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

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[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.

Best Answer

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

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

Related Solutions

[Math] Find the smallest positive integer n such that there are exactly four non-isomorphic abelian groups of order n

The number of non-isomorphic Abelian groups depends on the partition number of the exponents in the prime factorization. Let $P(k)$ be the number of ways to partition $k$ (for example, $P(4) = 5$, since we have $4, 3+1, 2+2, 2+1+1$, and $1+1+1+1$). Then for $n=p_1^{e_1} \ldots p_r^{e_r}$, there are $$\prod_i P(e_i)$$ non-isomorphic Abelian groups of order $n$. You missed $2+2$ among the ways to partition $4$, corresponding to $\mathbb Z/(2^2\mathbb Z) \times \mathbb Z/(2^2\mathbb Z) = \mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$.

You can use this formula to answer the original question. The first few partition numbers are:

P(1) = 1    1
P(2) = 2    2, 1+1
P(3) = 3    3, 2+1, 1+1+1
P(4) = 5    4, 3+1, 2+2, 2+1+1, 1+1+1+1
P(5) = 7    5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1

and they obviously increase from here. Now, if there are $4$ Abelian groups of order $n$, then we must have $e_1=e_2=2$, and we choose as bases the two smallest primes. Thus the number we are looking for is $2^23^2=36$.

Classifying abelian groups of order $56$.

Must be isomorphic to one of the other because 4 doesn't divide 14

Best Answer

Related Solutions

[Math] Find the smallest positive integer n such that there are exactly four non-isomorphic abelian groups of order n

Classifying abelian groups of order $56$.

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