Let $R$ be a principal ideal domain, $p \in R$ a prime element and $M$ a finitely generated $p$-torsion module of the form $M = R/(p^{e_1}) \oplus \cdots \oplus R/(p^{e_t})$.
Let $_pM = \{m \in M: p m = 0\}$.
Using the fact that $_pM$ is a $R/(p)$-vector space of dimension $t$, I now want to determine all abelian groups with 625 elements except for isomorphism, that have exactly 24 elements of order 5. (Where the order of an element $m$ in an abelian group $G$ is defined as the smallest $n \in \mathbb{N}$, so that $n m = 0$, and $\infty$, in case that such an $n \in \mathbb{N}$ does not exist.)
Thanks in advance. I'm not very used to working with these constructions.
$\newcommand{\Z}{\mathbb{Z}}$You want your group $M$ to be $$ M = \Z/(5^{e_1}) \oplus \cdots \oplus \Z/(5^{e_t})$, $$ with $e_{1} \ge e_{2} \ge \dots \ge e_{t} > 0$, and $\sum e_{i} = 4$, so that $M$ has order $5^{4} = 625$. But then note that each summand $\Z/(5^{e_i})$ contributes a factor $5$ to ${}_{5}M$. Since you want ${}_{5}M$ to have order $25$ ($24$ elements of order $5$, plus zero), you must have $t = 2$, and now you have to solve a simple counting problem.