The $5$-divisible integer group determinants for the elementary abelian group of order 25

Abstract

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 $25$.