Number of non-isomorphic abelian groups of a given fixed order

Question

Let G be a finite abelian group with |G| = $p_1^{n_1}...p_k^{n_k}$ in its prime factorized form.

Also, let $p(n)$ be the number of unique partitions of $n$, where we call $\sum_{i=1}^l k_i = n$ a partition of n with $k_1 \leqslant ... \leqslant k_l$ all positive integers.

Now, I want to find the number of possible groups which G might be isomorphic to. It seems likely that the fundamental theorem of finite abelian groups can be used here.

Along those lines, is it as simple as $\prod_{i=1}^l p(n_i)$, or am I missing something?

Answer 1

Your proposed formula is indeed correct. The numbers of finite abelian groups of fixed order are given by OEIS A000688.

Answer 2

You can also prove it by considering $|G|=p^n$, then the general case will follow repeatedly using the fact that $\Bbb Z_a\times\Bbb Z_b\cong\Bbb Z_{ab}\iff\gcd(a,b)=1$ (consider distinct primes). An abelian group of order $p^n$ may be written $$\Bbb Z_{p^{x_1}}\times \Bbb Z_{p^{x_2}}\times \dots\times \Bbb Z_{p^{x_k}}$$

With the requirement that $$x_1+x_2+\dots+x_k=n\tag{1}$$ In order to not have duplicates we can also require that $0\leq x_1\leq x_2\leq\dots\leq x_k$. The number of solutions to $(1)$ is now the partition-function, $p(n)$. (I've given a more detailed answer here)