Bourbaki

Exercise 5, § 7

Lie Groups and Lie Algebras, III, 392

Bourbaki marks this one as harder than the rest.

(a) Let Γ\Gamma be a compact subgroup of G=GL(n,Qp)G = \mathbf{GL}(n, \mathbf{Q}_p). Show that there exists a conjugate of Γ\Gamma which is contained in GL(n,Zp)\mathbf{GL}(n, \mathbf{Z}_p). (If TT is a lattice of Qpn\mathbf{Q}_p^n with respect to Zp\mathbf{Z}_p (Commutative Algebra, Chapter VII, § 4, Definition 1), show that the stabilizer of TT in Γ\Gamma is open in Γ\Gamma and hence of finite index and ∑γ∈ΓγT\sum_{\gamma \in \Gamma} \gamma T is a lattice which is stable under Γ\Gamma.)

(b) Deduce that GfG_f is the union of the conjugates of GL(n,Zp)\mathbf{GL}(n, \mathbf{Z}_p).

(c) Deduce that every finite subgroup Φ\Phi of GL(n,Qp)\mathbf{GL}(n, \mathbf{Q}_p) has order a divisor of an(p)a_n(p), where an(p)a_n(p) is defined by

an(p)=(pn−1)(pn−p)…(pn−pn−1)if p≠2an(p)=2n2(2n−1)(2n−2)…(2n−2n−1)if p=2. a_n(p) = (p^n - 1)(p^n - p) \ldots (p^n - p^{n-1}) \quad \text{if } p \neq 2 \\ a_n(p) = 2^{n^2}(2^n - 1)(2^n - 2) \ldots (2^n - 2^{n-1}) \quad \text{if } p = 2.
(Using (a), it can be assumed that Φ⊂GL(n,Zp)\Phi \subset \mathbf{GL}(n, \mathbf{Z}_p). Then use Exercise 4.)

(d) Show that every finite subgroup of SL(n,Qp)\mathbf{SL}(n, \mathbf{Q}_p) has order a divisor of sn(p)s_n(p), where sn(p)=an(p)p−1s_n(p) = \frac{a_n(p)}{p-1} for p≠2p \neq 2 and sn(2)=an(2)2s_n(2) = \frac{a_n(2)}{2}.

Solution

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Transcribed from the printed volume. Not translated.