Comparison of no. of abelian groups of order $p^r$ with that of order $q^r$

Question

[J.B. Fraleigh, Exercises 11, problem 37] Let $p$ and $q$ be distinct prime numbers. How does the number of abelian groups of order $p^r$ compare with the number of abelian groups of order $q^r$ ?

For instance, I take $p = 2$, $q = 3$ and $r=3$. Then, for $p = 2, r = 3$, we have $\mathbb Z_8$, $\mathbb Z_4 × \mathbb Z_2$ and $\mathbb Z_2 × \mathbb Z_2 × \mathbb Z_2$. Similarly for $q = 3, r = 3$, we have $\mathbb Z_{27}$, $\mathbb Z_9 × \mathbb Z_3$ and $\mathbb Z_3 × \mathbb Z_3 × \mathbb Z_3$.

It seems that they have same no. of abelian groups, since it depends on $r$. But I'm not sure if I'm correct. Any help or hint would be appreciated.

Answer 1

According to the Fundamental Theorem of Finite Abelian Groups, these must be equal.

See https://proofwiki.org/wiki/Fundamental_Theorem_of_Finite_Abelian_Groups

Answer 2

Yes, the classification theorem for finite abelian groups tells you that those of order $p^r$ have sum decomposition into $p$-power cyclic groups. You can find a corresponding group of order $q^r$ by taking each such summand and replace just the base, i.e., we have a bijection $$ \bigoplus_{i}\Bbb Z/p^{n_i}\Bbb Z\mapsto \bigoplus_{i}\Bbb Z/q^{n_i}\Bbb Z$$ (where $\sum n_i=r$ and we e.g. normalize by requiring $0<n_1\le n_2\le \ldots $)

Answer 3

The number of such groups equals the number of partitions of $r$, $p(r)$, by FTFAG (in both cases).