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A question says: prove or prove or disprove that there are only 2 non-isomorphic abelian groups of order 2009.

I think that it is true because... I split up $2009$ into $7 \times 7 \times 41$ and so this gives the groups $$\mathbb{Z}_{2009}$$ $$\mathbb{Z}_{7} \oplus \mathbb{Z}_{7\times 41}$$ where the first order divides the next by the fundamental theorem of finite abelian groups (and we can prove these are not isomorphic by using order of elements) but then I thought about the group $\mathbb{Z}_{7}\oplus \mathbb{Z}_{7} \oplus \mathbb{Z}_{41}$. What actually IS this group? By FTFAG 7 doesn't divide 41 so surely it can't be another decomposition. And 7,7 is not coprime so it can't be isomorphic to $\mathbb{Z}_{2009}$. But maybe I'm wrong? Argh!

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up vote 5 down vote accepted

Take a look here. The group $\mathbb{Z}_{7\times 41}$ is isomorphic to $\mathbb{Z}_7\oplus\mathbb{Z}_{41}$ because $7$ and $41$ are coprime, and therefore $\mathbb{Z}_7\oplus \mathbb{Z}_{7\times 41}$ is isomorphic to $\mathbb{Z}_7\oplus \mathbb{Z}_7\oplus\mathbb{Z}_{41}$.

Note that, similarly, $\mathbb{Z}_{2009}\cong \mathbb{Z}_{49}\oplus \mathbb{Z}_{41}$ because $49$ and $41$ are coprime.

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Ah of course, thank you :) – user26069 Apr 26 '12 at 17:40
No problem, glad to help :) – Zev Chonoles Apr 26 '12 at 17:41

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