It is well known that every group of exponent $n=2$ is abelian. I remember having seen that this is also the case for $n=3$. (can someone give a proof). How does this generalize to any $n$ or to any prime $p$.

1$\begingroup$ I'm not sure I agree with your claim about exponent 3. See mathoverflow.net/questions/31797/… in particular the comments of Pete Clark and the answer of Francesco Polizzi $\endgroup$– Yemon ChoiJul 16 '10 at 5:43

5$\begingroup$ Wadim, "exponent" and "order" are not synonyms in finite group theory. The order of a finite group is its cardinality, while the exponent is the least $n$ with $x^n$ the identity for all $x$ in the group. $\endgroup$– Robin ChapmanJul 16 '10 at 7:20

9$\begingroup$ The group $G$ of upper triangular matrices over $\mathbb{F}_3$ with diagonal $(1,1,1)$ is of exponent $3$. $\endgroup$– Martin BrandenburgJul 16 '10 at 8:53

2$\begingroup$ @marwalix: The result you "remember having seen" for n=3 might have been Jacobson's result for rings, not groups. See mathoverflow.net/questions/29590/… $\endgroup$– Doug ChathamJul 16 '10 at 12:26

2$\begingroup$ @marwalix: For every finite group of exponent $3$, $3$ divides the order of $G$. So what you say is true but vacuous. @everybody: I am frankly surprised at the number of comments on this question: there's no researchlevel math issue here. $\endgroup$– Pete L. ClarkJul 17 '10 at 0:33
The group defined by $\langle x,y,z; x^3 = y^3 = z^3 = 1, yz = zyx, xy = yx, xz = zx\rangle$ has order 27, exponent 3 and is nonabelian.
(Checking exponent 3 basically comes down to ensuring that $(yz)^3 = (y^2z)^3 = (yx^2)^3 = 1$. Or by using Gap.)

$\begingroup$ See also the comments at mathoverflow.net/questions/31797/… $\endgroup$ Jul 16 '10 at 17:12

$\begingroup$ I should probably point out that this can be generalised to every $p>2$. That is, there always exists a group of order $p^{2n+1}$ with exponent $p$ such that $Z(G) = p$. Look up the classification of Extraspecial $p$groups for more details. $\endgroup$– ADLJul 21 '10 at 8:18
Well, any finitely generated group G of exponent 3 is finite by a classical theorem of Burnside. And since the order of every element is 3, the order of G must be a power of 3 by Cauchy's theorem. It follows that G is a finite nilpotent group. A similar argument shows that the same is true for any finitely generated group of exponent 4. This is unknown for 5, and false for 6.
Correction: it seems (see the answer of Primoz above) that any group of exponent 3 is nilpotent, altough it can be infinite if it not finitely generated.

$\begingroup$ See also the comments at mathoverflow.net/questions/31797/… $\endgroup$ Jul 16 '10 at 17:12
Every group of exponent 3 is nilpotent of class at most 3, and this bound is best possible. The question whether finitely generated groups of exponent $n$ are finite is also known as the Burnside problem. There is an excellent historical overview of this problem, along with a list of relevant references.
The result I have seen is that every finite group $G$ of exponent $3$ such as $3$ does not divide $o(G)$ is abelian.
It generalises the following way : every nabelian group (such as $(xy)^n=x^n y^n$) that has got no element (other than $1$) whose order divides $n(n1)$ is abelian. One can refer to J.L Alperin A classification of nabelian groups in Canadian Journal of Mathematics (1969)

$\begingroup$ I do not understand the sentence "every finite group $G$ of exponent $3$ such as $3$ does not divide $o(G)$". If $G$ is a finite group of exponent $3$ then it has order $3^n$, and so $3$ divides $o(G)$, so no groups satisfy this condition. Right? (I realise this is over 10 years old, so sorry for dredging it up!) $\endgroup$– ADLJul 22 '21 at 8:32