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ERDőS #274 · PARTIAL

Erdős problem 274 — wave w040

Date: 2026-07-29 (UTC)

Claim labels

source are named.

Source-access and bibliographic facts are marked [source-checked] rather than being mathematical claims.

Step 0: mandatory live-page check

The live page was fetched through the Bright Data browser at erdosproblems.com/274, followed by its discussion thread, on 2026-07-29. [source-checked]

The verbatim problem statement is:

If \(G\) is a group then can there exist an exact covering of \(G\) by more than one cosets of different sizes? (i.e. each element is contained in exactly one of the cosets)

The live page displays:

Thus none of the mandatory stop conditions applies. [source-checked]

The page's accompanying formulation is the Herzog–Schönheim conjecture: if \(a_1G_1,\ldots,a_kG_k\) are finitely many cosets of subgroups of \(G\), with the finite indices \([G:G_i]\) pairwise distinct, then they cannot partition \(G\). For finite \(G\), distinct coset sizes and distinct indices are equivalent conditions. [a]

The page lists the following known results:

  1. Sun proved the conjecture when every participating subgroup \(G_i\) is

subnormal in \(G\); this includes the abelian case. [b: Sun 2004]

  1. Margolis and Schnabel proved it for every group of order \(<1440\).

[b: Margolis–Schnabel 2019]

The three live comments say:

  1. On 14 March 2026, Alfaiz listed the additional cases of groups possessing a

Sylow tower (Berger–Felzenbaum–Fraenkel), groups whose orders satisfy the prime-factorization conditions of Ginosar–Schnabel, and simple and symmetric groups (Garonzi–Margolis).

  1. On 29 October 2025, Alfaiz pointed to arXiv:1803.03569 for the

\(<1440\) theorem; the site says it was updated in response.

  1. On 7 October 2025, Sean Eberhard pointed out that the abelian case is not

open and cited Sun's subnormal-subgroup theorem; the site says it was updated in response.

The site explicitly warns that comments are user-supplied and unverified, so I checked the mathematical claims against the papers below. [source-checked]

Primary-source literature audit

  1. Zhi-Wei Sun, *On the Herzog–Schönheim conjecture for uniform covers of

groups*, J. Algebra 273 (2004), 153–175, DOI 10.1016/S0021-8693(03)00526-X00526-X). Its abstract states the uniform-cover result when all participating subgroups are subnormal; a partition is the multiplicity-one case. [b, source-checked]

  1. Marc Berger, Alexander Felzenbaum, and Aviezri Fraenkel, *Remark on the

multiplicity of a partition of a group into cosets*, Fund. Math. 128 (1987), 139–144, DOI 10.4064/fm-128-3-139-144. The paper treats finite pyramidal groups, which include finite supersolvable groups; later papers restate this as the Sylow-tower case. [b, source-checked]

  1. Yuval Ginosar and Ofir Schnabel, *Prime Factorization Conditions Providing

Multiplicities in Coset Partitions of Groups*, J. Comb. Number Theory 3(2) (2011), 75–86 (University of Haifa record). I checked the author-uploaded full text. Its exact relevant statements are:

\[ \prod_{p\mid |G|}\left(1+\frac1{p-1}\right)\le 2. \]

[b, source-checked]

  1. Leo Margolis and Ofir Schnabel, *The Herzog–Schönheim Conjecture for small

groups and harmonic subgroups*, arXiv:1803.03569, DOI 10.1007/s13366-018-0419-1. The abstract and Theorem A say order smaller than 1440 (strict inequality). [b, source-checked]

  1. Fusun Akman and Papa Sissokho, Transversal Coset Partitions of Groups,

Beitr. Algebra Geom. 66 (2025), 417–441 (author-hosted paper). Theorem 6 proves that a counterexample cannot use only \(2,\ldots,7\) distinct subgroups, hence any counterexample needs at least eight cosets. [b, source-checked]

