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By theorem, every cyclic group is abelian, but does all abelian group are cyclic? I mean, is there a abelian group that is not cyclical?

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closed as off-topic by Dietrich Burde, Claude Leibovici, Strants, Xam, user190080 Feb 28 '18 at 17:57

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    $\begingroup$ What are some finite groups you know? There are very small counterexamples you should be aware of. $\endgroup$ – lulu Feb 28 '18 at 13:08
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    $\begingroup$ Try to find an abelian noncyclic group of order 4. What can you say about the order of the non-identity elements? $\endgroup$ – almagest Feb 28 '18 at 13:09
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    $\begingroup$ Possible duplicate of Abelian group that is non-cyclic $\endgroup$ – освящение Feb 28 '18 at 13:28
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The real numbers without zero is an abelian group under multiplication. It is not cyclic.

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    $\begingroup$ Or you could use the real numbers with zero, under addition. $\endgroup$ – MJD Feb 28 '18 at 13:26
  • $\begingroup$ @MJD very true. I find it easier to justify with multiplication. If we have a generator, then it is either less than 1, equal to 1, or greater than 1. In each case powers of the generator retain this property. Thus, the powers don’t generate all of the reals. $\endgroup$ – Joel Feb 28 '18 at 17:19
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No, consider the abelian group $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ consisting out of the elements $\{(0,0), (1,0), (0,1), (1,1)\}$. The group operation is entrywise addition and reduction modulo $2$. The non-identity elements have order $2$, and hence there is no element of order $4$.

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$\mathbb Z_m×\mathbb Z_n$ is abelian but not cyclic when $gcd(m,n)\gt1$...

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