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

[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.

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

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

• What does $FTFAG$ mean ?? Aug 27, 2018 at 19:37
• Fundamental Theorem of Finite Abelian Groups.
– user403337
Aug 27, 2018 at 19:39
• Ohhhh, I get it. Thanks !! Aug 27, 2018 at 19:39

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 $0<n_1\le n_2\le \ldots$)