  1. Martino Garonzi and Leo Margolis, *The Herzog–Schönheim conjecture for

simple and symmetric groups*, arXiv:2509.25118v2, dated 5 May 2026. Theorems 1.2 and 1.3 prove the symmetric and simple cases. Remark 5.1 explains that their reciprocal-index method cannot cover all direct products of nonabelian simple groups, since that class contains examples with unbounded reciprocal-index sum. [b, source-checked]

Targeted searches for "A5 x A5", "A_5 \times A_5", "A5 direct powers", and "direct product" + "Herzog-Schönheim" found no primary source explicitly treating all powers \(A_5^r\). This is a literature-search miss, not a claim of novelty. [c, source-checked]

New verified result

Theorem

For every integer \(r\geq 1\), the Herzog–Schönheim conjecture holds for the finite group

\[ A_5^r. \]

Equivalently, every nontrivial partition of \(A_5^r\) into cosets contains two cosets whose subgroups have the same index. [b: modulo the standard elementary theorem that \(A_5\) is simple; all remaining steps are [a]]

For \(r=2\), this is a group of order \(3600\), already outside the \(<1440\) theorem. It is neither simple nor symmetric and is nonsolvable, so the simple, symmetric, and Sylow-tower results do not apply. Its order is

\[ |A_5^2|=2^4 3^2 5^2. \]

The Ginosar–Schnabel three-prime theorem does not apply because \(p_2=3\), and their general test gives

\[ \left(1+\frac1{2-1}\right) \left(1+\frac1{3-1}\right) \left(1+\frac1{5-1}\right) =\frac{15}{4}>2. \]

Thus the result is not merely an instance of any theorem scope listed on the live page. [a, with theorem scopes checked in the cited sources]

Proof

1. Reciprocal-index criterion

For a finite group \(G\), let

\[ {\cal I}(G)=\{[G:H]:H\leq G\} \quad\text{and}\quad J(G)=\sum_{n\in{\cal I}(G)}\frac1n, \]

where indices are counted without multiplicity and \(1=[G:G]\) is included.

If \(G\) had a coset partition \(g_iH_i\) with pairwise distinct indices \(n_i=[G:H_i]\), counting elements would give

\[ \sum_i\frac1{n_i}=1. \]

All \(n_i\) are proper-subgroup indices, while \(1\) is a further, distinct member of \({\cal I}(G)\). Consequently \(J(G)\geq2\). Therefore

\[ J(G)<2 \quad\Longrightarrow\quad G\text{ is Herzog–Schönheim}. \]

This is the elementary criterion also used by Garonzi–Margolis. [a]

2. Exact index-product lemma

For any finite groups \(G\) and \(K\),

\[ {\cal I}(G\times K) ={\cal I}(G)\,{\cal I}(K) :=\{mn:m\in{\cal I}(G),\,n\in{\cal I}(K)\}. \]

Indeed, take \(L\leq G\times K\), put \(V=\pi_K(L)\leq K\), and set

\[ U=\{g\in G:(g,1)\in L\}\leq G. \]

The restriction \(\pi_K:L\to V\) is onto with kernel \(U\times\{1\}\), so

\[ |L|=|U||V| \quad\text{and hence}\quad [G\times K:L]=[G:U][K:V]. \]

Conversely, the product subgroup \(U\times V\) realizes every product of two indices. [a]

By induction,

\[ {\cal I}(A_5^r)={\cal I}(A_5)^r \]

as a set of products (duplicates are counted once). [a]

3. The base index set

The subgroup orders of \(A_5\) are exactly

\[ 1,2,3,4,5,6,10,12,60. \]

A short hand check is as follows. Lagrange's theorem leaves these orders and \(15,20,30\). Simplicity excludes order \(30\). A subgroup of order \(20\) or \(15\) would give a nontrivial coset action \(A_5\to S_3\) or \(A_5\to S_4\); simplicity would make that action injective, impossible because \(|A_5|=60>|S_3|,|S_4|\). Every remaining order occurs: cyclic subgroups give orders \(2,3,5\); the double transpositions fixing one point give \(V_4\); \(\langle(123),(12)(45)\rangle\cong S_3\); \(\langle(12345),(25)(34)\rangle\cong D_{10}\); and a point stabilizer is \(A_4\). [b: standard simplicity of \(A_5\); otherwise a]

