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I am currently trying to find all abelian groups of order $4900$ up to isomorphism and I wanted to apply the fundamental theorem of finite abelian groups in the form "there are $d_i$, s.t. $d_i \mid d_{i+1}$ and $G \cong \Bbb Z_{d_1} \times\Bbb Z_{d_2} \times ... \times \Bbb Z_{d_s}$.

Now it is clear that $4900 = 2^2 \cdot 5^2 \cdot 7^2$. I have found the isomorphic types:

$$\begin{align} G& \cong\Bbb Z_{4900}, \\ G &\cong\Bbb Z_{2} \times \Bbb Z_{2450}, \\ G &\cong\Bbb Z_{5} \times\Bbb Z_{980}, \\ G &\cong\Bbb Z_{7} \times\Bbb Z_{700}. \end{align}$$

I don't know whether there are more - is there any way to definetly answer this in some kind of algorithmic way?

Shaun
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max_121
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    If $m$ and $n$ are coprime, do you know how to assemble the list of all abelian groups of order $mn$ (in invariant factor form as you have described) from the lists of all abelian groups of order $m$ and all abelian groups of order $n$? Between that and the fact that listing all abelian groups of order $p^2$ ($p$ prime) is simple, that provides an algorithm. Hint: you have found four of the eight such groups. – Greg Martin Jul 24 '21 at 21:50
  • In my course we showed that if m and n are coprime, than Z_m x Z_n is isomorphic to Z_mn. However I don't know how to list them – max_121 Jul 24 '21 at 21:53
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    There is no triple $d_1\mid d_2\mid d_3$ with $d_1d_2d_3=4900,$ because if $p\mid d_1,$ then $p^3\mid 4900.$ But there are many more pairs $d_1d_2.$ – Thomas Andrews Jul 24 '21 at 22:00
  • I just found the remaining types, they should be $Z_{10}\times Z_{490}$, $Z_{14}\times Z_{350}$, $Z_35\times Z_{140}$, $Z_{70}\times Z_{70}$. However, I still don't understand how to proof that these are all types and how to go about finding them all with another order – max_121 Jul 24 '21 at 22:02
  • @GregMartin What was your algorithmic approach to listing all abelian groups of order $mn$ from the two other lists? And how did you know that there must be eight such groups? – max_121 Jul 24 '21 at 22:05
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    https://math.stackexchange.com/questions/786525/abelian-groups-of-order-n https://math.stackexchange.com/questions/226351/abelian-groups-of-order-2000 https://math.stackexchange.com/questions/779082/all-finite-abelian-groups-of-order-1024 https://math.stackexchange.com/questions/1270737/finite-abelian-groups-of-order-100 https://math.stackexchange.com/questions/2180691/abelian-groups-of-order-105 – Asinomás Jul 24 '21 at 22:31
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    It might be easier to use "the other" classification, where $G$ is isomorphic to a direct sum of cyclic groups of order equal to a prime power. Then, group them into the groups of order $2^k$, $5^k$, $7^k$ and for each of 2,5,7, you have a choice of two ways to get to a total power of 2. – Daniel Schepler Jul 24 '21 at 23:59
  • Let me second that. Personally, I find it easier to make sure I've obtained all possibilities by using the primary divisor decomposition (into cyclic groups of prime power order) rather than the invariant factor decomposition (which is the one you are using). The latter can then be obtained from the former in essentially the same way that the fundamental theorem is usually proven. – Arturo Magidin Jul 25 '21 at 00:39
  • @DanielSchepler same here, however in my course only the classification I showed above is "allowed"/should be used. Therefore I wanted an easy and direct way to find all the possibilites using the invariant factor decomposistion. If this question comes up, I will probably use the primary divisor decomposition and using this list try to find the factor decomposition – max_121 Jul 25 '21 at 11:32

1 Answers1

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In general, there are two classification for abelian groups. One is isomorphic to a direct sum of cyclic groups of order equal to a prime power, and the other is isomorphic to a direct sum of cyclic groups of order equal to a positive integer.

Notice that $4900=2^2\times 5^2\times 7^2$.

Let $G$ be isomorphic to a direct sum of cyclic groups of order equal to a prime power. In this case, we denote all isomorphism classes by arrays $(49,25,4)$,$(49,25,2,2)$,$(49,5,5,4)$, $(49,5,5,2,2)$,$(7,7,25,4)$,$(7,7,25,2,2)$,$(7,7,5,5,4)$,$(7,7,5,5,2,2)$. That is $$\begin{align}G&\cong \mathbb{Z}_{49}\times \mathbb{Z}_{25}\times \mathbb{Z}_{4}\\ G&\cong \mathbb{Z}_{49}\times \mathbb{Z}_{25}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}\\ G&\cong \mathbb{Z}_{49}\times \mathbb{Z}_{5}\times \mathbb{Z}_{5}\times \mathbb{Z}_{4}\\ G&\cong \mathbb{Z}_{49}\times \mathbb{Z}_{5}\times \mathbb{Z}_{5}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}\\ G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{7}\times \mathbb{Z}_{25}\times \mathbb{Z}_{4}\\ G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{7}\times \mathbb{Z}_{25}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}\\ G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{7}\times \mathbb{Z}_{5}\times \mathbb{Z}_{5}\times \mathbb{Z}_{4}\\ G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{7}\times \mathbb{Z}_{5}\times \mathbb{Z}_{5}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}\end{align} $$

If we apply the fundamental theorem of finite abelian groups in the form "there are $d_i$, s.t. $d_i\mid d_{i+1}$ and $G\cong \mathbb{Z}_{d_1}\times \mathbb{Z}_{d_2}\times \cdots \times \mathbb{Z}_{d_s}$", then we rearrange the above results using $\mathbb{Z}_{nm}\cong \mathbb{Z}_m\times \mathbb{Z}_n$, where $\gcd(n,m)=1$. Thus we get $$\begin{align}G&\cong \mathbb{Z}_{4900}\\ G&\cong \mathbb{Z}_{2}\times \mathbb{Z}_{2950}\\ G&\cong \mathbb{Z}_{5}\times \mathbb{Z}_{980}\\ G&\cong \mathbb{Z}_{10}\times \mathbb{Z}_{490}\\ G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{700}\\ G&\cong \mathbb{Z}_{14}\times \mathbb{Z}_{350}\\ G&\cong \mathbb{Z}_{35}\times \mathbb{Z}_{140}\\ G&\cong \mathbb{Z}_{70}\times \mathbb{Z}_{70}\end{align} $$

DXT
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