All finite abelian groups of order 1024 List all finite abelian groups of order 1024.
Attempt: The prime decomposition of 1024 is 1024 = 2^10.
So 
Z_1024 = Z_2 x Z_512
   = Z_2 x Z_2 x Z_256

   = Z_2 x Z_2 x Z_2 x Z_128

   = Z_2 x Z_2 x Z_2 x Z_2 x Z_64

   = Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_32

   = Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_16

   = Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_8

   = Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_4

   = Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2 x Z_2

Can anyone please help me verify this are all the finite abelian groups of order 1024.
Thank you for the help.
 A: They're not. Essentially there are as many finitely many abelian groups of order $2^n$ as there are partitions of $n$. In other words, if you can write
$$
n = a_1 + \cdots + a_k, \quad k \ge 1, \quad a_i \ge 1
$$
Then the group
$$
\mathbb Z /2^{a_1}\mathbb Z \times \cdots \times \mathbb Z / 2^{a_k}\mathbb Z
$$
is obviously an abelian group, and it has order $2^{a_1} \cdots 2^{a_k} = 2^{a_1 + \cdots + a_k} = 2^n$. You can use induction on $k$ to show that "essentially distinct" partitions give rise to distinct groups (by essentially, I mean that we should consider $4 = 1+3 = 3+1$ as the same partition of $4$, where as $2+2 = 3+1$ are distinct ; if two partitions of $n$ are just a re-arrangement of the finite sum up to permuting terms, they give rise to the same abelian group). One way not to worry about unicity of the partition is to simply assume that $a_i \le a_{i+1}$. 
For those who know or are interested, this is a corollary of the classification of finitely generated $R$-modules when $R$ is a PID in terms of elementary divisors ; see Dummit & Foote's Abstract Algebra, Chapter 12.1.
If you want to prove any of this, feel free to ask for more help. 
Hope that helps,
