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I used the fundamental theorem of finite abelian groups.

$\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5}$

$\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{25}$

$\mathbb{Z}_{2} \times \mathbb{Z}_{4} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5}$

$\mathbb{Z}_{2} \times \mathbb{Z}_{4} \times \mathbb{Z}_{25}$

$\mathbb{Z}_{8} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5}$

$\mathbb{Z}_{8} \times \mathbb{Z}_{25}$

My question is:

Isn't $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5}$ isomorphic to the group of order $200$ by the theorem? Should such a product be included in the list?

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    You know, the groups you listed are not exactly of order $30$. They have order $200$. – Mark Oct 11 '18 at 16:56
  • See this question for an answer. – Dietrich Burde Oct 11 '18 at 16:59
  • @Mark My bad, I had two parts of the question on one page, one for order $30$ and one for order $200$, which I mixed up! But my question is still the same. –  Oct 11 '18 at 17:08
  • Well, you got it right then. From the fundamental theorem it follows that there are $6$ abelian groups of order $200$ up to isomorphism. I don't really understand what is the question. – Mark Oct 11 '18 at 17:10
  • @Mark May I ask what "up to isomorphism" means? The question asks to find non-isomorphic groups, but isn't $\mathbb{Z}{2} \times \mathbb{Z}{2} \times \mathbb{Z}{2} \times \mathbb{Z}{5} \times \mathbb{Z}_{5}$ isomorphic to the group? –  Oct 11 '18 at 17:13
  • Groups $G$ and $H$ are called isomorphic if there exists a bijection $\varphi:G\to H$ which has the property $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$ for all $g_1,g_2\in G$. So to find all groups "up to isomorphism" means you need to find one group from each isomorphism class. Now there is the Chinese Remainder Theorem which tells us that $\mathbb{Z_{mn}}$ is isomorphic to $\mathbb{Z_m}\times\mathbb{Z_n}$ if and only if $\gcd(m,n)=1$. It follows that the groups you found are not isomorphic to each other. – Mark Oct 11 '18 at 17:18
  • On the other hand you could for example replace $\mathbb{Z_2}\times\mathbb{Z_2}\times\mathbb{Z_2}\times\mathbb{Z_5}\times\mathbb{Z_5}$ with the group $\mathbb{Z_2}\times\mathbb{Z_{10}}\times\mathbb{Z_{10}}$ because these two groups are isomorphic. (follows from chinese remainder theorem once again) – Mark Oct 11 '18 at 17:19

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