Erdős Problem #274

Erdős Problem #274

1) ROUND-2 OBJECTIVE

Path (C), ending in a full proof of the original problem: prove the finite abelian version that appears in Erdős's cited source, and separate it from the stronger modern webpage version for arbitrary groups.

This is the right direction because Round-1 already established only a special case (finite $p$-groups), while the original finite abelian question is in fact fully settled by a known theorem.

2) Round-1 FOUNDATION USED

I use the following Round-1 facts.

1. Round-1 proved the statement for finite $p$-groups by a direct base-$p$ argument.
2. Round-1 identified the remaining gap correctly: the full arbitrary-group Herzog--Schönheim conjecture was not proved there.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a precise source correction.

Erdős explicitly wrote the question in the form:

Thus the original Erdős version is finite abelian, not arbitrary group.

The tool that closes this case is Sun's theorem on uniform covers by cosets of subnormal subgroups:

Since every subgroup of an abelian group is normal, hence subnormal, this theorem applies immediately.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was to pass from special cases to the full original finite abelian statement.

I will do that in three steps.

1. Restate the original finite abelian question precisely.
2. Invoke Sun's theorem in its exact form.
3. Verify all hypotheses: an exact partition is a nontrivial uniform cover of multiplicity $1$, and every subgroup in an abelian group is normal.

5) WORK (ROUND-2)

Original finite abelian statement.

Let $G$ be a finite abelian group. Suppose \[ G=\bigcup_{i=1}^k a_iH_i \qquad (k>1) \] is a disjoint union of left cosets of subgroups $H_i\le G$.

Write \[ n_i=[G:H_i]. \] Because $G$ is finite, \[ |a_iH_i|=|H_i|=\frac{|G|}{n_i}. \] Therefore the coset sizes are pairwise distinct if and only if the indices $n_i$ are pairwise distinct.

So the problem is equivalent to asking whether a partition of a finite abelian group into cosets can have pairwise distinct indices.

Sun's theorem.

Sun proved the following: if a group $G$ admits a nontrivial uniform cover \[ \{a_iG_i\}_{i=1}^k \] with finite indices \[ [G:G_1]\le \cdots \le [G:G_k], \] and if all the subgroups $G_i$ are subnormal in $G$, then the indices cannot be pairwise distinct.

Verification of the hypotheses.

Our partition \[ G=\bigcup_{i=1}^k a_iH_i \] is a uniform cover of multiplicity $1$: every element of $G$ lies in exactly one listed coset. Since $k>1$, it is nontrivial.

Because $G$ is abelian, every subgroup $H_i$ is normal in $G$, hence subnormal in $G$.

Also, $G$ is finite, so each index $[G:H_i]$ is finite.

Thus Sun's theorem applies directly and shows that the indices \[ [G:H_1],\dots,[G:H_k] \] cannot be pairwise distinct.

By the size-index relation above, the coset sizes cannot be pairwise distinct either.

Conclusion.

A finite abelian group cannot be partitioned into more than one coset of pairwise different sizes.

This proves the original Erdős version completely.

What remains open.

The stronger webpage version, \[ \mbox{``If $G$ is an arbitrary group, can there exist an exact covering by cosets of different sizes?''} \] is the general Herzog--Schönheim conjecture and remains open in full generality.

6) ADVERSARIAL VERIFICATION

Potential weak points and their resolution:

1. Finite versus infinite groups. The original Erdős source is finite abelian, and the proof above uses finiteness to identify size with $|G|/[G:H_i]$. So there is no hidden cardinal-arithmetic issue.
2. Does a partition count as a uniform cover? Yes. "Uniform cover" means every element is covered the same number of times; a partition is exactly the multiplicity-$1$ case.
3. Does subnormality hold? Yes. In an abelian group every subgroup is normal, hence automatically subnormal.
4. Did we accidentally prove the arbitrary-group statement? No. The final paragraph explicitly separates the solved finite abelian case from the still-open general-group version.

7) FINAL

FULL SOLUTION --- FULL PROOF.

The original Erdős problem (finite abelian groups) is solved negatively: such a partition into cosets of distinct sizes does not exist.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 100%

9) REFERENCES

1. P. Erdős, Some of my favourite problems in number theory. This source states the question for a finite abelian group.
2. T. F. Bloom, Erdős Problem #274. Current webpage statement and current status check.
3. Z.-W. Sun, On the Herzog--Schönheim conjecture for uniform covers of groups, J. Algebra 273 $2004$, 153--175.
4. Round-1 output for Problem #274 (complete proof for finite $p$-groups and gap identification for the general case).

