How to show that $\#(\mathbb Z^n/N) = \det(A)$ where $N$ is the subgroup of $\mathbb Z^n$ generated by columns of a square matrix $A$

Asked Jan 30 '19 at 07:58

Active Jan 30 '19 at 21:42

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$A$ is a $n\times n$ matrix with integer coefficients. I am trying to understand a) and b) claims.
a) $Z^n/N$ is finite $\Leftrightarrow$ $det(A)\neq 0$
b) #$(Z^n/N)=|det(A)|$

Looking at the linked answer, I can reduce $A$ to normal form by multiplying by unimodular matrices and this multiplication does not change $|det(A)|$. I understand that the $det(A)=d_1d_2..d_r$ where $d_i|d_{i+1}$ and $d_1,..d_r$ are entries of normal form. But I don't understand why $Z^n/AZ^n \cong \bigoplus_{i=1}^{n} Z/d_iZ$? The explanations I have seen in MSE are in the context of rings and I don't know them yet. Any hint to explain the isomorphism using group theory language is appreciated. Thanks.

edited Jan 30 '19 at 18:55

asked Jan 30 '19 at 07:58

trickymaverick

3,372

This is posted many times on MSE. Use search! – user26857 Jan 30 '19 at 08:16
I have tried that and could not find anything unfortunately. – trickymaverick Jan 30 '19 at 08:20
https://math.stackexchange.com/questions/881454/corollary-to-smith-normal-form – user26857 Jan 30 '19 at 08:44
https://math.stackexchange.com/questions/2416648/how-to-show-that-the-kernel-of-a-map-f-mathbbzm-to-g-is-free-of-rank-m-w – user26857 Jan 30 '19 at 08:45
https://math.stackexchange.com/questions/2648447/card-mathbbzn-m-mathbbzn-textrmdetm/2648596#2648596 – user26857 Jan 30 '19 at 08:45
Thanks for those links. – trickymaverick Jan 30 '19 at 08:55

How to show that $\#(\mathbb Z^n/N) = \det(A)$ where $N$ is the subgroup of $\mathbb Z^n$ generated by columns of a square matrix $A$

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