Bourbaki

Exercise 27, § 2

General Topology, III, 304

Bourbaki marks this one as harder than the rest.

A group GG is said to have the property P(n)P(n) (n an integer > 0) if, for each system (s1,…,sn)(s_1, \ldots, s_n) of nn elements of GG, there is a permutation σ∈Sn\sigma \in S_n, other than the identity permutation (but depending on s1,…,sns_1, \ldots, s_n) such that sσ(1)⋯sσ(n)=s1⋯sns_{\sigma(1)} \cdots s_{\sigma(n)} = s_1 \cdots s_n.

a) Show that the non-commutative group S3S_3 satisfies P(4)P(4).

b) Let GG be a connected Hausdorff topological group. Show that if GG satisfies P(n)P(n) for some integer n>1n > 1, then GG is commutative. [Show that GG satisfies P(n−1)P(n - 1): given n−1n - 1 elements s1,…,sn−1s_1, \ldots, s_{n-1} of GG, reduce to the case where there is a neighbourhood UU of ee such that, for each x∈Ux \in U, we have s1⋯sn−1x=s1⋯sixsi+1⋯sn−1s_1 \cdots s_{n-1} x = s_1 \cdots s_i x s_{i+1} \cdots s_{n-1} for some index i=i(x)i = i(x); deduce that the centralizer in GG of one of the elements sj+1⋯sn−1s_{j+1} \cdots s_{n-1} (0≤j≤n−20 \leq j \leq n - 2) is both open and closed and therefore equal to GG. Hence the result if j≥1j \geq 1; if j=0j = 0, show that

s1s2⋯sn−1=s2⋯sn−1s1]. s_1 s_2 \cdots s_{n-1} = s_2 \cdots s_{n-1} s_1 ].
Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.