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I'm reading this article and I don't understand what the authors are doing in Theorem 3, part 1: The authors wanted to compute the cardinality of a $\mathbb{Z}$-module $\Lambda$. They computed the relation matrix $A$ for the module $\Lambda$ (i.e. $\Lambda = coker(A)$):

$A = \begin{bmatrix} l & 0 \\ 0 & m \\ n & n \end{bmatrix}$,

with $l,m,n \ge 1$. They conclude that $\Lambda= \frac{\mathbb{Z}}{d_1\mathbb{Z}} \times \frac{\mathbb{Z}}{d_2\mathbb{Z}}$ with $d_1 = \gcd(l,m,n)$ and $d_2 = \frac{\gcd(lm,ln,mn)}{d_1}$ (and hence that the cardinality $|\Lambda|$ equals to $d_1d_2$).

I don't see where the $\gcd$ comes from and why this statement is true.

Giacomo Bascapè
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    Presumably, using the Smith Normal Form, which lets you just read off the index of the corresponding subgroup, hence the cardinality of $\Lambda$. – Arturo Magidin Aug 05 '22 at 17:52
  • I don't see how the Smith Normal Form gives me the answer: I know that if $A$ is an invertible matrix (and hence square) then $det(A= | \Gamma|$. In our case the matrix is rectangular. – Giacomo Bascapè Aug 16 '22 at 14:33
  • As I said, the Smith Normal Form lets you read off the index of the subgroup that corresponds to the matrix, and the index of the subgroup is the dimension of the cokernel. Invertibility has nothing to do with it. The gcds you mention arise naturally in the Smith Normal form, and I would guess that you end up with a Smith Normal Form that looks like $\left(\begin{array}{cc}d_1&0\0&d_2\0&0\end{array}\right)$, from which you read off the cardinality of the corresponding cokernel. – Arturo Magidin Aug 16 '22 at 14:36
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    See here. – Arturo Magidin Aug 16 '22 at 14:55
  • That said, doing an example didn't come out with the value described. I tried $\ell=42$, $n=70$, $m=30$, so $d_1=2$, $d_2= 210$, and what I got was $\mathbb{Z}/14\mathbb{Z}\times\mathbb{Z}/30\mathbb{Z}$; same size, but not the same invariant factors. – Arturo Magidin Aug 16 '22 at 15:10
  • @ArturoMagidin Thank you for the explanation. I tried your example with the website "http://www.numbertheory.org/php/smith.php" and this is what I've found: $A= \begin{bmatrix} 42 & 0 \ 0 30 \ 70 & 70 \end{bmatrix}$ then $SNF(A)= \begin{bmatrix}= 2 & 0 \ 0 & 210 \ 0 & 0 \end{bmatrix}$. – Giacomo Bascapè Aug 16 '22 at 17:07
  • Then I must have done something wrong when I tried to compute the Smith Normal form. From the SNF(A) you have, you can read off that the index is $2\times 210$, with invariant factors $2$ and $210$, so you get that the quotient, $\Lambda$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/210\mathbb{Z}$... – Arturo Magidin Aug 16 '22 at 17:13
  • Oh, I know what I did wrong. $14\nmid 30$, so I wasn't done... – Arturo Magidin Aug 16 '22 at 17:15

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