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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!

Zev Chonoles
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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 $n=|G|$, $|G|+1$. – Arturo Magidin Jan 27 '12 at 22:25
  • @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. – Zev Chonoles Jan 27 '12 at 22:37
  • 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). Then $n$ and $n+1$ work. – user1729 Aug 14 '13 at 10:51

3 Answers3

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

yoyo
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A trivial example is $a^0b^0=(ab)^0$ and $a^1b^1=(ab)^1$, which hold for $\forall a,b\in G$ with $G$ being any non-Abelian groups.

A non-trivial example is dihedral groups $D_4$ of order $8$: It will be obvious that $a^4 b^4=(ab)^4$ and $a^{5} b^{5}=(ab)^{5}$ hold, once you prove $\forall g\in D_n, g^n=e$.

In fact, Lagrange's theorem says that, for any finite group $G$, there exists $n\in\{\text{divisors of }|G|\}$ such that $g^n=e$ for all $g\in G$. So $g^{|G|}=e$ holds for any finite groups.

Misho
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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 $i=2$.

user1729
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ssk
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