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(a) Define what it means for an abelian group to be finitely generated. Explain the terms elementary divisors and rank of $G$ and describe the structure theorem for finitely generated abelian groups.

(b) Consider the integral matrix $R:=\left[\begin{array}{cccc}2&2&2&2\\4&4&8&5\\6&12&12&8\\4&10&8&6\end{array}\right]$.

Determine the structure of the abelian group given by generators and relations $$A_r:=\left\langle a_1,a_2,a_3,a_4\mid R\circ a = 0\right\rangle.$$

I know this is a long question, But I am really struggling to find a method online that is clear and can be applied again to another example. I put part a in to give some context of the question but I am pretty happy with the definitions, part b is the problem.

A method and answer for part b would be amazing, thanks.

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  • $\begingroup$ Do you know the Smith Normal (Rational) form of a matrix? If you don't, you may want to google it. $\endgroup$ – DonAntonio May 3 '14 at 0:41
  • $\begingroup$ Some notes that might help you are at rutherglen.science.mq.edu.au/ccooper/Groups/… $\endgroup$ – Gerry Myerson May 3 '14 at 12:51
  • $\begingroup$ Have you had a chance to look at those notes? $\endgroup$ – Gerry Myerson May 5 '14 at 13:22
  • $\begingroup$ Just saw this, looks good, will have a go now. $\endgroup$ – user140152 May 6 '14 at 0:23
  • $\begingroup$ Have a look at this question and my answer to it for a similar example. $\endgroup$ – Alexander Gruber May 6 '14 at 22:51
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For each matrix $A\in \mathrm{Mat}_{4}(\mathbb{Z})$, the map $$\begin{array}{rcl} \mathbb{Z}^4 & \to & \mathbb{Z}^4\\ \left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]&\mapsto & A\cdot\left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]\end{array}$$ is a group homomorphism of Abelian groups. It is moreover a group isomorphism if and only if $A\in \mathrm{GL}_4(\mathbb{Z})$, which corresponds to $\det(A)=\pm 1$.

You are trying to get the subgroup of $\mathbb{Z}^4$ given by $$\left\{\left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]\in \mathbb{Z}^4\ \left|\ R\cdot\left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]=0\right\}\right.$$ which is the kernel of the group homomorphism given by $$\begin{array}{rcl} \mathbb{Z}^4 & \to & \mathbb{Z}^4\\ \left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]&\mapsto & R\cdot \left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]\end{array}$$ Applying an automorphism of $\mathbb{Z}^4$ at the target does not change the kernel. Applying an automorphism at the source changes the kernel, but only by an isomorphism, so if you want to compute the isomorphism class of your group, you can multiply $R$ at the right and left by elements of $\mathrm{GL}(4,\mathbb{Z})$. In particular, you can add any multiple of any row of $R$ to another one, and the same for columns. You can also exchanges lines or columns and multiply one line or one column by $-1$.

At the end you get a diagonal matrix which is called Smith normal form. There is a short algorithm but you can do it easily by hand. For instance, here subtracting the first column to the others gives $$\left[\begin{array}{cccc}2&2&2&2\\4&4&8&5\\6&12&12&8\\4&10&8&6\end{array}\right]\to \left[\begin{array}{cccc}2&0&0&0\\4&0&4&1\\6&6&6&2\\4&6&4&2\end{array}\right].$$ You can then add multiples of the first line to the others and get $$\left[\begin{array}{rrrr}2&0&0&0\\4&0&4&1\\6&6&6&2\\4&6&4&2\end{array}\right]\to \left[\begin{array}{rrrr}2&0&0&0\\0&0&4&1\\0&6&6&2\\0&6&4&2\end{array}\right].$$ Then you exchange the second and fourth column and remove four times this one to the second to get $$\left[\begin{array}{rrrr}2&0&0&0\\0&0&4&1\\0&6&6&2\\0&6&4&2\end{array}\right]\to \left[\begin{array}{rrrr}2&0&0&0\\0&1&4&0\\0&2&6&6\\0&2&4&6\end{array}\right]\to \left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&2&-2&6\\0&2&-4&6\end{array}\right].$$ The last steps are the following $$ \left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&2&-2&6\\0&2&-4&6\end{array}\right]\to\left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&0&-2&6\\0&0&-4&6\end{array}\right]\to\left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&0&-2&6\\0&0&0&-6\end{array}\right]\to\left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&0&-2&0\\0&0&0&-6\end{array}\right].$$ Hence your group is isomorphic to the quotient of $\mathbb{Z}^4$ by the kernel of $$\begin{array}{rcl} \mathbb{Z}^4 & \to & \mathbb{Z}^4\\ \left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]&\mapsto & \left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&0&2&0\\0&0&0&6\end{array}\right]\cdot\left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]\end{array}$$ and is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/6\mathbb{Z}$.

If you want the generators of your group you can also take them back by the isomorphisms. You can also first do operations on the rows, so that you do not change the kernel, and obtain a triangular matrix.

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  • $\begingroup$ so what would the generators be? 2,1,2,6? $\endgroup$ – user140152 May 12 '14 at 19:16
  • $\begingroup$ In the last coordinates, the group corresponds to $(a_1,a_2,a_3,a_4)\mid 2a_1=0,a_2=0,2a_3=0,6a_4=0$. So the generators are $(1,0,0,0)$, $(0,0,1,0)$ and $(0,0,0,1)$, and have order $2$, $2$ and $6$ respectively. If you want the generatours of the group at begin, you need to write what are the matrix used for right multiplication and apply these to the three vectors. (Just keep track of all isomorphisms given by the steps of the algorithm) $\endgroup$ – Jérémy Blanc May 12 '14 at 20:24

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