowed to multiply rows or columns by fractions. How do I do it?


1 Answer 1


You want to find row/column operation for which your initial matrix looks like this

\begin{equation*} \begin{pmatrix}a_1 & 0 \\0&a_2 \\0&0 \end{pmatrix}\end{equation*}

and $a_1\mid a_2$. According to Wikipedia this numbers are unique up to multiplication by a unit, and they can be found by using the formula $a_i =\frac{d_i(A)}{d_{i-1}(A)}$, where $d_i(A)$ is the greatest common divisor of all $i\times i$ minor of you matrix.

Moreover according to this thread you are only allowed to

  1. interchange two rows or two columns,
  2. multiply a row or column by $\pm1$ (which are the invertible elements in $\mathbf{Z}$),
  3. add an integer multiple of row to another row (or an integer multiple of a column to another column).

Let us compute $a_1$. As you can see the greatest common divisor of all $1\times 1$ minor of your matrix is $2$, so we will have $a_1 =2$ once the algorithm is over.

To find $a_2$ you need to compute all the $2\times 2$ minor of your matrix, and you should get $d_2(A)= GCD(-24,108,96)=12$, so $a_2 = \frac{12}{2}=6$.

Now we should find operations that respects 1.,2.,3. such that the final matrix is


This is done in the following way (I can try it by yourself, and you should get the same solution also with other operations, because $a_1,a_2$ are unique up to multiplication by unit):

\begin{align*} \begin{pmatrix} 6&-6 \\-6&-12\\4&-8\end{pmatrix} &\overset{Ir-IIIr; \, IIr + IIIr}{\leadsto} \begin{pmatrix} 2&2\\-2&-20\\4&-8\end{pmatrix} \\ \overset{IIc-Ic}{\leadsto}\begin{pmatrix}2&0\\-2&-18\\4&-12\end{pmatrix} &\overset{IIr+Ir;\, IIIr-2Ir}{\leadsto} \begin{pmatrix}2&0\\0&-18\\0&-12\end{pmatrix} \\ \overset{IIr-IIIr}{\leadsto}\begin{pmatrix} 2&0\\0&-6\\0&-12\end{pmatrix}&\overset{IIIr-2IIr}{\leadsto}\begin{pmatrix}2&0\\0&-6\\0&0\end{pmatrix} \end{align*}

Now you can mulitply by $-1$ the second row.


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