Bourbaki

Exercise 15, § 5

Algebra, I, 143

Bourbaki marks this one as harder than the rest.

Let Γ\Gamma be an rr-ply transitive group of permutations of a set E with nn elements; for a permutation σ∈Γ\sigma \in \Gamma distinct from the identity permutation, let n−sn - s be the number of elements of E invariant under σ\sigma. If s>rs > r, show that there exists a permutation τ∈Γ\tau \in \Gamma such that σ−1τστ−1\sigma^{-1} \tau \sigma \tau^{-1} is distinct from the identity permutation and the number of elements of E it leaves invariant is ≥n−2(s−r+1)\geq n - 2(s - r + 1) (use the decomposition of σ\sigma into its component cycles). If s=rs = r, show similarly that there exists τ∈Γ\tau \in \Gamma such that σ−1τστ−1\sigma^{-1} \tau \sigma \tau^{-1} is a cycle of order 3. Deduce that, if r≥3r \geq 3 and Γ\Gamma does not contain the alternating group An\mathfrak{A}_n, then s≥2r−2s \geq 2r - 2 for every permutation in Γ\Gamma (use Exercise 10). Conclude finally that, if Γ\Gamma is not identical with An\mathfrak{A}_n or Sn\mathfrak{S}_n, then r≤n/3+1r \leq n/3 + 1.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.