Let $N$ be the order of an abelian group. The prime factorization is given by $N=\prod_{i=1}^{n}p_{i}^{e_{i}}$ with $p_{1}< p_{2}< \dots <p_{n}$ and $e_{i}\geq 1$. Let $\pi(n)$ denotes the number of partitions of an integer $n$.

Show that the number of isomorphism classes of abealian groups of order $N$ is equal to $\prod_{i=1}^{n} \pi (e_{i})$.

Good, i have managed to show that for order $N=p^k$ the number of isomorphism classes is equal to $\pi(k)$. In this case $e_{1}=k$. How to show the statement for any integer $N$ that can be written as a product of prime powers? Can anybody help me with this problem, please? Thank you very much!


You're almost there. As you've noticed all the groups of order $N=p^n$ is on the form: $$\mathbb{Z}_{p^{a_1}}\times \mathbb{Z}_{p^{a_2}}\times\dots\times \mathbb{Z}_{p^{a_n}}$$ where $0\leq a_1\leq a_2\leq\dots\leq a_n$ are non-negative integers, and $$a_1+a_2+\dots+a_n=n\tag{1}$$ The number of solutions to $(1)$ is the partition function $\pi(n)$, so there are $\pi(n)$ isomorphism classes of order $p^n$.

Suppose that $N=p_1^{e_1}p_2^{e_2}\cdot\ldots\cdot p_k^{e_k}$, then repeatedly using $\mathbb{Z}_{ab}\cong\mathbb{Z}_a\times\mathbb{Z}_b\iff\gcd(a,b)=1$ we can write any group of order $N$ as $$H_1\times H_2\times\dots\times H_k$$

where $H_i=\mathbb{Z}_{p_i^{a_{1,i}}}\times \mathbb{Z}_{p_i^{a_{2,i}}}\times\dots\times \mathbb{Z}_{p_i^{a_{e_i,i}}}$ with the condition $a_{1,i}+a_{2,i}+\dots +a_{e_i,i}=e_i$. There are $\pi(e_i)$ possibilities for each $H_i$, and $H_i\cong H_j\Rightarrow i=j$ by considering the order of $H_i,H_k$, so in total there are $$\prod_{i=1}^{k}\pi(e_i)$$ isomorphism classes of order $N$.


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