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let p and q be distinct prime numbers. how does the number (up to isomorphism) of abelian groups of order p^r compare with the number (up to isomorphism) of abelian groups of order q^r?

I am just not sure where to start here.

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    $\begingroup$ If you look at the structure of such groups, you will see that the numbers are the same. $\endgroup$ – André Nicolas Sep 26 '14 at 6:37
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For an abelian group $G$ of order $p^r$ there exists a unique partition of $r$ $$r_1\ge r_2 \ge \ldots \ge r_m\\ r_1 + \ldots + r_m = r $$ so that $$G \simeq \mathbb{Z}/p^{r_1}\times \ldots \times \mathbb{Z}/p^{r_m}$$

Therefore the number of abelian groups of order $p^r$ up to isomorphism equals the number of partitions of $r$.

The same thing holds for the number of abelian groups of order $q^r$ up to isomorphism.

Hence these numbers are equal.

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Hint: both will be equal as power of primes are same.

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