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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?

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    $\begingroup$ 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. $\endgroup$
    – Greg Martin
    Commented Jul 24, 2021 at 21:50
  • $\begingroup$ 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 $\endgroup$
    – max_121
    Commented Jul 24, 2021 at 21:53
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    $\begingroup$ 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 $d_1d_2.$ $\endgroup$
    – Thomas Andrews
    Commented Jul 24, 2021 at 22:00
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    $\begingroup$ math.stackexchange.com/questions/786525/… math.stackexchange.com/questions/226351/… math.stackexchange.com/questions/779082/… math.stackexchange.com/questions/1270737/… math.stackexchange.com/questions/2180691/… $\endgroup$
    – Asinomás
    Commented Jul 24, 2021 at 22:31
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    $\begingroup$ 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. $\endgroup$
    – Daniel Schepler
    Commented Jul 24, 2021 at 23:59

1 Answer 1

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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 $$\begin{align}G&\cong \mathbb{Z}_{4900}\\ G&\cong \mathbb{Z}_{2}\times \mathbb{Z}_{2950}\\ G&\cong \mathbb{Z}_{5}\times \mathbb{Z}_{980}\\ G&\cong \mathbb{Z}_{10}\times \mathbb{Z}_{490}\\ G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{700}\\ G&\cong \mathbb{Z}_{14}\times \mathbb{Z}_{350}\\ G&\cong \mathbb{Z}_{35}\times \mathbb{Z}_{140}\\ G&\cong \mathbb{Z}_{70}\times \mathbb{Z}_{70}\end{align} $$

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