So I want to check if I have the right idea here. I am told and abelian group $G$ is generated by elements $a,b,c$ such that $3a + 6b + 3c = 9b + 9c = -3a + 3b + 6c = $ From my understanding, I proceed by constructing a matrix with the coefficients of these relations as so:

$$ X = \left( \begin{matrix} 3 & 0 & -3 \\ 6 & 9 & 3 \\ 3 & 6 & 6 \end{matrix} \right) $$

And then, by getting the principal ideal representation for the fitting ideals $Fit_1(X), \;Fit_2(X),\; Fit_3(X)$ we can put the matrix in Smith-Normal Form.

$$Fit_1(X) = (3),\; Fit_2(X) = (27), \; Fit_3(X) = (0)$$

$$ \Rightarrow X' = \left( \begin{matrix} 3 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 0 \end{matrix} \right) $$

Then this tells us that:

$$ G \cong \frac{\mathbb Z}{(3)} \times \frac{\mathbb Z}{(9)} \times \frac{\mathbb Z}{(0)} $$

$$ \Rightarrow G \cong \mathbb Z_3 \times \mathbb Z_9 \times \mathbb Z $$

I just wanted to clarify on the last bit since normally I'm asked to do this for non-singular matrices, meaning I get three non-zero values on the Smith-Normal Form Matrix, and so I have the product of three finite abelian groups.

Here I have assumed that since one of the fitting ideals is $(0)$ then it corresponds to a vector that never returns to zero, i.e. that there should be an infinite abelian group in the product that makes up $G$ and hence we get the $\mathbb Z$ term.

I just wanted to check whether or not that is the right way to think about it. Thanks in advance for any help you may be able to offer.

  • 1
    $\begingroup$ It looks just fine to me. Singularity of the matrix shows that in the cyclic decomposition of the group there are non-torsion elements in the group, and that $\;\Bbb Z^r\;$ , with $\;r=n-\text{rank}\,X\;$ , appears in the decomposition. $\endgroup$ – DonAntonio May 19 '17 at 11:06

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