Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to find the structure of the abelian group: $$G=\frac{\mathbb{Z}^{3}}{\langle (2,0,10),(0,4,8),(4,-4,12) \rangle}$$ The Smith normal form of the matrix associated to $G$ is: $$P= \left( \begin{array}{ccc} 2 & 0 & 0 & 0\\ 0 & 4 & 0 & 0\\ 0 & 0 & 0 & 0\end{array} \right),$$ which is correct (verified it with software). Thus the decomposition of $G$ as a direct sum of cyclic groups is $\mathbb{Z}_{2} \oplus \mathbb{Z}_{4} \oplus \mathbb{Z}$ yes? why the answer is $\mathbb{Z}_{2} \oplus \mathbb{Z}_{4} \oplus \mathbb{Z} \oplus \mathbb{Z}$, i.e why the extra summand $\mathbb{Z}$?. I thought that we only look at the elements of the diagonal of $P$ which in this case there are only three: $2,4,0$.

What am I doing incorrect? Can you please explain?

share|improve this question
3  
What makes you think the answer is $\mathbb{Z}_2\oplus\mathbb{Z}_4\oplus\mathbb{Z}\oplus\mathbb{Z}$? That group requires at least four generators, and since $G$ is a quotient of a 3-generator group, it can be generated by 3 elements. It cannot equal the group you claim it equals. (Note also that $(4,-4,12) -2(2,0,10)=(0,-4,-8)$, so you can drop one of the generators of the normal subgroup you are moding out by). –  Arturo Magidin Jun 9 '11 at 4:36
3  
I don't get it. We have a subgroup of $\mathbf{Z}^3$ generated by three elements. Shouldn't we be looking at the Smith normal form of a 3x3 matrix instead of 3x4? –  Jyrki Lahtonen Jun 9 '11 at 5:12

1 Answer 1

You are correct that the decomposition of $G$ is $\mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}$. If the computer is telling you that the answer is $\mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z} \oplus \mathbb{Z}$, the most likely explanation is that the matrix for $G$ was somehow transposed at some point during your calculation. For example, if you originally entered the transpose of the original matrix for $G$, then you would get the transpose of the Smith normal form.

Note that there isn't an established convention for which direction a matrix presentation for an abelian group should go. In some books (and in some computer programs), the rows correspond to generators and the columns correspond to relations, while in other books (and computer programs) this convention is reversed.

share|improve this answer
    
Thanks! –  kev Jun 9 '11 at 7:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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