TITLE: Fundamental group of a Root System and determinant of the Cartan matrix
QUESTION [8 upvotes]: This is the first time I am posting, so I hope I didn´t get the formatting wrong.
 I am currently reading J. E. Humphreys "Introduction to Lie Algebras and Representation Theory" and got stuck at chapter 13 "Abstract theory of weights".
Let E be an arbitrary euclidean space and $\Phi$ a root system in E. 
 First there are defined the group $\Lambda=\{\lambda \in E \ \vert \ \left< \lambda, \alpha \right> \in \mathbb{Z} \ \forall \alpha\in \Phi\}$ and the subgroup    $\Lambda_{r}$, called root lattic, generated by $\Phi$. It is $\left<\lambda, \alpha \right>=\frac{2(\lambda, \alpha)}{(\alpha,\alpha)}$ and (,) denotes the positiv definit symmetric bilinear form on E.
 Then the fundamental group of  $\Phi$ is defined as the quotient $\Lambda/\Lambda_{r}$.
 My question is, why the determinant of the Cartan matrix should equal the index of $\Lambda_{r}$ in $\Lambda$ and what that says about the root system. 
The book gives the following arguments:
Let $\Delta=\{\alpha_{1}, \dots, \alpha_{l}\}$ be a base of $\Phi$, then the dual base $\lambda_{1}, \dots, \lambda_{l}$, i.e. $\left<\lambda_{i},\alpha_{j}\right>=\delta_{ij}$, is a basis of the lattice $\Lambda$. The Cartan matrix expresses a chage of basis, so in order to write the $\lambda_{j}$ as linear combiation of the $\alpha_{i}$ one simply has to invert the Cartan matrix. This part I think, I understand. What I do not understand is the following statement: The determinant of the the Cartan matrix is the only denominator involved, so it measures the index of $\Lambda_{r}$ in $\Lambda$.
I fail to see the connection.
Thank you for helping me.

REPLY [4 votes]: This follows from the classification of finitely generated modules over
a PID, or from the (not very) special case of the stacked basis theorem,
or the corresponding calculations with matrices.
Consider the lattices as modules over $\mathbb Z$. The weight lattice is isomorphic to $\mathbb Z^n$ where $n$ is the dimension of the weight space,
and the root lattice is some submodule of this.  This depends on the
entries in the Cartan matrix being integers.  The stacked basis theorem
says that we can change the basis for the full lattice so that integer
multiples of the basis elements form a basis for the sublattice.
Translate this to integer matrices.  Change of basis converts the
Cartan matrix into a diagonal matrix with the multiples for the stacked
basis on the diagonal.  The order of the quotient is obviously the
absolute value of the determinant of this.  Change of basis doesn't
change the determinant except possibly to multiply it by $-1$, since
the change of basis matrix has integer elements and is invertible.  In
fact, there is no change of sign provided we make the natural choice
of positive multiples.  The determinant of the Cartan matrix is strictly
positive since the matrix is positive definite.