Number of Abelian Groups of Order 256 I am trying to find the number of abelian groups of order 256. Is the following correct?
We may write $256=2^8$ we then know that this may be represented in the form:
$C_{n_1}\times.....\times C_{n_s}$ where $n_i|n_{i+1}$ and $n_1.....n_s=|G|$ So this may be represented as:
$C_2\times C_2\times C_2\times C_2\times C_2\times C_2\times C_2\times C_2, C_2\times C_2\times C_2\times C_2 \times C_2\times C_2 \times C_4.....$
Which I belive comes to 22 groups if you keep going in this way, is there a faster way?
Cheers
 A: You can find the number of abelian groups of order $256$ by partitioning $8$: that is, find the number of ways you can sum positive integers to equal $8$: If we partition the number 8, we'll see that there are $22$ distinct ways to sum positive integers to equal $8$ , and hence $22$ abelian groups of order $256$.
Prior to Edit of question:
I'm afraid your prime factorization is off: $256 = 2^8$.
If you meant to find abelian groups of order $252 = 2^2 \times 3^2 \times 7$, yes, then there are four distinct abelian groups of order $252$: and your list is correct, except for a missing factor of $C_3$ in your third group.
A: $2^4:$ $$G_1=C_2\times C_2\times C_2\times C_2$$ $$G_2=C_4\times C_4$$ $$G_3=C_8\times C_2$$ $$G_4=C_{16}$$ $3^2:$ $$G_5=C_3\times C_3$$ $$G_6=C_9$$
$7:$ $$G_7=C_7$$ Now try to built the desired abelian groups of that order. Here I assume $|G|=2^4\times 3^2\times 7$. Some of them are $$G=G_1\times G_5\times G_7\cong C_{14}\times C_6\times C_{6}\times C_2$$ $$G=G_1\times G_6\times G_7\cong C_{18}\times C_6\times C_{14}\times C_2\times C_2$$ $$G=G_2\times G_5\times G_7\cong C_{12}\times C_{12}\times C_7\cong C_{12}\times C_{84}$$ $$\vdots$$
