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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

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    $\begingroup$ Where did you get that factorization? $256 = 2^8$. $\endgroup$ – Tobias Kildetoft Apr 26 '13 at 15:11
  • $\begingroup$ Your calculation suggests you were looking at $252 = 2^2 \times 3^2 \times 7$ instead? $\endgroup$ – Lord_Farin Apr 26 '13 at 15:13
  • $\begingroup$ @TobiasKildetoft Haha I have no idea that is pretty horrendous, I'll edit it now and apply the same method, thanks! $\endgroup$ – user73957 Apr 26 '13 at 15:13
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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.

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  • $\begingroup$ Even solving multiple problems at once! :-) +1 $\endgroup$ – Amzoti Apr 27 '13 at 0:15
  • $\begingroup$ Thanks for the answer, sorry about my failed factorisation at the start but you answered both questions which was helpful as well so thanks! :) $\endgroup$ – user73957 Apr 27 '13 at 8:01
  • $\begingroup$ You're welcome! ;-) $\endgroup$ – Namaste Apr 27 '13 at 23:35
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$2^4:$ $$G_1=C_2\times C_2\times C_2\times C_2$$ $$G_2=C_4\times C_4$$ $$G_3=C_8\times C_2$$ $$G_4=C_{16}$$ $3^2:$ $$G_5=C_3\times C_3$$ $$G_6=C_9$$ $7:$ $$G_7=C_7$$ 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 $$G=G_1\times G_5\times G_7\cong C_{14}\times C_6\times C_{6}\times C_2$$ $$G=G_1\times G_6\times G_7\cong C_{18}\times C_6\times C_{14}\times C_2\times C_2$$ $$G=G_2\times G_5\times G_7\cong C_{12}\times C_{12}\times C_7\cong C_{12}\times C_{84}$$ $$\vdots$$

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