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I have been studying Abstract Algebra, and all the finite groups that we have studied so far have also been cyclic. So, is it true that all finite groups are cyclic?
If yes, what is the theorem? If no, please provide a counterexample.

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    $\begingroup$ Not by a long shot. Look up symmetric and dihedral groups. Most finite groups are not cyclic or even commutative. $\endgroup$
    – KCd
    Sep 30, 2014 at 16:55
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    $\begingroup$ Smallest counterexample: $\mathbb{Z}_2 \times \mathbb{Z}_2$. en.wikipedia.org/wiki/Klein_four-group $\endgroup$
    – E W H Lee
    Sep 30, 2014 at 16:59
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    $\begingroup$ The symmetric group $S_3$ is finite, of order $6$ but not is cyclic. $\endgroup$
    – user59969
    Sep 30, 2014 at 17:02
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    $\begingroup$ You may coincidentally be observed only finite groups of prime order. In this case, his statement was true, but in general is false. $\endgroup$
    – user59969
    Sep 30, 2014 at 17:04
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    $\begingroup$ I find it surprising that your teacher did not show you the group of permutations of three things as soon as groups were defined. $\endgroup$
    – Lubin
    Sep 30, 2014 at 17:11

2 Answers 2

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No. Try and find a generator for $$\mathbb{Z}_2 \times \mathbb{Z}_2$$ Shouldn't take long, there's only four possibilities. Next try:

$$\mathbb{Z}_6 \times \mathbb{Z}_2$$

In general, a finite group will be cyclic if and only if it is isomorphic to some direct product of cyclic groups with relatively prime orders. For example:

$$\mathbb{Z}_2 \times \mathbb{Z}_3, \ \ \text{ and } \ \ \mathbb{Z}_\mathbb{11} \times \mathbb{Z}_4$$

are cyclic. But of course, not all finite groups are Abelian (commutative) let alone cyclic.

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For abelian take klein 4 group and for non abelian take dihedral group of order 8.

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