Smallest $n$ for non-isomorphic groups for external product of groups

Question

I was solving the following question from the book Contemporary Abstract Algebra by Joseph A. Gallian:

What is the smallest positive integer n such that there are exactly four nonisomorphic Abelian groups of order n? Name the four groups.

According to me I came up with the number $n=16$, where I get following four non-isomorphic groups and all are abelian too

$$Z_{16}, Z_4 \oplus_4, Z_8\oplus Z_2, Z_2\oplus Z_2\oplus Z_2\oplus Z_2$$.

But in the book answer is $n=36$. Where I am going wrong. Kindly help.

Answer 1

The number of different isomorphic abelian groups of a size $N=p_1^{e_1}\cdots p_n^{e_n}$ where the $p_i$ are distinct primes is given by $P(e_1)\cdot P(e_2) \cdots P(e_n)$ where $P(i)$ is the number of partitions of $i$.

Thus, you are looking for the smallest number so that you can get 4 from these partitions. P(1)=1, $P(2)=2$ since you can have 1+1 or 2, P(3)=3, P(4)=5, and from there they are larger. Thus you want to use groups of size two distinct prime factors each occurring twice. This means the smallest should be of size $2^2\cdot 3^2=36$.