Converse characterization of Property R graphs

Proposed solution: the converse holds (every star configuration is realized)

Claim. Ritter's Conjecture 1 (arXiv:math/0308013) is true, in the following sharpened form: a finite digraph Γ is isomorphic to the Property R graph T_G of a finite group G iff Γ is a nonempty disjoint union of stars S̄j (star with j leaves and a self-loop at the center), j ≥ 0. In particular, for any m_1 ≥ … ≥ m_k ≥ 1 there is a finite G with T_G ≅ S̄{m_1−1} ⊔ … ⊔ S̄_{m_k−1}. (A bare, unlooped isolated vertex — Ritter's "S_{−1}" — can never occur: every vertex of T_G has out-degree exactly one, so the "isolated points" of his Theorem 4 are precisely the S̄_0 components.)

Step 1 — reduction to a multiset problem. The edge map is Q(G,N) = N·G′ (preimage of (G/N)′ is the join of N with the derived subgroup). So T_G is the functional graph of the closure operator N ↦ N ∨ G′ on the normal subgroup lattice. Consequences (two lines each): edges only land on the closed vertices {a : G′ ≤ a ≤ G}; those carry self-loops (⟺ abelian quotient); two-step paths collapse. Hence T_G = ⨆{a ⊇ G′} S̄{m_a−1} with weights m_a = #{N ⊴ G : N·G′ = a}, and the isomorphism type of T_G is exactly the multiset W(G) = {m_a}. The conjecture becomes: every finite nonempty multiset of positive integers is W(G) for some finite G.

Step 2 — building blocks. (a) All-ones multiset {1,…,1} (k entries): G = C_{2^{k−1}} (abelian ⇒ all loops; k subgroups). (b) Wreath towers: for nonabelian simple X_1,…,X_s let P = X_1 ≀ (X_2 ≀ (… X_s)) with regular actions. Then P is perfect, centerless at every proper quotient, and its normal subgroups form a chain 1 = J^0 < J^1 < … < J^s = P (the base of each stage is the unique minimal normal subgroup; the proof uses only "normal subgroups of X^Λ are subproducts" + transitivity + C_W(base) = 1). This already gives every singleton {m}: a perfect group with exactly m normal subgroups. (c) Diagonal automorphisms: automorphisms σ_t of the layers assemble into a "diagonal" automorphism D(σ_1,…,σ_s) of P, and D(σ_1,…,σ_s) is inner iff it is the identity, provided each σ_t is trivial or non-inner. (Key point: an inner automorphism conjugates the base's direct factors by a left translation of the index set; comparing with the coordinate permutation of the diagonal automorphism at the identity index forces the translation to be trivial.) Taking layers X = PSL(2, 3^{2^e}) with field automorphisms φ of order 2^e (whose nontrivial powers are never inner, since ⟨φ⟩ injects into Out) gives cyclic twist groups of any 2-power order per layer.

Step 3 — classification of Normal(P ⋊ C). If P is perfect with chain normal structure and centerless proper quotients, and a finite abelian C acts so that any element inducing an inner automorphism on some P/J^ℓ induces the identity there, then the normal subgroups of G = P ⋊ C are exactly J^ℓ ⋊ D for D ≤ K_ℓ := {c : c acts trivially on P/J^ℓ}; moreover of the center P ⋊ D is #{ℓ : D ≤ K_ℓ}.

Step 4 — assembly. Given m_1 ≥ … ≥ m_k ≥ 2-or-1 with m_1 ≥ 2: take C = C_{2^{k−1}} with subgroup chain D_0 < … < D_{k−1}, s = m_1 − 1 layers, and a non-decreasin with fiber sizes #g^{-1}(i) = m_{i+1} − m_{i+2} (convention m_{k+1} = 0). Give layer t the twist of order 2^{k−1−g(t−1)}. Then K_ℓ = D_{g(ℓ)}, and the weight at D_i is #{ℓ : telescoping. So W(G) = {m_1,…,m_k}. ∎

Smallest illustrative example. T_G = S̄_2 ⊔ S̄_1 for G = (PSL(2,9) ≀ A_5) ⋊ C_2, the involution acting by the field automorphism of PSL(2,9) diagonally on all 60 base co A_5. Its five normal subgroups are 1, B = PSL(2,9)^60, B⋊C_2, P, G; centers P (leaves 1, B) and G (leaf B⋊C_2).

Verification. The Step-1 structure theory is (i) machine-checked in Lean (Q = commutator ⊔ N against the quotient definition; idempotency; star collapse; loop ⟺ abelian quotient; abelian and simple realizations), and (ii) verified computation definition of Q for S_3, S_4, S_5, A_5×C_2, C_12, SL(2,3), GL(2,3), Q_8, D_4 — reproducing the GAP tables in Ritter's paper (e.g. T_GL(2,7) = (S̄_0 S̄_2)²). Full write-up wit Steps 2–4 available; happy to share and to have any step challenged.