# Set of non-isomorphic abelian groups of a large order

Define a set

$A:=\{G: G$ is abelian group; $\operatorname{order}(G)=10.000$, no two groups are isomorphic $\}$.

What is the largest size of $A$?

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Hint: How many abelian groups of order $p^4$ exist? ($p$ is a prime number) Finally use the decomposition in primary components. –  Andrea Aug 15 '11 at 17:56
There is only two abelian groups of order $p^{4}$, namely, $\mathbb Z_{p}\times \mathbb Z_{p}\times \mathbb Z_{p}\times \mathbb Z_{p}$, and $\mathbb Z_{p^{4}}$. So, if $|G|=2^{4}5^{4}$, then $G$ is isomorphic to $$$\mathbb Z_{2}$$^{4}\times$$\mathbb Z_{5}$$^{4}$, $\mathbb Z_{2^{4}}\times$$\mathbb Z_{5}$$^{4}$, $$$\mathbb Z_{2}$$^{4}\times\mathbb Z_{5^{4}}$, or $\mathbb Z_{2^{4}}\times\mathbb Z_{5^{4}}$. –  William Aug 15 '11 at 18:00
Oh I forgot the cases $\mathbb Z_{p^{2}}\times \mathbb Z_{p^{2}}$, $\mathbb Z_{p^{2}}\times \mathbb Z_{p}\times \mathbb Z_{p}$, and $\mathbb Z_{p^{3}}\times \mathbb Z_{p}$, so we can proceed as above! –  William Aug 15 '11 at 18:08
@William: What happened to $\mathbb{Z}_{p^2}\times\mathbb{Z}_{p^2}$, and to $\mathbb{Z}_{p^2}\times\mathbb{Z}_p\times \mathbb{Z}_p$? –  Arturo Magidin Aug 15 '11 at 18:12
Largest size of $A$? Does $A$ have a smaller size? Or more than one size for that matter? –  lhf Aug 15 '11 at 23:49

This is an easy application of the Fundamental Theorem of Finitely Generated Abelian Groups. Since $|G|=2^4\times 5^4$, then $G\cong G_2\times G_5$, where $G_p$ is the $p$-part of $G$.
Thus, the problem reduces to asking how many different abelian groups of order $p^4$ there are. Since an abelian $p$-group must be a direct sum/product of cyclic groups of prime power order, the answer is that there are as many abelian groups of order $p^n$ as there are partitions of $n$.
So there are as many possibilities for $G_2$ as there are partitions of $4$, and also as many possibilities for $G_5$ as there are partitions of $4$, and hence the number will be the product of these two, the square of the number of partitions of $4$.
$4=1+1+1+,2+2,1+3,4$, so we have $\#$$A$$=4.4=16$. –  William Aug 15 '11 at 18:20
@William: And $2+1+1$ doesn't count, then? (By the way, it's better to use either \times or \cdot for multiplication of numbers; otherwise, it looks like you are saying that $\frac{44}{10}=16$). –  Arturo Magidin Aug 15 '11 at 18:24
Sure, we will have 2+1+1- I already mentioned all of them above in my first and second comments-typo!, so its $25$ not $16$. –  William Aug 15 '11 at 18:32