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Question: Give a representative of each Isomorphism class of Abelian group of order 225. Which ones are cyclic?

By the Fundamental theorem of finite abelian group:

$\left | G \right |=225=3^{2}5^2$

Now, the possible isomorphism are

$G\cong \mathbb{Z}_{3}\otimes \mathbb{Z}_{3}\otimes \mathbb{Z}_{5}\otimes \mathbb{Z}_{5}, \mathbb{Z}_{9}\otimes \mathbb{Z}_{25}, \mathbb{Z}_{9}\otimes \mathbb{Z}_{5}\otimes \mathbb{Z}_{5}, \mathbb{Z}_{3}\otimes \mathbb{Z}_{3}\otimes \mathbb{Z}_{25}$

Here's my confusion:

My solution indicates the ONLY cyclic groups to be $\mathbb{Z}_{9}\otimes \mathbb{Z}_{25}$, but from the theorem of external direct product of cyclic group, this cannot be the only cyclic group since

for all cyclic groups G and H, $G \otimes H$ are cyclic IFF $GCD\left ( \left | H \right |,\left | G \right | \right )$ are coprime.

Any help is appreciated.

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  • $\begingroup$ All your $\otimes$ (\otimes) should be $\oplus$ (\oplus). $\endgroup$ – PseudoNeo Jul 25 '16 at 10:11
  • $\begingroup$ And why would that mean it was not cyclic? Those integers are indeed coprime (also, you should use either $\times$ or $\oplus$ for direct product, rather than $\otimes$). $\endgroup$ – Tobias Kildetoft Jul 25 '16 at 10:11
  • $\begingroup$ Yes agreed. I've edited the critical portion. $\endgroup$ – Mathematicing Jul 25 '16 at 10:13
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There's only one cyclic group of any order. $9$ and $25$ are coprime. $\mathbb Z_9\oplus\mathbb Z_{25}$ is cyclic.

I can only guess that $\oplus$ or $\times$ is what you mean, but one of them must be because $\otimes$ is nonsensical in this case, as $\mathbb Z_9\otimes\mathbb Z_{25}=0$.

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  • $\begingroup$ But Gcd(3,3,5,5) are co prime too $\endgroup$ – Mathematicing Jul 25 '16 at 10:18
  • $\begingroup$ @Mathematicing Your question seems to be asking for possible cyclic groups. You'd be right if it was asking for all possible abelian groups. $\endgroup$ – florence Jul 25 '16 at 10:19
  • $\begingroup$ @Mathe $3$ and $3$ are not coprime. $\endgroup$ – Matt Samuel Jul 25 '16 at 10:19
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    $\begingroup$ @Mathe what's relevant is that the numbers be pairwise coprime. You would need $\gcd(3,3)=\gcd(3,5)=\gcd(3,5)=\gcd(3,5)=\gcd(3,5)=\gcd(5,5)=1$, which is not true (the duplicates are all pairs). $\endgroup$ – Matt Samuel Jul 25 '16 at 10:22
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    $\begingroup$ @Mathematicing There is nothing to expand in that context. The definition is not related to this, you just seem to keep forgetting the "pairwise" part, which was already spelled out above. $\endgroup$ – Tobias Kildetoft Jul 25 '16 at 11:13

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