It follows that

\[ {\cal I}(A_5)=D=\{1,5,6,10,12,15,20,30,60\}. \tag{1} \]

[b]

The standalone checker independently constructs all 60 even permutations, enumerates all 59 subgroups without using this classification, and recovers the same order and index sets. [d]

4. The full exponent semigroup

Write an index as \(2^a3^b5^c\). The exponent vectors of the nine elements of \(D\) are

\[ \begin{array}{c|c} d& (a,b,c)\\ \hline 1&(0,0,0)\\ 5&(0,0,1)\\ 6&(1,1,0)\\ 10&(1,0,1)\\ 12&(2,1,0)\\ 15&(0,1,1)\\ 20&(2,0,1)\\ 30&(1,1,1)\\ 60&(2,1,1). \end{array} \]

The additive semigroup \(S\) generated by these vectors is exactly

\[ S=\{(a,b,c)\in\mathbb Z_{\geq0}^3: \max(0,b-c)\leq a\leq2(b+c)\}. \tag{2} \]

Necessity follows because each generator satisfies

\[ a\geq0,\qquad b\leq a+c,\qquad a\leq2(b+c), \]

and these inequalities are preserved by addition. [a]

For sufficiency, fix \(b,c\), and choose \(0\leq t\leq\min(b,c)\). Use:

\(0,1,\) or \(2\) (indices \(15,30,60\));

\(1\) or \(2\) (indices \(6,12\));

\(0,1,\) or \(2\) (indices \(5,10,20\)).

For this \(t\), every integer in

\[ [\,b-t,\;2(b+c-t)\,] \]

is obtained. These intervals overlap as \(t\) runs from \(0\) through \(\min(b,c)\), and their union is exactly

\[ [\,\max(0,b-c),\;2(b+c)\,]. \]

This proves (2). Extra factors equal to the index \(1\) may be inserted at will. [a]

For every finite \(r\), the exponent support of \({\cal I}(A_5^r)\) is a finite subset of \(S\). [a]

5. Exact uniform reciprocal bound

Put \(L=\max(0,b-c)\). From (2),

\[ \begin{aligned} \sum_{(a,b,c)\in S}\frac1{2^a3^b5^c} &=\sum_{b,c\geq0}\frac1{3^b5^c} \sum_{a=L}^{2(b+c)}2^{-a}\\ &=\sum_{b,c\geq0}\frac1{3^b5^c} \left(2^{1-L}-2^{-2(b+c)}\right). \tag{3} \end{aligned} \]

Let \(x=1/3\) and \(y=1/5\). Splitting the first term of (3) into \(b\leq c\) and \(b>c\) gives

\[ \begin{aligned} T_{\leq} &=\frac{2}{1-x}\left(\frac1{1-y}-\frac{x}{1-xy}\right) =\frac{75}{28},\\ T_{>} &=2\frac{x/2}{1-x/2}\frac1{1-xy} =\frac37. \end{aligned} \]

The final term of (3) is

\[ U=\frac1{1-x/4}\frac1{1-y/4}=\frac{240}{209}. \]

Therefore the reciprocal mass of the entire infinite semigroup is

\[ T_{\leq}+T_{>}-U =\frac{75}{28}+\frac37-\frac{240}{209} =\frac{11463}{5852} =2-\frac{241}{5852} <2. \tag{4} \]

[a]

Since every finite index set \({\cal I}(A_5^r)\) is contained in this semigroup,

\[ J(A_5^r)\leq\frac{11463}{5852}<2 \qquad(r\geq1). \]

The reciprocal-index criterion proves the theorem uniformly for every finite \(r\); there is no unproved limiting or finiteness step. [a, modulo the base classification in Step 3]

