All abelian groups of order $4900$ 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?
 A: 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
$$\begin{align}G&\cong \mathbb{Z}_{4900}\\
G&\cong \mathbb{Z}_{2}\times \mathbb{Z}_{2950}\\
G&\cong \mathbb{Z}_{5}\times \mathbb{Z}_{980}\\
G&\cong \mathbb{Z}_{10}\times \mathbb{Z}_{490}\\
G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{700}\\
G&\cong \mathbb{Z}_{14}\times \mathbb{Z}_{350}\\
G&\cong \mathbb{Z}_{35}\times \mathbb{Z}_{140}\\
G&\cong \mathbb{Z}_{70}\times \mathbb{Z}_{70}\end{align}
$$
