#1162: exact order counts in a p-orbit slice

The target (erdosproblems.com/1162, OPEN). Problem 1162 asks for an asymptotic formula for the number of subgroups of the symmetric group S_n, and for a statistical theorem about their orders. Roney-Dougal and Tracey proved in 2025 that

|Sub(S_n)|=2^{n²/16+o(n²)}.

We do not solve the remaining global questions. Instead, we exactly count by order the natural p-orbit family used in their lower-bound construction, then extract its second-order term and its internal order distribution.

The exact count. Fix a prime p and write n=pm+q. Let T_p(n;m,r) count the p-subgroups H≤S_n having exactly m nontrivial orbits, each of size p, exactly q fixed points, and order p^r. If {d choose r}_p denotes the Gaussian binomial coefficient, then

T_p(n;m,r)
= n! / (q! m! [p(p-1)]^m)
  * sum_(j=0)^m (-1)^j C(m,j)
    {m-j choose r}_p.

Why the formula is exact. On each nontrivial p-point orbit, the image of H is a regular cyclic group C_p. The faithful action on all m orbits identifies H with a vector subspace of F_p^m whose projection onto every coordinate is nonzero. Inclusion-exclusion over the m coordinate hyperplanes gives the alternating Gaussian-binomial sum.

The factor before the sum counts the orbit frameworks: select the fixed points, partition the rest into unordered p-sets, and choose one regular cyclic subgroup on each set. There is no overcount, because H itself recovers its point-orbits and its projected cyclic group on each orbit.

One order carries the leading exponent. At middle rank, almost every vector subspace has full coordinate support. Combining the Gaussian-binomial asymptotic with Stirling's formula shows that, for even n, the single exact subgroup order 2^floor(n/4) occurs

2^(n^2/16 + (n/2)log_2(n)
   - (n/2)log_2(e) + O(1))

times inside this family. Thus one exact order already realizes the known n²/16 leading exponent, and this slice has an explicit positive (n/2)log₂n term. As a finite check, at n=20 the formula gives exactly 52,502,131,339,708,050 such subgroups of order 32.

A restricted order law. Choose uniformly from the same p-orbit family on n=pm points, allowing every rank, and let R_m be the base-p logarithm of the subgroup order. If m=2s, then for each fixed integer k,

Pr(R_m−s=k) → p^−k² / Σ_j∈Zp^−j².

For m=2s+1, the exponent −k² is replaced by −k(k−1), producing two equal central modes. These are discrete theta laws and imply R_m=m/2+O_Pr(1). They apply only to this explicitly defined family, not to a uniformly random subgroup of S_n.

Verification and hostile audit. Exact-integer programs check the formula through small vector spaces and construct the actual permutation groups in selected cases through S₉. A separate hostile audit rebuilt the bijection without the author code, enumerated the complete 156-subgroup lattice of S₅, checked 2,730 groups in the S₈ p=2 slice and 4,200 groups in the S₉ p=3 slice, and independently derived every sign in the asymptotics and both parity laws.

Scope. This does not determine the unknown total for S₁₉, improve the global leading exponent, provide a matching global second-order upper bound, or settle the order law for a uniformly random subgroup. The live page had no claimed proof or current worker, and a dated literature search did not find this order-refined formula; that is a scoped search report, not a claim that unindexed prior work cannot exist.