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Let $G$ be an abelian group with generators $x, y, z$ and $t$ subject to the following relations:

$\begin{align*} 4x - 4y + 18z + 18t &= 0\\ 2x + 4z + 10t &= 0\\ x - 3y + 12z + 6t &= 0. \end{align*}$

My thought was to try to represent this as the matrix:

$$\begin{bmatrix}4 & -4 & 18 & 18\\ 2 & 0 & 4 & 10\\ 1 & -3 & 12 & 6 \end{bmatrix}$$

and calculate its Smith normal form. We know that $d_0 = 1$ and then $d_1$ will be the greatest common divisor of all the entries which is $1$ and then finally $d_2 = \operatorname{gcd}(8, -16, 108, -6, 12, -96) = 2.$

Then the Smith normal form of the matrix would be:

$$\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 2 & 0\\ \end{bmatrix}$$

which would mean that $G \cong \mathbb{Z}/\langle 1 \rangle \oplus \mathbb{Z}_\langle 1 \rangle \oplus \mathbb{Z}/\langle 2 \rangle \cong \mathbb{Z}_2$. Is this correct? Any help is appreciated.

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A quick check on the program Magma tells me that the Smith Normal Form of your matrix is different to the one you have found.

I'm not familiar with the method you seem to be using to calculate it, the more usual approach uses row and column operations. There's a good description and example of it in this question, it might be worth following the process there and seeing what you come out with.

Also, you still have an empty column in your Smith Normal Form, so you'll need to add something else to the sum in $G$. There's an example you might find helpful in this question.

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I ended up figuring out the answer, I made a few mistakes in my computation in the question. We need to compute the determinant of all the $i \times i$ minors and define $d_i$ to be the $\operatorname{gcd}$ of them for a fixed $i$. We also define $d_0 = 1$, then the diagonal entries of the SNF of the matrix will be $d_i/d_{i-1}$. We have $d_0 = 1$, $d_1$ which is just the $\operatorname{gcd}$ of all the entries is $2$ and then $d_2 = 2$ and then finally $d_3 = 20$. We can see this by computing all of the determinants but I don't want to do that here.

Then this yields the SNF:

$$\begin{bmatrix} \frac{d_1}{d_0} & 0 & 0 & 0\\ 0 & \frac{d_2}{d_1} & 0 & 0\\ 0 & 0 & \frac{d_3}{d_2} & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 10 & 0 \end{bmatrix}.$$

Then in order to compute $G$ we mod by the image of this matrix so we have $\mathbb{Z}^4/\langle (1, 2, 10, 0) \rangle \cong \mathbb{Z}_2 \oplus \mathbb{Z}_{10} \oplus \mathbb{Z}$.

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    $\begingroup$ Yep, this matches what Magma told me $\endgroup$ – Dave Aug 6 at 19:04

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