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I am trying to learn abstract algebra on my own. Unfortunately I am confused and not sure how to proceed with the following question.

I want to find all abelian groups of order 63. By theorem of finitely generated abelian groups that since 63 = 9*7 is $\mathbb{Z}_9\times\mathbb{Z}_7$ and $\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_7$ are abelian. But what about $\mathbb{Z}_{36}$? Is it abelian? It is cyclic, so it should be abelian?

I am sorry, but I am trying to learn these things on my own, it'd be great if someone could give me a complete explanation.

Thanks.

Kind regards

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    $\begingroup$ Do you mean "But what about $\Bbb Z_{63}$?" Don't worry, it's isomorphic to $\Bbb Z_9\times \Bbb Z_7$. $\endgroup$ – Arthur Jun 10 '16 at 19:57
  • $\begingroup$ Yes, $Z_36$ is abelian. And yes, it is also cyclic. Note that any group of the form $Z_n$ is always abelian and cyclic since adding numbers is obviously commutative. They are cyclic since they can all be generated by $1$. $\endgroup$ – athul777 Jun 10 '16 at 19:58
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    $\begingroup$ Just to add to Arthur's comment, it's because whenever $gcd(a, b)=1$, we have that $\mathbb{Z}_a \times \mathbb{Z}_b$ is isomorphic to $\mathbb{Z}_{ab}$. That's because $(1, 1)$ is a generator for $\mathbb{Z}_a \times \mathbb{Z}_b$ $\endgroup$ – JasonM Jun 10 '16 at 20:00

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