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I am looking for an example of a group $G$ where the equality $(ab)^{n}=a^{n}b^{n}$ holds for two consecutive integers $n$, but $G$ is not an abelian group. I've started do some calculations in the group $D_{4}$ (I gave up!) Do you know where I can find that example? Is it possible find such example without doing a lot of calculations?

Thanks for your help!

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    How about any nonabelian group, with $n=0$ and $n=1$? If you don't like $n=0$, then take $G$ finite and $n=|G|$, $|G|+1$.2012-01-27
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    @spohreis: Your previous title did not mean what you wanted - by putting "$G$ is not abelian" at the end, you made it sound as if it were a consequence of the preceding phrase, which produces a statement which is not correct.2012-01-27
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    A similar, but less trivial, approach to Arturo's would be to take a group of fixed exponent $n$, and $n<|G|$ (see, for example, [here](http://math.stackexchange.com/questions/454170/for-given-prime-number-p-neq-2-construct-a-non-abelian-group-with-exponent/454172#454172)). Then $n$ and $n+1$ work.2013-08-14

2 Answers 2

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$D_4$ should work with $r^4=1$ commuting with $s^3=s$ and $r^3$ commuting with $s^2=1$

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Consider the symmetric group of order $3$ , $S_3$, and then the result holds for $i=0$ and $i=1$ (equivalently, for $i=6$ and $i=7$). Note that the result does not hold for $i = 2$.

In fact there is another theorem which states that a group is abelian if the result holds, for only $i=2$.