# Listing all the albelian groups of order 900

A past exam question says: Using the fundamental theorem (of finitely generated albelian groups), list all the abelian groups of order 900 where no two groups in your list should be isomorphic but any group of order 900 must be isomorphic to a group in your list.

The list I have so far is

$$\mathbb{Z}_{900}$$ $$\mathbb{Z}_{3} \oplus \mathbb{Z}_{300}$$ $$\mathbb{Z}_{5} \oplus \mathbb{Z}_{180}$$ $$\mathbb{Z}_{15} \oplus \mathbb{Z}_{60}$$ $$\mathbb{Z}_{30} \oplus \mathbb{Z}_{30}$$

I have a started with the method of writing down all the factors of 900 but making sure that the factors aren't relatively prime, this is because if they were they would be isomorphic to $\mathbb{Z}_{900}$. I'm pretty sure I haven't got all the groups yet but I was just checking; is my method right? Is there a systematic method that I'm missing to get all the groups? (Also I haven't actually used the fundamental theorem and can't actually see how to use it.)

Thanks!

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You are using the theorem when you say that your list is complete! Indeed if there was another one, then it would be... Anyhow the method you are using seems right! – Giovanni De Gaetano Apr 26 '12 at 10:35
The fundamental theorem usually includes the precision that the expression is unique if one demands that $d_1 | d_2 | \dots d_k$, that is, each order of the cyclic direct summands divide the next order. In particular, if you have three or more summands, the cube of the order of the smallest summand has to divide 900 in your case. – Phira Apr 26 '12 at 10:35
@Giovanni De Gaetano So how would you use the theorem? – user26069 Apr 26 '12 at 10:58

Nothing wrong with the answer by m_l. The other way to state the structure theorem is the one in Phira's comment. This gives you $C_{900}$, $C_2\times C_{450}$, $C_3\times C_{300}$, $C_5\times C_{180}$, $C_6\times C_{150}$, $C_{10}\times C_{90}$, $C_{15}\times C_{60}$, and $C_{30}\times C_{30}$.

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And you get from m_l's list to this one by merging coprime factors as in Nicky Hekster's comment. – lhf Apr 26 '12 at 13:32

The systematic approach is to write down the factorisation $900 = 2^23^25^2$. Then, for each prime factor, find all partitions of its exponent. Combine all these partitions to get the following list (I write $C_n$ instead of $\mathbb{Z}_n$ because I'm lazy):

$\begin{array}c C_4 \times C_9 \times C_{25} ( \cong C_{900} ) ,\\ C_2 \times C_2 \times C_9 \times C_{25} ,\\ C_4 \times C_3 \times C_3 \times C_{25} ,\\ C_4 \times C_9 \times C_{5} \times C_5 ,\\ C_2 \times C_2 \times C_3 \times C_3 \times C_{25} ,\\ C_2 \times C_2 \times C_9 \times C_{5} \times C_5 ,\\ C_4 \times C_3 \times C_3 \times C_{5} \times C_5 ,\\ C_2 \times C_2 \times C_3 \times C_3 \times C_{5} \times C_5. \end{array}$

The structure theorem for finitely generated abelian groups, also known as fundamental theorem, confirms that these are indeed all abelian groups of order $900$ and that they are pairwise non-isomorphic.

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What do you mean by 'pairwise' non-isomorphic? – user26069 Apr 26 '12 at 11:00
This means that no two of these are isomorphic to one another. – m_l Apr 26 '12 at 11:11
Ah OK, also, am I right in thinking that $C_6 \times C_{10} \times C_{15}$ is isomorphic to $C_2 \times C_2 \times C_3 \times C_3 \times C_{5} \times C_5$? – user26069 Apr 26 '12 at 11:20
Yes, correct, use $C_m \times C_n$ is isomorphic to $C_{mn}$ whenever $m$ and $n$ are relatively prime ($gcd(m,n)=1$). – Nicky Hekster Apr 26 '12 at 11:29