Welcome to MathOverflow

Question

2

3

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 $p$.

flag	asked Jul 16 2010 at 5:39 marwalix 43●5

1

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 – Yemon Choi Jul 16 2010 at 5:43

1

Wadim, I don't follow. See the link in my comment and also en.wikipedia.org/wiki/Exponent_%28group_theory%29 – Yemon Choi Jul 16 2010 at 6:04

5

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. – Robin Chapman Jul 16 2010 at 7:20

7

The group $G$ of upper triangular matrices over $\mathbb{F}_3$ with diagonal $(1,1,1)$ is of exponent $3$. – Martin Brandenburg Jul 16 2010 at 8:53

2

@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 research-level math issue here. – Pete L. Clark Jul 17 2010 at 0:33

show 13 more comments

AL · Accepted Answer · 2010-07-16 11:36:28Z UTC

10

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 non-abelian.

(Checking exponent 3 basically comes down to ensuring that $(yz)^3 = (y^2z)^3 = (yx^2)^3 = 1$. Or by using Gap.)

answered Jul 16 2010 at 11:36

AL
1,246●6●13

	See also the comments at mathoverflow.net/questions/31797/… – Yemon Choi Jul 16 2010 at 17:12
	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 Extra-special $p$-groups for more details. – AL Jul 21 2010 at 8:18

Ashot Minasyan · Answer 2 · 2010-07-19 20:25:35Z UTC

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.

Primoz · Answer 3 · 2010-07-17 06:47:43Z UTC

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.

marwalix · Answer 4 · 2010-07-16 22:19:26Z UTC

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 n-abelian group (such as $(xy)^n=x^n y^n$) that has got no element (other than $1$) whose order divides $n(n-1)$ is abelian. One can refer to J.L Alperin A classification of n-abelian groups in Canadian Journal of Mathematics (1969)

Exponent of a group

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

4 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.