Solution: Before proceeding for a solution to the problem, let us see the number of non-isomorphic abelian group of order $p^4$, where $p$ is a prime. We consider all partitions of $4$. Each partition will give an abelian group of order $p^5$. We have \begin{align*} & 4 & & \mathbb{Z} _{p^4} \\ & 3 + 1 & & \mathbb{Z} _{p^3} \oplus \mathbb{Z} _p \\ & 2 + 2 & & \mathbb{Z} _{p^2} \oplus \mathbb{Z} _{p^2} \\ & 2 + 1 + 1 & & \mathbb{Z} _{p^2} \oplus \mathbb{Z} _{p} \oplus \mathbb{Z} _{p} \\ & 1 + 1 + 1 + 1 & & \mathbb{Z} _{p} \oplus \mathbb{Z} _{p} \oplus \mathbb{Z} _{p} \mathbb{Z} _{p} . \end{align*} So there are $5$ non-isomorphic abelian groups of order $p^4$.

In order to determine the number of non-isomorphic groups of order $2^2 \cdot 3^3 \cdot 5^4$, we consider the partition of each powers. So we have

• The number of non-isomorphic abelian groups of order $2^2$ is parton of $2$, that is, $2$.
• The number of non-isomorphic abelian groups of order $3^3$ is parton of $3$, that is, $3$.
• The number of non-isomorphic abelian groups of order $5^4$ is parton of $4$, that is, $5$.

Therefore, total number of non-isomorphic abelian group of order $2^2\cdot 3^3\cdot 5^4$ is $30$.