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This question already has an answer here:

I am aware of some theorem that says that if $M$ is non singular, $\det(M) \neq 0$, then: $$\mathrm{card}(\mathbb{Z}^n/M \mathbb{Z}^n)= |\det(M)|.$$ How does one prove this? Figured if I put in the context of the rational canonical form, this would help, but cant piece it together. Thanks.

This result is mentioned in the first answer of this question: Cardinality of a Quotient Ring.

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marked as duplicate by Watson, José Carlos Santos, Shailesh, Namaste abstract-algebra Feb 16 '18 at 0:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ What if $M=2I$? $\endgroup$ – Lord Shark the Unknown Feb 13 '18 at 5:10
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    $\begingroup$ And that is isomorphic to $\Bbb Z/2^n\Bbb Z$? $\endgroup$ – Lord Shark the Unknown Feb 13 '18 at 5:14
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    $\begingroup$ If $M\in GL_n(\mathbb{Z})$ then $\det(M)=\pm1$. $\endgroup$ – eloiprime Feb 13 '18 at 5:17
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    $\begingroup$ If $M \in GL_n(\mathbb Z)$ then $\mathbb Z=M \mathbb Z$ and the result is trivial. $\endgroup$ – N. S. Feb 13 '18 at 5:18
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    $\begingroup$ If you do not have any idea about the answer to Lord's question, think about it. «No idea» is very close to being disrespectful as an answer to his question, by the way... It was obviously a hint. $\endgroup$ – Mariano Suárez-Álvarez Feb 13 '18 at 5:44
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The standard method for computing the abelian invariants of $\mathbb{Z}^n/M\mathbb{Z}^n$ is to put the matrix $M$ into Smith Normal Form.

This is done by applying a sequence of unimodular transformations to $M$, and these can be effected by pre- or post-multiplying $M$ by a unimodular matrix over $\mathbb{Z}$. These unimodular matrices all have deteminant $\pm 1$, so they do not change $|\det M|$.

At the end of the process, the transformed matrix $M$ is diagonal with entries $d_1,d_2,\ldots,d_n$ (where each $d_i|d_{i+1}$). The determinant of the matrix is now $d_1d_2 \cdots d_n$, and the algorithm proves that $\mathbb{Z}^n/M\mathbb{Z}^n \cong \oplus_{i=1}^n \mathbb{Z}/d_i\mathbb{Z}$, of which the order is also $d_1d_2 \cdots d_n$. QED.

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Let $A = \mathbb{Z}^n$, so that $\operatorname{End}(A) = M_n\left(\mathbb{Z}\right)$ and $\operatorname{End}(A \otimes \mathbb{Q}) = M_n\left(\mathbb{Q}\right)$. Define $f:\operatorname{End}(A) \to \mathbb{Z}$ by \begin{align*} f(g) &= \begin{cases} |A/gA| & \text{if $\det g \not= 0$}; \\ 0 & \text{if $\det g = 0$}. \end{cases} \end{align*} Extend it to a map $f:\operatorname{End}(A\otimes \mathbb{Q}) \to \mathbb{Q}$ by setting $f(g) = |\alpha|^{-n} f(\alpha g)$ for any nonzero $\alpha\in \mathbb{Z}$ with $\alpha g\in \operatorname{End}A$ (and check that it's well-defined). For nonsingular $g, h\in \operatorname{End}(A)$, the map $A/hA \to A/ghA$ of abelian groups defined by $x \to gx$ is injective. Thus its image $gA/ghA$ is isomorphic to $A/hA$. But $$(A/ghA) / (gA/ghA) = A/gA$$ by the third isomorphism theorem, and taking cardinalities shows that $f(gh) = f(g)f(h)$. It follows that $f(gh) = f(g)f(h)$ for arbitrary nonsingular $g, h\in \operatorname{End}(A\otimes \mathbb{Q})$. Now use the fact that the nonsingular matrices $T_{ij, \lambda} = 1 + \lambda\delta_{ij}$ for $\lambda\in \mathbb{Q}$ generate $GL_n(\mathbb{Q})$ to compute $f$.

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