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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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    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Jul 28 '10 at 21:47
  • $\begingroup$ (There are lots of examples: the most famous ones are the Burnside groups B(m,3), which you'll find described at en.wikipedia.org/wiki/…; the group in the first comment is B(2,3)) $\endgroup$ – Mariano Suárez-Álvarez Jul 28 '10 at 21:49
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    $\begingroup$ @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. $\endgroup$ – bryn Jul 29 '10 at 6:57
  • $\begingroup$ By the way, your statement becomes true if you change 3 by 2. $\endgroup$ – falagar Jul 29 '10 at 14:54
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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

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    $\begingroup$ Urgh! Why do they write proofs in such a complicated way?! $\endgroup$ – Mariano Suárez-Álvarez Jul 29 '10 at 16:25
  • $\begingroup$ Yeah! This is what I had in mind! Thanks! :) $\endgroup$ – user218 Jul 29 '10 at 18:07
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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 $B(2,3)$.

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I wrote a short proof here .

Steve

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