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I am revising for my upcoming university exams and I have a past exam question that I am finding particularly challenging...

a) Consider the integral matrix

$$R=\begin{bmatrix} 2 & 2 & 4 & 2\\ 4 & 4 & 8 & 5\\ 6 & 12 & 12 & 8\\ 4 & 10 & 8 & 6\\\end{bmatrix}$$

Determine the structure of the abelian group given by generators and relations

$$A_R := \langle a_1,a_2,a_3,a_4 \mid R.a=0 \rangle$$

b) Determine the number of elements of order 2 in $A_R$.

Any help with this would be very much appreciated!

Thanks in advance!

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  • $\begingroup$ You can do $\Bbb Z^n/R\Bbb Z^n\cong \Bbb Z^n/D\Bbb Z^n$ with $D$ diagonal via Smith Normal Form. $\endgroup$ – blue Apr 19 '14 at 15:55
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You know that invertible matrix operations do not change the range of the matrix, thus $$A_R = \mathrm{cok}(R \colon \mathbb Z^4 \to \mathbb Z^4)$$ remains the same. Subtract row one once from row two and four and twice from row three. You end up with the matrix $$ \begin{pmatrix} 2 & 0 & 0 & 0 \\ 4 & 0 & 0 & 1 \\ 6 & 6 & 0 & 2 \\ 4 & 6 & 0 & 2\end{pmatrix} $$ Now you make the obvious column operations to end up with

$$ \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 6 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ The result is clearly $$ A_R = \mathbb Z \oplus \mathbb Z/2 \mathbb Z \oplus \mathbb Z/6\mathbb Z $$

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  • $\begingroup$ Thanks a lot this is very helpful, could you please explain to me the "cok" notation, thanks again for the help!! $\endgroup$ – Justin Apr 27 '14 at 17:33
  • $\begingroup$ Sure. There are several levels of insight. As a first step, you may view it as a short hand for $\mathbb Z^4 / \mathrm{im} R$, where $R$ is the homomorphism associated to the matrix $A_R$. If you are used to some notions of category theory, you may understand it by its universal property as a special a colimit. This is useful to make the similarities between several categories (like Modules, Algebras and everything that has enought 'commutative structure') precise. Here the notion of abelian category the right concept. $\endgroup$ – Felix Boes Apr 29 '14 at 7:53
  • $\begingroup$ 4 years late to the party, I am trying to understand where the $\mathbb{Z}$ in $A_R\simeq \mathbb{Z} \times \mathbb{Z}_2\times\mathbb{Z}_6$ came from while comparing to my notes. Am I correct in thinking that it is because there are 4 generators and 3 zeros on the main diagonal? For a moment I thought it might be because of the 1 in the fourth column but it seems that the one in the fourth column would give rise to a phantom $\times \{e\}$. Thanks! $\endgroup$ – Ruth May 28 '18 at 5:13

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