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I am a student trying to learn some abstract algebra this summer, and I recently proved (as an exercise) that if $G$ is a group where every element has order 2, then $G$ is abelian. I was wondering could we make a similar conclusion about groups where every element has order 3, namely I am asking if $G$ is a group where all elements have order 3, then $G$ is abelian. I think that this is not true, but I cannot think of a counterexample.

The only groups that I can think of which have all elements order 3 are the groups $(\mathbb{Z}/3\mathbb{Z})^n$, but these are abelian. Any help is appreciated. Thanks!

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Related:… . I remember being told that this is true but hard, but don't take my word for it. – Qiaochu Yuan Jun 10 '12 at 1:31
@Qiaochu: You might be remembering – Cam McLeman Jun 10 '12 at 1:41
@Qiaochu: I think you are misinterpreting the question. – Arturo Magidin Jun 10 '12 at 1:44
@Student81: Note that, technically, it is impossible for all elements of a group to have order $3$ (or order $2$, for that matter): the identity always has order $1$. But we can ask that all nonidentity elements have order $3$, or that all elements have exponent $3$. Just a nitpick, though, since it is clear what it is you are asking. – Arturo Magidin Jun 10 '12 at 1:45
Hard but true is that all finitely generated groups of exponent 3 are actually finite. – Cam McLeman Jun 10 '12 at 1:46

2 Answers 2

up vote 14 down vote accepted

The standard example is the Heisenberg group. Consider the group of all matrices of the form $$\left(\begin{array}{ccc} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{array}\right),$$ where $x,y,z\in\mathbb{Z}/3\mathbb{Z}$. It is not hard to verify that this is a group, that every one of its 27 elements is of exponent $3$, and that it is not abelian. Replacing $\mathbb{Z}/3\mathbb{Z}$ with $\mathbb{Z}/p\mathbb{Z}$ for odd prime $p$ shows that a similar result cannot hold for any prime other than $p=2$.

This is an example of smallest possible order: a finite group in which every element is of exponent $3$ must have order $3^n$ for some $n$ (a consequence of Cauchy's Theorem), and every group of order $3^2$ is abelian.

There is another nonabelian group of order $27$, but in that group there is an element of order $9$: $$\langle a,b\mid a^9 = b^3 = 1, ba = a^4b\rangle.$$

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Thanks! This is very helpful! – Student81 Jun 10 '12 at 1:51
Then if we consider the Heisenberg group over $\mathbb{Z}/2\mathbb{Z}$, we still have a non-abelian group, but this time there are elements whose orders do not divide $2$ (therefore no counter-example to the exercise result mentioned by the asker). – Jeppe Stig Nielsen Jul 8 at 22:18

No, it isn't true, but if you're beginning with this stuff perhaps you won't fully understand the example: the semidirect product of a non-cyclic group of order $9$ by a group of order $3$ has all its non-unit elements of order 3...

You can read here,3) an exposition about this one as a group of unitriangular matrices.

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+1. Smart one. :) – Babak S. Jun 10 '12 at 5:15

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