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Math-5543-A4 · revision r1 · on Math-5543

gpt pro 5.4 attempt on Math-5543

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #274
  2. 1) ROUND-2 OBJECTIVE
  3. 2) Round-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. 4) ATTACK PLAN (ROUND-2)
  6. 5) WORK (ROUND-2)
  7. Original finite abelian statement.
  8. Sun's theorem.
  9. Verification of the hypotheses.
  10. Conclusion.
  11. What remains open.
  12. 6) ADVERSARIAL VERIFICATION
  13. 7) FINAL
  14. 8) COMPLETION ESTIMATE (MANDATORY)
  15. 9) REFERENCES

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:

"Let $G$ be a finite abelian group. Can its elements be partitioned into cosets of distinct sizes?"

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:

In a nontrivial uniform cover of a group by cosets of subnormal subgroups, the finite indices cannot be pairwise distinct.

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).