Are Cyclic Groups with x^n = 1 the Only Finite Groups?

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Is it true that cyclic groups with [tex]x^n = 1[/tex] the only finite groups (with order n)?

I've been experimenting with a few groups and I think this is true but I'm not sure.thanks
 
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Have you looked at the permutation groups Sn, dihedral groups Dn (symmetries of regular polygons), etc.?

In fact, almost all finite groups are non-abelian, so the cyclic groups are just very few in number.
 
CompuChip said:
In fact, almost all finite groups are non-abelian, so the cyclic groups are just very few in number.
There are as many cyclic groups ([itex]\aleph_0[/itex]) as finite groups. By, "almost all finite groups are non-abelian", do you mean [itex]\lim_{n\rightarrow \infty}a_n/g_n=0[/itex], where [itex]a_n[/itex] is the number of abelian groups of order [itex]\leq n[/itex] and [itex]g_n[/itex] is the number of groups of order [itex]\leq n[/itex]? (This doesn't seem likely.)
 
Given any n, there certainly exists a cylic group of order n.

It is also true that if p is a prime number then then the only group of order p is the cylic group.

But if n is NOT prime, then there exist other, non-cyclic, groups of order n.

The simplest example is for n= 4. The "Klein 4-group" is not cyclic.

The Klein 4-group has 4 members, e, a, b, c satifying
ee= e, ea= a, eb= b, ec= c (e is the identity)
ae= a, aa= e, ab= c, ac= b
be= b, ba= c, bb= e, bc= a
ce= c, ca= b, cb= a, cc= e

The fact that every element is its own inverse proves this group is not cyclic.

Any group of order 4 is isomorphic either to the cyclic group of order 4 or to the Klein four-group.
 
Martin Rattigan said:
...(This doesn't seem likely.)
Then again it doesn't seem unlikely. Has it been proved?

The non-abelian groups of orders 2n will probably arrange it by themselves (e.g. there are over ten times as many non-abelian groups of order 1024 as there are other groups up to and including 1024) so this may not be too difficult to prove.
 
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