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Suppose a finite group has the property that for every $x, y$, it follows that

\begin{equation*} (xy)^3 = x^3 y^3. \end{equation*}

How do you prove that it is abelian?

Edit: I recall that the correct exercise needed in addition that the order of the group is not divisible by 3.

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    You don't, as the group is not necessarily abelian! The group of upper triangular 3-by-3 matrices with ones along the diagonal and coefficients in the three-element field $\mathbb Z/3\mathbb Z$ has exponent three, so your equation holds, but it is not abelian. – Mariano Suárez-Álvarez Jul 28 '10 at 21:47
  • (There are lots of examples: the most famous ones are the Burnside groups B(m,3), which you'll find described at http://en.wikipedia.org/wiki/Burnside%27s_problem#Bounded_Burnside_problem; the group in the first comment is B(2,3)) – Mariano Suárez-Álvarez Jul 28 '10 at 21:49
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    @Mariano, why don't you give that as the answer and then it can be accepted? Otherwise it looks as though nobody has answered the question. – bryn Jul 29 '10 at 06:57
  • By the way, your statement becomes true if you change 3 by 2. – falagar Jul 29 '10 at 14:54
  • I wrote a short proof here . Steve –  Aug 04 '10 at 08:17

3 Answers3

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You don't, as the group is not necessarily abelian! The group of upper triangular 3-by-3 matrices with ones along the diagonal and coefficients in the three-element field $\mathbb {Z}/3\mathbb{Z}$ has exponent three, so your equation holds, but it is not abelian.

There are lots of examples: the most famous ones are the Burnside groups $B(m,3)$: the group I described above is $B(2,3)$.

Mariano Suárez-Álvarez
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On the other hand, if the order of your group is not a multiple of 3 then it must be abelian!

You can read a proof here

Jorge Miranda
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Both proofs so far are for finite groups. However, the problem (with the complete assumptions) holds for not-necessarily-finite groups, provided that the group have no element of order $3$.

Here is a proof:

From $(ab)^3 = a^3b^3$, we immediately conclude that $(ba)^2 = a^2b^2$ by cancellation. We also conclude that cubes commute with squares, since $b^2a^2=(ab)^2 = (ab)^3(ab)^{-1}= a^3b^3b^{-1}a^{-1} = a^3b^2a^{-1}$, hence $b^2a^3=a^3b^2$.

Now, consider the cube of a commutator, $$\begin{align*} [a,b]^3 &= (a^{-1}b^{-1}ab)^3\\ &= a^{-3}b^{-3}a^3b^3\\ &= a^{-3}b^{-3}b^2a^3b\\ &= a^{-3}b^{-1}a^3b\\ &= [a^3,b]. \end{align*}$$ In particular, for any square we have $[a,x^2]^3 = [a^3,x^2]=1$, because cubes commute with squares. But since $G$ has no elements of order $3$, we conclude that $[a,x^2]=1$. Thus, we conclude that every square in $G$ is actually central.

But that means that $(ab)^2 = b^2a^2 = a^2b^2$ for all $a,b\in G$. And this condition is well known to imply that the group is abelian: $abab=aabb$ yields $ba=ab$.

In particular, this holds for a finite group whose order is not divisible by $3$.

Arturo Magidin
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