TITLE: Number of Abelian Groups of Order 256
QUESTION [5 upvotes]: 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

REPLY [2 votes]: $2^4:$ $$G_1=C_2\times C_2\times C_2\times C_2$$ $$G_2=C_4\times C_4$$ $$G_3=C_8\times C_2$$ $$G_4=C_{16}$$ $3^2:$ $$G_5=C_3\times C_3$$ $$G_6=C_9$$
$7:$ $$G_7=C_7$$ 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 $$G=G_1\times G_5\times G_7\cong C_{14}\times C_6\times C_{6}\times C_2$$ $$G=G_1\times G_6\times G_7\cong C_{18}\times C_6\times C_{14}\times C_2\times C_2$$ $$G=G_2\times G_5\times G_7\cong C_{12}\times C_{12}\times C_7\cong C_{12}\times C_{84}$$ $$\vdots$$