# Subgroups of prescribed power-of-two order in symmetric groups

Submitted by tienxion. 4 September 2026. A partial result relevant to Erdős 1163. This note
counts actual subgroups, not conjugacy classes, and does not determine
the order distribution of a uniformly chosen unrestricted subgroup.
Developed with OpenAI GPT-6 assistance, including independent agent
proof audits. No claim of novelty or prior expert endorsement is made.

Let
\[
a_{n,j}=\#\{H\leq S_n:|H|=2^j\}.
\]
Uniformly for integers \(\lfloor n/4\rfloor\leq j\leq\lfloor n/2\rfloor\),
\[
\log_2 a_{n,j}\geq\frac{n^2}{16}
+\left(\frac{7n}{8}-\left|j-\frac{3n}{8}\right|\right)\log_2 n-O(n).
\tag{1}
\]
The groups supplying this bound have nilpotency class at most two,
exponent dividing four, and orbits of size at most eight. Combining
(1) with the upper bound for **all** 2-subgroups in Roney-Dougal–Tracey,
[Theorem 2](https://arxiv.org/html/2503.05416v1), gives the uniform conclusion
\[
\log_2 a_{n,j}=\frac{n^2}{16}+O(n\log n).
\tag{2}
\]
Thus every prescribed order in this interval attains the full quadratic
counting exponent. The remaining error can still change relative
probabilities substantially; (2) does not imply equidistribution.

## An explicit eight-point group

On \(\mathbb F_2^2\times\mathbb F_2\), let \(B_0\) consist of the flips
\[
(i,e)\longmapsto(i,e+b_i),\qquad \sum_i b_i=0.
\]
Let \(V=\mathbb F_2^2\) translate the first coordinate, and put
\(E=B_0\rtimes V\). It has order 32 and acts transitively: translations
move between fibers, while even flips can interchange the two points
of any chosen fiber. A nonzero translation exchanges the four indices
in two pairs. For even-weight \(b\), the two pair sums agree, so
\(b+v(b)\) is zero or the all-one vector \(\mathbf1\); both values occur.
Consequently
\[
E'=\langle\mathbf1\rangle\cong C_2,
\qquad E/E'\cong\mathbb F_2^4.
\]
The square formula \( (b,v)^2=(b+v(b),0) \) shows that all squares
lie in the central derived group, proving the
class and exponent assertions. This classical degree-eight factor
appears in Kovács–Praeger,
[*Finite permutation groups with large abelian quotients*, §2](https://archives.maths.anu.edu.au/people/Kovacs/K070.pdf).

## Construction at every prescribed order

Assume \(n\geq16\), and write
\[
m=\lfloor n/2\rfloor,\quad f=\lfloor m/2\rfloor,
\quad a=m-2f\in\{0,1\},\quad\delta=n-2m\in\{0,1\}.
\]
Take \(k=f\), except when \(m=2f+1\) and \(j=m\), in which case take
\(k=f+1\). Then \(k\geq4\), \(k(m-k)=\lfloor m^2/4\rfloor\), and
\(z=j-k\) lies between zero and \(f\). Set
\[
c=\min(z,f-z),\quad b=f-2c,\quad d=z-c.
\]
These are nonnegative integers satisfying
\[
a+2b+4c=m,\qquad c+d=z,\qquad d\in\{0,b\}.
\]
Partition the labels into \(\delta\) singletons, \(a\) pairs, \(b\)
four-point blocks, and \(c\) eight-point blocks. Use \(C_2\) on each
pair and \(E\) on each eight-point block. On every four-point block
use the regular Klein four-group if \(d=0\), or the transitive
dihedral group of order eight if \(d=b\). When \(b=0\), this
distinction is immaterial. Transport one fixed model to each labelled
block using its increasing ordering.

Their direct product \(P\) satisfies
\[
P/P'\cong\mathbb F_2^m,\qquad |P'|=2^z.
\]
Every quotient factor has dimension at most four. Choose a linear
surjection from \(\mathbb F_2^4\) onto each factor. The image of their
combined map surjects onto every factor and has dimension at most four;
extend it to a four-dimensional subspace \(W\leq\mathbb F_2^m\).

For each \(k\)-dimensional subspace \(U\) containing \(W\), take its
full preimage \(H\) in \(P\). This group contains \(P'\) and projects
fully onto each quotient factor, so it projects fully onto each orbit
group. Its orbits are exactly the chosen blocks and
\(|H|=2^{k+z}=2^j\). Distinct \(U\)'s give distinct subgroups; different
partitions give different orbit decompositions. Therefore
\[
a_{n,j}\geq
\frac{n!}{\delta!\,2^a a!\,(4!)^b b!\,(8!)^c c!}
{m-4\brack k-4}_2.
\tag{3}
\]
All counted groups inherit the class and exponent restrictions from \(P\).

The Gaussian coefficient satisfies
\[
{R\brack s}_2
=\prod_{i=0}^{s-1}\frac{2^R-2^i}{2^s-2^i}
\geq 2^{s(R-s)}.
\]
Thus the logarithm of the last factor in (3) is at least
\((k-4)(m-k)=n^2/16-O(n)\), uniformly in \(j\).
Stirling's formula gives
\[
\log_2\frac{n!}{\delta!\,2^a a!\,(4!)^b b!\,(8!)^c c!}
=(n-b-c)\log_2 n+O(n)
\]
uniformly even if \(b\) or \(c\) is zero: use
\(b\log b+c\log c=(b+c)\log n+O(n)\), with \(0\log0=0\).
Finally,
\[
b+c=f/2+|z-f/2|,
\qquad
n-b-c=7n/8-|j-3n/8|+O(1).
\]
Substitution proves (1). The cited published upper bound proves (2).

The upper bound and the eight-point group are prior results. The
contribution of this note is the explicit prescribed-order construction
and its uniform estimate; whether this refinement is already recorded
in the literature has not been established.