Proposition 3

With the assumptions and notation of Th. 3, let H be the set of pseudo-reflections belonging to G and distinct from 1. Assume that K is of characteristic 0. Then $\operatorname{Card}(H) = \sum_{i=1}^{l} (k_i - 1)$.

By Prop. 3 of the Appendix, we can assume that K is algebraically closed. For any $g \in G$, let $\lambda_1(g), \ldots, \lambda_l(g)$ be its eigenvalues. Since every $g \in G$ is diagonalizable (Appendix, Prop. 2), $g = 1$ if and only if all of the $\lambda_i(g)$ are equal to 1, and $g \in H$ if and only if the number of $\lambda_i(g)$ equal to 1 is $l - 1$ (we then denote by $\lambda(g)$ the eigenvalue distinct from 1). Prop. 1 of no. 1 and Lemma 3 prove that

in $K[[T]]$, and hence in $K(T)$. Consequently, we have in $K(T)$

which can be written

It follows that $q - \prod_{i=1}^{l} (1 + T + \cdots + T^{k_i-1})$ vanishes for $T = 1$, so $q = k_1 k_2 \ldots k_l$, which we knew already by the Cor. of Th. 3. This granted, let Q(T) be the polynomial $(1 - T)^{-1}(q - \prod_{i=1}^{l} (1 + T + \cdots + T^{k_i-1}))$. Differentiating the equality $(1 - T)Q(T) = q - \prod_{i=1}^{l} (1 + T + \cdots + T^{k_i-1})$ and putting $T = 1$, we see that $-Q(1)$ is the value for $T = 1$ of

so

Returning to (17), we have on the other hand

Thus,

Now the elements of G that leave fixed the points of a given hyperplane leave stable a complementary line of the hyperplane (Appendix, Prop. 2), and so form a cyclic subgroup G' of G by Algebra, Chap. V, § 11, no. 1, Th. 1. Let t be the order of G'; the values of $\lambda(g)$ for $g \in G'$ are $\theta, \theta^2, \ldots, \theta^{t-1}$ with $\theta$ a primitive tth root of unity. We have $\frac{1}{1-\theta^i} + \frac{1}{1-\theta^{t-i}} = 1$, so $\sum_{g \in G', g \neq 1} \frac{1}{1-\lambda(g)} = \frac{1}{2}(t-1) = \frac{1}{2}\mathrm{Card}(H \cap G')$. The equality (18) thus proves the proposition.