Quotient group and adjoint matrix The exercise 1211 in "Problems and Solutions in Mathematics" by Ta-Tsien:
Let $M$ be an $n \times n$ matrix of integers. Suppose that $M$ is invertible when viewed as a matrix of rational numbers.
Show that $\mathbb{Z} / M \mathbb{Z}^n$ is finite.
The proposed answer is: 
It is obvious from the facts that the map: $g:\mathbb{Z}^n/ M \mathbb{Z}^n \rightarrow \mathbb{Z}^n / |M| \mathbb{Z}^n$, $\delta(\overline{X}) = \overline{M^*X}$ is injective, and the fact that $|\mathbb{Z}^n / |M| \mathbb{Z}^n| = (|\det M|)^n$.
$|M|$ is the determinant of $M$ and $M^*$ is the adjoint of $M$.
I think there is a typo and $\delta$ and $g$ designate the same function. Can someone details why this function is injective, and why $|\mathbb{Z}^n / |M| \mathbb{Z}^n| = (|\det M|)^n$ ?
 A: Let $k$ be any positive integer. Is it obvious for you that $\Bbb{Z}^n/k\Bbb{Z}^n$ is the same as $(\Bbb{Z}/k\Bbb{Z})^n$? If yes, then $|\Bbb{Z}^n/k\Bbb{Z}^n|=k^n$, so the last equality applies with $k=|M|$ (the symbol $|M|$ usually denotes the determinant of $M$).
Now, denote by $k=\det M$. It is well known that $M^*M=MM^*=kI$, where $I$ denotes the identity matrix. In particular this implies that $M^*M\Bbb{Z}^n=k\Bbb{Z}^n$.
Consider the composition of the maps
$$\Bbb{Z}^n \longrightarrow \Bbb{Z}^n \longrightarrow \Bbb{Z}^n/k\Bbb{Z}^n$$
where the first map is multiplication by $M^*$, the second is the canonical projection.  Clearly the kernel of this map is
$$\{ v \in \Bbb{Z}^n: M^*v \in k\Bbb{Z}^n\} = \{ v \in \Bbb{Z}^n: M^*v \in M^*M\Bbb{Z}^n\} = M\Bbb{Z}^n$$
where the second inequality follows from the fact that $M^*$ has rank $n$. Moreover this map is surjective.
So by the first isomorphism theorem
$$\Bbb{Z}^n/M\Bbb{Z}^n \cong \Bbb{Z}^n/k\Bbb{Z}^n$$
and actually the isomorphism is exactly the map $g$ (or $\delta$).
