发表机构
Department of Mathematics and Statistics, St. Xavier’s College（圣泽维尔学院数学与统计系）
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中元素阶之和的不等式，通过奇平方根计数将其转化为负相关，并排除两种自然方法，计算验证至 n=60。
AI 中文摘要
对于有限群 $G$，令 $\psi(G)$ 为其元素阶之和。商群 $S_n/A_n$ 的阶为 2，因此 $\psi(S_n/A_n)=3$。因此，与 $\psi(S_n)$ 比较的量是 $3\psi(A_n)$。计算表明，对于所有 $n\ge3$，有 $3\psi(A_n)>\psi(S_n)$。我们将此不等式归结为平方根奇偶性的问题。设 $r(\beta)$ 为满足 $\sigma^{2}=\beta$ 的奇置换 $\sigma$ 的个数。我们证明奇置换的阶之和等于 $2\sum_{\beta\in A_n}r(\beta)o(\beta)$。我们还证明 $\sum_{\beta\in A_n}r(\beta)$ 等于 $|A_n|$。该不等式表明 $r$ 与阶函数在 $A_n$ 上呈负相关，等价地说，奇置换的平均阶小于偶置换平均阶的两倍。然后我们排除了两种自然的方法：不存在从奇置换到 $A_n$ 的注入能在每一点将阶减半，最小的失败发生在 $n=12$；通过 Landau 函数的阈值论证也失败。报告了直到 $n=60$ 的计算结果。
英文摘要
For a finite group $G$ let $ψ(G)$ be the sum of the orders of its elements. The quotient $S_n/A_n$ has order two. Hence $ψ(S_n/A_n)=3$. The quantity to be compared with $ψ(S_n)$ is therefore $3ψ(A_n)$. Computation indicates that $3ψ(A_n)>ψ(S_n)$ for every $n\ge3$. We reduce this inequality to a question on the parity of square roots. Let $r(β)$ be the number of odd permutations $σ$ with $σ^{2}=β$. We show that the sum of the orders of the odd permutations equals $2\sum_{β\in A_n}r(β)o(β)$. We also show that $\sum_{β\in A_n}r(β)$ equals $|A_n|$. The inequality then says that $r$ and the order function are negatively correlated on $A_n$. It says equally that the average order of an odd permutation is less than twice the average order of an even one. We then rule out two natural approaches. No injection from the odd permutations into $A_n$ can halve the order at every point. The smallest failure occurs at $n=12$. A threshold argument through Landau's function fails as well. Computations up to $n=60$ are reported.
