25阶初等阿贝尔群的5-可整除整数群行列式
The $5$-divisible integer group determinants for the elementary abelian group of order 25
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中文总结 AI 辅助
本文针对25阶初等阿贝尔群C₅×C₅，证明其5-可整除群行列式的反向包含关系，结合已有结果完成25阶群的Taussky-Todd整数群行列式问题分类。
中文摘要 AI 辅助
设G=C₅×C₅，S(G)为其群行列式的整数值集合。已有研究确定S(G)中与5互素的数值，并证明所有5-可整除数值都被5⁸整除。本文证明反向包含关系5⁸ℤ⊆S(G)，因此S(G)={m∈ℤ:m≡±1或±7 mod25}∪5⁸ℤ。证明采用一般移位准则及三个显式多项式，其群行列式分别为5⁸、2·5⁸和5⁹。结合已知的C₂₅分类结果，这完成了所有25阶群的Taussky-Todd整数群行列式问题。
英文摘要
Let $G=C_5\times C_5$, and let $S(G)$ denote the set of integer values of its group determinant. Previous work determines the values in $S(G)$ coprime to $5$ and proves that every $5$-divisible value is divisible by $5^8$. We prove the converse inclusion $5^8\mathbb{Z}\subseteq S(G)$. Consequently, $S(G)=\{m\in\mathbb{Z}:m\equiv\pm1\text{ or }\pm7\pmod{25}\}\cup5^8\mathbb{Z}$. The proof uses a general shift criterion and three explicit polynomials whose group determinants are $5^8$, $2\cdot5^8$, and $5^9$. Together with the known classification for $C_{25}$, this completes the Taussky--Todd integer group determinant problem for all groups of order $25$.
