Why is $Z^n/AZ^n \cong Z/d_1Z\oplus Z/d_2Z..$ where $d_i$ are the entries of normal form of A?

Question

In trying to show that $Card(Z^n/AZ^n)=det(A)$ which has been answered earlier here $\mathrm{card}(\mathbb{Z}^n/M\mathbb{Z}^n) = |\det(M)|$? , the answer refers to the isomorphism $Z^n/AZ^n \cong Z/d_1Z\oplus Z/d_2Z..$. Can you point me to a theorem that would lead to this direct sum decomposition? I know that a finitely generated Abelian group can be factored into cyclic groups and I suppose $Z^n/AZ^n$ is finitely generated. But I am not sure what the generator set for $Z^n/AZ^n$ is? A hint is appreciated. Thanks.

Answer 1

Suppose $A=P \mathrm{diag} (d_1,...,d_n) Q$ with $P,Q\in GL_n(\mathbb{Z})$.

Then $A\mathbb{Z}^n = P\mathrm{diag} (d_1,...,d_n) Q\mathbb{Z}^n = P\mathrm{diag} (d_1,...,d_n) \mathbb{Z}^n$ because $Q$ is invertible.

Now it's easy to check that when $f: M\to M$ is an automorphism of an abelian group $M$, and $N$ is a subgroup of $M$, then $M/N\simeq M/f(N)$. Indeed, consider $f:M\to M$, $p$ the projection onto $M/f(N)$ and $q$ the one onto $N$: then $p\circ f$ is $0$ when restricted to $N$, so that $p\circ f$ factors uniquely through $Q$: we get $\overline{f} : M/N\to M/f(N)$. Use $f^{-1}$ to build a map in the opposite direction, and then check (it's easy) that the two maps are invere.

Thus $\mathbb{Z}^n/A\mathbb{Z}^n = \mathbb{Z}^n/P(\mathrm{diag}(d_1,...,d_n)\mathbb{Z}^n) \simeq \mathbb{Z}^n/\mathrm{diag}(d_1,...,d_n)\mathbb{Z}^n$ because $P$ is an automorphism of $\mathbb{Z}^n$. Now $\mathrm{diag}(d_1,...,d_n)\mathbb{Z}^n$ is just $d_1\mathbb{Z}\oplus ... \oplus d_n\mathbb{Z}$, and so $\mathbb{Z}^n/A\mathbb{Z}^n \simeq \displaystyle\bigoplus_{i=1}^n\mathbb{Z}/d_i\mathbb{Z}$