Erdős Problem 274: a checked attempt

GPT_6_Astra_Ultra

25 September 2026

Statement and scope

Can a group be partitioned into more than one coset of pairwise different sizes? For infinite groups the standard Herzog--Schönheim formulation uses pairwise distinct subgroup indices. We study finite partitions $G=\bigsqcup_{i=1}^r a_iH_i$ with finite indices $d_i=[G:H_i]$ and show why distinct indices are impossible in several cases. The general assertion is unresolved here. Source: https://www.erdosproblems.com/274.

We extend the earlier GPT-5.2 prime-power counting argument, including a reduction that applies even when $G$ is infinite. The arguments for cyclic quotients and small numbers of parts are included in full; deeper results about all subnormal subgroups are not imported as supplied proofs.

Finite-index partitions reduce to finite quotients

The core $N_i=\bigcap_{g\in G}gH_ig^{-1}$ is the kernel of the action on $G/H_i$, so it is normal of finite index at most $d_i!$. Thus $N=\bigcap_iN_i$ is normal of finite index and lies in every $H_i$. Each coset $a_iH_i$ is a union of $N$-cosets. Passing to $G/N$ preserves the disjoint partition and every index $d_i$. Conversely a finite quotient's coset partition lifts to $G$ with the same indices.

In particular counting in this quotient proves \[ \sum_{i=1}^r\frac1{d_i}=1. \tag{1} \] In a nontrivial partition every $d_i\ge2$. This reduction requires finite indices; it makes no unsupported claim about a core of finite index for arbitrary infinite-index subgroups.

Prime-power indices and a coprimality obstruction

If all indices are powers of the same prime $p$, write $d_i=p^{e_i}$ and $E=\max e_i\ge1$. Multiplying $1$ by $p^E$ and reducing modulo $p$ shows \[ \#\{i:e_i=E\}\equiv0\pmod p. \] The largest index therefore occurs at least $p$ times. This includes every finite $p$-group, and also every infinite group partition whose finite indices are powers of one prime.

Two disjoint cosets cannot have coprime finite indices. It suffices to work in the finite quotient above. If $[G:H]=d$ and $[G:K]=e$ with $\gcd(d,e)=1$, then $[G:H\cap K]$ is divisible by both $d,e$ and is at most $de$. The upper bound follows from the injection of $G/(H\cap K)$ into $G/H\times G/K$. Thus $[G:H\cap K]=de$ and \[ |HK|=\frac{|H||K|}{|H\cap K|}=|G|. \] Consequently $HK=G$. For any left cosets $aH,bK$, write $a^{-1}b=hk$ with $h\in H,k\in K$; then $ah=bk^{-1}$ belongs to their intersection.

These facts settle partitions with two or three parts. Two distinct integers at least two cannot satisfy $1$ for two terms. For three distinct indices $d_1<d_2<d_3$, $1$ forces $d_1=2$ and then $d_2=3,d_3=6$: if $d_1\ge3$ the reciprocal sum is at most $1/3+1/4+1/5<1$, while after $d_1=2$, $d_2\ge4$ gives at most $1/4+1/5<1/2$. The candidate $(2,3,6)$ is excluded by the coprimality obstruction. This also illustrates why the Egyptian-fraction identity alone is insufficient.

All finite cyclic groups

For $G=\mathbb Z/N\mathbb Z$, an index-$d_i$ coset is precisely one residue class $a_i\pmod{d_i}$, where $d_i\mid N$ and $0\le a_i<d_i$. The partition identity is the polynomial equality \[ \sum_i z^{a_i}\sum_{j=0}^{N/d_i-1}z^{jd_i} =\sum_{j=0}^{N-1}z^j. \tag{2} \] Let $D=\max d_i>1$ and take a primitive $D$th root of unity $\zeta$. The right side vanishes. Every term with $d_i<D$ also vanishes, since $\zeta^{d_i}\ne1$ and $\zeta^N=1$. Terms with $d_i=D$ equal $(N/D)\zeta^{a_i}$. Thus $2$ implies \[ \sum_{i:d_i=D}\zeta^{a_i}=0. \] There must be at least two such terms. Hence the largest index repeats in every nontrivial cyclic coset partition. This also proves the assertion for finite partitions of $\mathbb Z$ into finite-index cosets by taking $N$ to be the least common multiple of the indices.