Exact computed table

The pure-Python checker multiplies the distinct base indices in (1) and sums with fractions.Fraction. [d]

| \(r\) | number of distinct subgroup indices | exact \(J(A_5^r)\) | |---:|---:|---:| | 1 | 9 | \(103/60\) | | 2 | 33 | \(3439/1800\) | | 3 | 82 | \(42113/21600\) | | 4 | 165 | \(8454647/4320000\) | | 5 | 291 | \(507640811/259200000\) | | 6 | 469 | \(22846919647/11664000000\) | | 7 | 708 | \(68542522337/34992000000\) | | 8 | 1017 | \(329005731837371/167961600000000\) | | 9 | 1405 | \(19740362730074663/10077696000000000\) | | 10 | 1881 | \(65801221273740761/33592320000000000\) |

These values increase toward, but remain below, the exact uniform bound \(11463/5852\). The proof uses the bound, not an extrapolation from this table. [a for the bound; d for the table]

For \(A_5^2\), the complete index set is

\[ \begin{split} \{&1,5,6,10,12,15,20,25,30,36,50,60,72,75,90,100,120,144,\\ &150,180,200,225,240,300,360,400,450,600,720,900,1200, 1800,3600\}. \end{split} \]

[d, also implied rigorously by the index-product lemma and (1)]

Reproduction and independent checks

Complete standalone code: erdos274_wavew040_verify.py

Run the standard-library-only certificate with:

python runs/erdos274_wavew040_verify.py

Run it together with the independent GAP check with:

python runs/erdos274_wavew040_verify.py --gap-cross-check

The executed combined check completed in about six seconds on this VM and reported:

|A5| = 60
subgroups of A5 = 59
subgroup counts by order =
  {1:1, 2:15, 3:10, 4:5, 5:6, 6:10, 10:6, 12:5, 60:1}
I(A5 x A5) has 33 indices
J(A5 x A5) = 3439/1800
full exponent-semigroup mass = 11463/5852
2 - full mass = 241/5852
GAP conjugacy classes of subgroups of A5 x A5 = 113
GAP J(A5 x A5) = 3439/1800
GAP CROSS-CHECK: PASS

The checker performs four genuinely separate checks:

  1. It constructs \(A_5\) from all even permutations and exhaustively enumerates

subgroups by repeatedly adjoining one element. This obtains all 59 subgroups and their orders without a group-theory database. [d]

  1. It checks the two-factor section data and the exact \(A_5^2\) index set.

[d]

  1. It verifies the semigroup characterization on a bounded exhaustive box,

evaluates (4) by exact rational arithmetic, and independently sums the box \(0\leq b,c\leq40\) term by term; the remaining difference from the closed form is \(8.25\times10^{-29}\). [d]

  1. Optional GAP 4.14 independently enumerates all 113 conjugacy classes of

subgroups of \(A_5^2\), recovering the same 33 indices and the same exact reciprocal sum. [d]

No computation exceeded a few CPU-seconds.

What this does and does not settle

The result proves an infinite, non-solvable family of finite groups, including arbitrarily large groups \(A_5^r\), but it does not settle Erdős problem 274 for arbitrary groups. [a]

The exact obstruction to extending this argument blindly is also clear. The criterion only sees the set of available subgroup indices. In broader families of direct products that reciprocal sum can be at least \(2\), and Garonzi–Margolis give direct-product families for which it is unbounded. Once \(J(G)\geq2\), index arithmetic alone cannot distinguish an Egyptian fraction from an actually disjoint family of cosets. A general proof then needs a structural lemma forbidding simultaneous disjoint realization of the candidate indices (a higher-order “harmonic subgroup” or intersection obstruction), not a larger version of the present enumeration. [a for the failure of the criterion; b for the unbounded family in Garonzi–Margolis]

PARTIAL: Proved, with a uniform exact bound and standalone verification, that every finite direct power \(A_5^r\) satisfies the Herzog–Schönheim conjecture; the general problem remains open.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger