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I came across this while studying:

How many abelian groups are there of order $p^5q^4$?

I know that if I was asked how many abelian groups of order $128$ there were, I could just say $128=2^7$ and I would just look at the number of integer partitions of $7$ (which is $15$).These $15$ groups would be isomorphic to the groups $\mathbb{Z_{128}}$,$\mathbb{Z_2} \oplus \mathbb{Z_{64}} , \mathbb{Z_2} \oplus \mathbb{Z_2} \oplus \mathbb {Z_{32}},$... and so on. In this instance, is there a similar process?

Thanks!

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    $\begingroup$ Surely you mean number of non-isomorphic groups. $\endgroup$
    – quid
    Dec 10, 2015 at 23:35

1 Answer 1

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Recall that an abelian group can be written as the direct sum of its maximal subgroups of prime power order.

Thus every group of order $p^5q^4$ is a direct sum of a group of order $p^5$ and one of order $q^4$.

The number of groups you seek is thus the number of abelian groups of order $p^5$ times the number of abelian groups of order $q^4$.

That is, as you said, the number of integer partitions of $5$ times the number of integer partitions of $4$.

That is of course assuming $p$ and $q$ are distinct primes. If they are the same prime you are just looking for groups of order $p^9$, and this gives integer partitions of $9$ as answer.

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  • $\begingroup$ Glad it was helpful. There is no rush with accepting. $\endgroup$
    – quid
    Dec 10, 2015 at 23:37

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