69:T74b,63 Sec.2.5 Primitive Groups Theorem 2.5.4 A transitive group P action on Ω is primitive if and only if Pa is maximal for ∀a ∈ Ω. Proof If Pa is not maximal, then there exists a subgroup H of P such that Pa < H < P. Define a subset of Ω by A = {aτ |τ ∈ H }. Then |A| ≥ 2 because of H > Pa . First, if A = Ω, then for ∀π ∈ P we can find an element σ ∈ H such that aπ = aσ . Thus πσ−1 ∈ Pa , which gives π ∈ H and H = P. Now if there is π ∈ P with A ∩ Aπ , ∅ hold, then there are σ1 , σ2 ∈ H such π that aσ1 = aσ2 π . Thus σ−1 1 ∈ Pa < H . Whence, π ∈ H , which implies that A = A . Therefore, A is an iprimitive block and P is imprimitive. Conversely, let A be an imprimitive block of P. By the transitivity of P on Ω, we can assume that a ∈ A. Define H = {π ∈ P|Aπ = A, π ∈ P}. Then H ≤ G . For b, c ∈ A, there is a π ∈ G such that bπ = c. Thus c ∈ A ∩ Aπ . Whence, A = Aπ and π ∈ H by definition. Therefore, H is transitive on A. Consequently, A = |H : Ha |. Now if π ∈ P, then a = aπ ∈ A ∩ Aπ . So A = Aπ and π ∈ H . Thereafter, Pa < H and Pa = Ha . Applying Theorem 2.1.1, we know that |Ω| = |P : Pa | and |A| = |H : Ha | = |H : Pa |. So Pa < H < P and Pa is not maximal in P. Corollary 2.5.1 Let P be a transitive group action on Ω. If there is a proper subset A ⊂ Ω, |A| ≥ 2 such that a ∈ A, aπ ∈ A ⇒ Aπ = A for π ∈ P, then P is imprimitive. Proof By Theorem 2.5.4, we only need to prove that Pa < P{A} < P, i.e., Pa is not maximal of P. In fact, Pa ≤ P{A} is obvious by definition. Applying the transitivity of P, for ∀b ∈ A there is an element σ ∈ P such that aσ = b. Clearly, σ ∈ P{A} , but σ < Pa . Whence, Pa < P{A}. Now let c ∈ Ω \ A. Applying the transitivity of P again, there is an element τ ∈ P6a:T7ad,64 Chap.2 Action Groups such that aτ = c. Clearly, τ ∈ G but τ < G{A} . So we finally get that Pa < P{A} < P, i.e., Pa is not maximal in P. Theorem 2.5.5