Bourbaki

Exercise 1, § 1

General Topology, VII

Let G be a discrete subgroup of rank pp in Rn\mathbf{R}^n and let (ai)1≤i≤p(\mathbf{a}_i)_{1 \leq i \leq p} be a free system of pp points of G. The proof of Theorem 1 of no. 1 shows that G is a subgroup of the group generated by the pp points d−1aid^{-1} \mathbf{a}_i, where dd is some integer. Show, without using Theorem 1, that there exists a free system of pp points bi=∑j=1pbijaj\mathbf{b}_i = \sum_{j=1}^p b_{ij} \mathbf{a}_j of G such that, if xi=∑j=1pxijai(1≤i≤p)x_i = \sum_{j=1}^p x_{ij} \mathbf{a}_i \quad (1 \leq i \leq p) is any free system of pp points of G, we have ∣det⁡(xij)∣≥∣det⁡(bij)∣>0|\det(x_{ij})| \geq |\det(b_{ij})| > 0. Hence give a proof of Theorem 1 which does not rest on the theory of invariant factors, by showing that the bi\mathbf{b}_i generate G. (Argue by contradiction: if z=∑i=1pzibi∈Gz = \sum_{i=1}^p z_i \mathbf{b}_i \in G and one of the ziz_i is not an integer, there exists a point u=∑i=1puibiu = \sum_{i=1}^p u_i \mathbf{b}_i of the group generated by zz and the bi\mathbf{b}_i such that 0<ui<10 < u_i < 1 for some index ii, and hence obtain a contradiction.)

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