# possible cyclic group from fundamental theorem of finite abelian

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.

• All your $\otimes$ (\otimes) should be $\oplus$ (\oplus). – PseudoNeo Jul 25 '16 at 10:11
• 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 $\otimes$). – Tobias Kildetoft Jul 25 '16 at 10:11
• Yes agreed. I've edited the critical portion. – Mathematicing Jul 25 '16 at 10:13

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 $\mathbb Z_9\otimes\mathbb Z_{25}=0$.
• @Mathe $3$ and $3$ are not coprime. – Matt Samuel Jul 25 '16 at 10:19
• @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). – Matt Samuel Jul 25 '16 at 10:22