5
$\begingroup$

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

$\endgroup$
3
  • 4
    $\begingroup$ Where did you get that factorization? $256 = 2^8$. $\endgroup$ Apr 26, 2013 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, 2013 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, 2013 at 15:13

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Even solving multiple problems at once! :-) +1 $\endgroup$
    – Amzoti
    Apr 27, 2013 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, 2013 at 8:01
  • $\begingroup$ You're welcome! ;-) $\endgroup$
    – amWhy
    Apr 27, 2013 at 23:35
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.