TITLE: Counting Groups
QUESTION [5 upvotes]: My task is to determine the number of Abelian groups of order $p_{1}^{4}p_{2}^{4}...p_{n}^{4}$, where each $p_{n}$ is a distinct prime.
My attempt:
$\forall p$, there are 5 possible non isomorphic groups of order $p^4$, namely $Z_{p^4}$, $Z_{p^3}\oplus Z$, $Z_{p^2}\oplus Z_{p^2}$, $Z_{p^2}\oplus Z_{p} \oplus Z_{p}$, and $Z_{p} \oplus Z_{p}\oplus Z_{p} \oplus Z_{p}$. We then can form a group of order $p_{1}^{4}p_{2}^{4}...p_{n}^{4}$ by choosing one of these groups for each $p_{n}$, and then forming the  direct product of those choices.  So is it $5^n$ possible groups? I sense that there are redundancies I'm not taking into account.

REPLY [3 votes]: For every $p$, there are 5 possible non isomorphic groups of order $p^4$, namely $Z_{p^4}$, $Z_{p^3}\oplus Z$, $Z_{p^2}\oplus Z_{p^2}$, $Z_{p^2}\oplus Z_{p} \oplus Z_{p}$, and $Z_{p} \oplus Z_{p}\oplus Z_{p} \oplus Z_{p}$. We then can form a group of order $p_{1}^{4}p_{2}^{4}...p_{n}^{4}$ by choosing one of these groups for each $p_{n}$, and then forming the  direct product of those choices.  Therefore, because there are 5 such choices for each $p_{n}$, there are $5^n$ possible direct products, and thus $5^n$ distinct (up to isomorphism) Abelian groups of order $p_{1}^{4}p_{2}^{4}...p_{n}^{4}$.