Why is the determinant equal to the index?

Solution 1:

$\newcommand\ZZ{\mathbb{Z}}$Suppose $\Gamma\subseteq\ZZ^n$ is a subgroup of finite index, so that it is in fact the image of some linear map $\ZZ^n\to\ZZ^n$ which is, in turn, given by multiplication by a matrix $A\in M_n(\ZZ)$.

According to the theory of the Smith normal form, there exist $P$, $Q\in\mathrm{GL}_n(\ZZ)$ such that $D=PAQ$ is diagonal. It is easy to see that the index of $\Gamma=A\ZZ^n$ is the same as the index of $D\ZZ^n$, which is easily seen to be the absolute value of the product of the diagonal entries in $D$, which is the same as $|\det D|$. Since $P$ and $Q$ have determinant $\pm1$, we conclude that the index of $\Gamma$ is $|\det A|$.