Remaining gap

The reductions and special cases above do not treat a general finite noncyclic group with mixed-prime indices and at least four parts. The full subgroup structure imposes constraints beyond $1$, but these elementary arguments do not show that one of the indices must repeat in that remaining case. No general resolution or improvement of the known subnormal case is claimed.

Completion rate estimate: 35%.

Problem #274

FORMAL RESTATEMENT

Let $G$ be a group and let $a_iH_i$ $(1\le i\le k)$ be left cosets of subgroups $H_i\le G$. An exact coset covering (or coset partition) is a finite family $\{a_iH_i\}_{i=1}^k$ such that \[ G \;=\; \bigsqcup_{i=1}^k a_iH_i \] (disjoint union).

Write $d_i=[G:H_i]$ for the (finite) index of $H_i$ in $G$; then $|a_iH_i|=|H_i|=|G|/d_i$ if $G$ is finite. The question asks whether a nontrivial coset partition can occur with all indices $d_i$ distinct (equivalently, all cosets have different sizes in the finite case).

Herzog--Schönheim conjecture (general form). If $\{a_iH_i\}_{i=1}^k$ is a coset partition of $G$ with all indices $[G:H_i]$ finite, then at least two of the indices are equal.

QUICK LITERATURE/CONTEXT CHECK

The conjecture was posed by Herzog and Schönheim $1974$. It is proved for finite nilpotent groups (hence for finite abelian groups) by Berger--Felzenbaum--Fraenkel $1986$. Many further partial results are known (e.g. for various classes of solvable/subnormal situations). As of 2025, the general conjecture for arbitrary groups remains open, but there has been recent progress for certain families (e.g. symmetric and nonabelian simple groups in a 2025 preprint by Garonzi--Margolis).

ATTACK PLAN

1. Prove a clean special case that is fully self-contained (here: $p$-groups) to illustrate the key obstruction.
2. Explain how this feeds into the known nilpotent/abelian results.
3. Summarize the remaining gap to the full conjecture and what kinds of tools seem relevant (group actions, representation theory, subgroup structure, normal series reductions).

WORK

A fully self-contained special case: finite $p$-groups.

Finite abelian (indeed finite nilpotent) groups.

A much stronger result is known:

Any coset partition of a finite nilpotent group into at least two cosets contains two cosets of the same index.

This was proved by Berger--Felzenbaum--Fraenkel $1986$ by translating the problem to a partition of a finite Cartesian product (a "parallelepiped") by "product sets" whose side lengths are prime powers, and then proving a combinatorial multiplicity theorem for such partitions.

Since every finite abelian group is nilpotent, this yields the statement quoted in the problem: in abelian groups one cannot have an exact coset covering by more than one coset of pairwise different sizes.

What remains open.

For general (possibly non-nilpotent, possibly infinite) groups the Herzog--Schönheim conjecture is still open in full generality. There are partial results for various classes of groups and subgroups; recent progress includes verification for broad families such as symmetric groups and nonabelian finite simple groups (as reported in a 2025 preprint by Garonzi--Margolis).

VERIFICATION

• The $p$-group proof (Theorem Theorem) is a complete standalone argument using only index counting and uniqueness of base-$p$ representation.
• The reduction "finite abelian $\Rightarrow$ finite nilpotent" is standard: finite abelian groups are nilpotent because they are direct products of their Sylow subgroups.
• The remaining open status for general groups is consistent with the current research literature (as of late 2025) which contains many partial cases but no general resolution.

FINAL: UNRESOLVED

The general Herzog--Schönheim conjecture for arbitrary groups is not resolved here. A complete proof is provided for the special case of finite $p$-groups, and the known (published) result for finite nilpotent groups is recorded, which covers all finite abelian groups.

COMPLETION ESTIMATE (honest)

A full resolution would require either:

• a general proof for all groups (likely via structural reductions to finite groups and then to composition factors, together with control of coset partitions under extensions/quotients), or
• an explicit counterexample: a group admitting a finite coset partition with all indices distinct.