发表机构
The University of Melbourne（墨尔本大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文研究对称群与交错群被共轭类平铺的问题，证明除对换类外所有非单位共轭类均不能平铺对称群，且交错群中避开单位元的正规子集不能平铺。
AI 中文摘要
我们研究了对称群与交错群被共轭类平铺的问题。对于$S_n$，共轭类由对换构成的情形最早由Rothaus和Thompson于1966年研究，至今仍未解决。我们证明，这一长期未决的情形是唯一未解决的情形：$S_n$的每一个其他非单位共轭类都不能平铺$S_n$。我们的证明结合了表示论方法与关于置换乘积的组合论证。我们还以更强的形式完全解决了交错群的相应问题：对于$n\geq5$，$A_n$中任何避开单位元的正规子集都不能平铺$A_n$。
英文摘要
We study tilings of symmetric and alternating groups by conjugacy classes. For $S_n$, the case where the conjugacy class consists of transpositions was first investigated by Rothaus and Thompson in 1966 and remains open. We prove that this long-standing case is the only unresolved case for symmetric groups: every other nonidentity conjugacy class of $S_n$ fails to tile $S_n$. Our proofs combine representation-theoretic methods with combinatorial arguments concerning products of permutations. We also completely resolve the corresponding problem for alternating groups in a stronger form: for $n\geq5$, no normal subset of $A_n$ that avoids the identity can tile $A_n$. \iffalse We study the problem of whether a conjugacy class tiles the symmetric group or alternating group. For $S_n$, the case where the conjugacy class consists of transpositions was first investigated by Rothaus and Thompson in 1966 and remains open. We prove that this long-standing case is the only unresolved case for symmetric groups: every other nonidentity conjugacy class of $S_n$ fails to tile $S_n$. The proof combines representation-theoretic methods with combinatorial arguments concerning products of permutations. We also completely resolve the corresponding problem for alternating groups in a stronger form: for $n\geq5$, no normal subset of $A_n$ that avoids the identity can tile $A_n$. \fi \iffalse In this paper, we study when a conjugacy class or a normal subset tiles the symmetric or alternating group. The case of transpositions was first investigated by Rothaus and Thompson, and in general, it remains open. We show that every other conjugacy class of a symmetric group does not tile. We also prove that normal subsets in $A_n$ ($n\ge 5)$ not containing the identity element cannot tile.
