Bourbaki

Exercise 17, § 4

Lie Groups and Lie Algebras, II, 192

Bourbaki marks this one as harder than the rest.

Suppose that G is nilpotent. Let P be a set of prime numbers and let Vp(G)\mathfrak{V}_p(G) be the set of normal subgroups of G of finite index whose index is a P-integer (cf. Exercise 14). Let TP(G)\mathscr{T}_P(G) be the topology defined on G by the filter base Vp(G)\mathfrak{V}_p(G) (General Topology, Chapter III, § 1, no. 2, Example). (a) Let N be a subgroup of G of finite index such that (G:N)(G:N) is a P-integer and let N' be the intersection of the conjugates of N. Show that (G:N′)(G:N') is a P-integer (use the fact that the group G/N' is the product of its Sylow groups); deduce that N is open with respect to TP(G)\mathscr{T}_P(G). (b) Let H be a subgroup of G of finite index. Show that TP(H)\mathscr{T}_P(H) coincides with the topology induced on H by TP(G)\mathscr{T}_P(G). (Same method.) (c) Let H be a normal subgroup of G such that G/H is isomorphic to Z\mathbf{Z} and let x be a representative in G of a generator of G/H. Let N∈VP(H)N \in \mathfrak{V}_P(H) and let N′=⋂n∈ZxnNx−1N' = \bigcap_{n \in \mathbf{Z}} x^n N x^{-1}. Show that N′∈VP(H)N' \in \mathfrak{V}_P(H) and that xN′x−1=N′x N' x^{-1} = N'. If N′′N'' is the subgroup generated by N' and x, show that (G:N′′)=(H:N′)(G:N'') = (H:N'). Deduce that TP(H)\mathscr{T}_P(H) is induced by TP(G)\mathscr{T}_P(G). (d) Suppose that G is finite. Show that, if H is a subgroup of G, there exists a sequence of subgroups

H=H0<H1<…<Hk=G H = H_0 < H_1 < \ldots < H_k = G
such that HiH_i is normal in Hi+1H_{i+1} for 0≤i<k0 \leq i < k and that Hi+1/HiH_{i+1}/H_i is finite or isomorphic to Z\mathbf{Z}. Deduce, using (b) and (c), that TP(H)\mathscr{T}_P(H) is induced by TP(G)\mathscr{T}_P(G).

(e) With the hypotheses of (d), let P′P' denote the complement of PP in the set of prime numbers. Show that the closure of HH under TP(G)\mathscr{T}_P(G) coincides with the P′P'-saturation of HH in GG (Exercise 14); in particular, GG is Hausdorff if and only if it is P′P'-torsion-free.

Solution

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