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Classify the abelian groups of order n, 2n, 4n

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What have you tried? –  egreg May 12 '13 at 9:46
    
i tried to write 81 = 3^4, 144 = (3^2).(2^4), 216 = (2^3).(3^4) but i don't where to to go next –  user77456 May 12 '13 at 9:54
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Here, you will want to use the theorem of structure of finite abelian groups, that is: Every abelian finite group is isomorphic to a group of the form:

$$\mathbb{Z}_{m_1}\times \mathbb{Z}_{m_2} \times\mathbb{Z}_{m_3} \times... \times\mathbb{Z}_{m_n}$$

In which $m_i$ divides $m_{i-1}$, and abviously the product of all $m_i$ has to be the order of the group.

So you have to find possible combinations of numbers $m_i$ that meet the above conditions. For example $81=3^4$, just find the combinations.

Now, a further result that will allow you to not repeat any group is this one:

If $G$ is a group of the form $G=G_1\times G_2$, then both of the next things are equivalent

  1. G is cyclic
  2. $\gcd(|G_1|,|G_2|)=1$

This means for example, if we have groups of order 18, then the group $\mathbb{Z}_2 \times \mathbb{Z}_9$ will be the same as $\mathbb{Z}_{18}$

With this results you should be able to classify all the abelian groups of those orders.

Sadly, I've studied this from books (really good books), written in spanish and I can't give references to the proves of these results in english, but I suppose that you will find them in any elementary group theory book. Maybe someone wants to comment with some references.

P.S.: As you're not directly asking anything, I haven't explicitly solved the exercise that looks a lot like homework. If you have any quetion regarding some of the results you don't understand, then ask it.

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For references in english, I think Lang or Dummit & Foote would have this, and much more. –  Jean-Claude Arbaut May 12 '13 at 9:39
    
@arbautjc Thank you for completing the answer –  MyUserIsThis May 12 '13 at 9:40
    
thanks for your response but you said mi divides mi−1 but 2 doesn't divide 9 as in Z2 x Z9 –  user77456 May 12 '13 at 9:50
    
The "combinations" of positive integers you mention are called "partitions" of a natural number:en.wikipedia.org/wiki/Partition_(number_theory) –  DonAntonio May 12 '13 at 10:13
    
@Theo You're right, but that just means you still can reduce it to a combination of that kind, in this case the cyclic group $\mathbb{Z}_{18}$ The second proposition is actually contained in the first one. –  MyUserIsThis May 12 '13 at 11:05
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