Abelian groups of order 96

Question

I'm not sure how to figure out systematically the number of abelian groups of a certain order. $96 = 2^5 \times 3$, and is equivalent to $Z/32Z \times Z/3Z$, then you can have $Z/4Z \times Z/8Z \times Z/3Z$ and then group the $2s$ in different ways, but I struggle to mix and match the powers of $2$ system in order to find out how many total abelian groups there are.

Answer 1

The Fundamental Theorem of Finite Abelian Groups states that every finite Abelian group is a direct product of cyclic groups of prime-power order. If a finite Abelian group $G$ has order $|G|=p_1^{n_1}p_2^{n_2}...p_t^{n_t}$ where $p_1,...,p_t$ are distinct primes then you can write $G$ as the direct product

$$ G = G_1 \times G_2 \times ... \times G_t$$

where each $G_i$ has prime-power order, $|G_i|=p_i^{n_i}$. Then each $G_i$ is a direct product of cyclic groups.

$$ G_i \cong Z_{p_i^{a_1}}\times Z_{p_i^{a_2}}\times ... \times Z_{p_i^{a_s}}.$$

where $\sum a_i = n_i$. To count the number of isomorphism classes of $G_i$, it is enough to count the number of integer partitions of $n_i$. In general, there is no formula for this, but you can do it by hand. For example, the following lists all the isomorphism classes of Abelian groups of order $32=2^5$ corresponding to the 7 partitions of the integer $5$.

$$ \begin{array}{rl} 5 & Z_{32}\\ 4+1& Z_{16}\times Z_2 \\ 3+2& Z_{8}\times Z_4 \\ 3+1+1& Z_{8}\times Z_2 \times Z_2\\ 2+2+1& Z_{4}\times Z_4 \times Z_2 \\ 2+1+1+1& Z_{4}\times Z_2 \times Z_2\times Z_2\\ 1+1+1+1+1& Z_2 \times Z_{2}\times Z_2 \times Z_2\times Z_2 \end{array} $$

Let $P(n)$ denote the number of partitions of the integer $n$. Then each $G_i$ has $P(n_i)$ isomorphism classes. To find the number of isomorphism classes of the group $G$, we should multiply the number of isomorphism classes of $G_1,G_2,...,G_t$. Thus it is the product $P(n_1)\cdot P(n_2) \cdot ... \cdot P(n_t)$.

In the case that $|G|=96=2^5 \cdot 3$, there are $P(5)\cdot P(1) = 7 \cdot 1$ isomorphism classes.