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Give an example of a non-abelian group, in which $(ab)^3=a^3b^3$ for every element $a,b$ in $G$.

I understand that such a group should be of order divisible by 3 (see Problem from Herstein on group theory). Also, it is easily seen that $(ab)^3=a^3b^3 \iff (ba)^2=a^2b^2$. But I can't come up with one single example.

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    $\begingroup$ Any nonabelian group of exponent 3, for example. $\endgroup$ Commented Feb 16, 2014 at 16:38

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Hint: try the group of triangular $3 \times 3$ matrices over the field with 3 elements, with 1's on the diagonal. This group has exponent 3, that is for every $g \in G$, it holds that $g^3=1$. Can you see that $|G|=27$?

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  • $\begingroup$ I'd just like to add: as evidence that $G$ is not abelian, note that $A = \pmatrix{ 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\}$ and $B = \pmatrix{ 1 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\}$ satisfy $AB \neq BA$. $\endgroup$ Commented Feb 16, 2014 at 16:50
  • $\begingroup$ Yes good remark - on the other hand, I was hoping that the OP would figure that (non-abelianess) out for him/herself. $\endgroup$ Commented Feb 16, 2014 at 17:04
  • $\begingroup$ Ah, I suppose I could have been more delicate there. I wanted to save OP the hunt for non-commuting matrices, which is particularly unpleasant if you don't know what to look for. $\endgroup$ Commented Feb 16, 2014 at 17:11

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