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In this question it is asked for an example of a group $G$ such that $(ab)^n=a^nb^n$ for all $a, b\in G$ holds for two consecutive integers $n\in\{m, m+1\}$, but $G$ is not an abelian group. The solution is easy: take $m=0$, or even $m=|G|$. I would be interested to see how long these chains can get.

What is the least $k$ such that there exists some integer $n$ and some non-abelian group $G=\langle a, b\rangle$ such that the following holds? $$(ab)^n=a^nb^n, (ab)^{n+1}=a^{n+1}b^{n+1}, \ldots, (ab)^{n+k}=a^{n+k}b^{n+k}$$

Note that if $k=2$ then simply taking $n=-1$ (so the chain $-1, 0, 1$) with $G$ arbitrary doesn't work, as $(ab)^{-1}=b^{-1}a^{-1}$ so if $(ab)^{-1}=a^{-1}b^{-1}$ and $a$ and $b$ necessarily commute. One can also prove that $a^n$ and $b$ commute (and so by symmetry $a$ and $b^n$ commute). Therefore, $|n|>1$ for all $k$.

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user1729
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  • Hmm, it might be possible to say something about this using the Hall-Petrescu formula. – Tobias Kildetoft Aug 14 '13 at 11:18
  • Oh - interesting! So...actually, any $p$-group for $p>>n$ and of nilpotency class two will work, no? – user1729 Aug 14 '13 at 11:22
  • I don't quite see how to get that. The extra term in that case will be of the form $c^{\binom{m}{2}}$ where $c$ is in the Derived subgroup and $m$ the exponent in question. But why should that exponent be divisible by the order of $c$? – Tobias Kildetoft Aug 14 '13 at 11:28
  • Sorry, I was just getting confused with derived length. (I -essentially- thought that the two was a three.) If the derived length is two then it turns out that $c=[y, x]$, according to Robinson's book. Which doesn't look promising. – user1729 Aug 14 '13 at 11:37
  • $(ab)^{n+k}=a^{n+1}b^{n+k}$ or $(ab)^{n+k}=a^{n+k}b^{n+k}$ ? – Boris Novikov Aug 14 '13 at 12:06
  • @BorisNovikov Thanks. Fixed it now. – user1729 Aug 14 '13 at 12:07
  • @YACP It would have been more useful if you had simply been less cryptic and more helpful with your first comment. – user1729 Aug 16 '13 at 08:58
  • @YACP But thanks for pointing my omission out anyway :-) – user1729 Aug 16 '13 at 12:23

1 Answers1

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The answer: $k=1$. Proof: Let $k\ge 2$, so $$ (ab)^{n}=a^{n}b^{n} $$ $$ (ab)^{n+1}=a^{n+1}b^{n+1} $$ $$ (ab)^{n+2}=a^{n+2}b^{n+2} $$ From 1st and 2nd $ab=(ab)^{-n}(ab)^{n+1}=b^{-n}ab^{n+1}$, i.e. $a=b^{-n}ab^{n}$. Similarly from 2nd and 3rd $a=b^{-n-1}ab^{n+1}$. Now we have $a=b^{-1}b^{-n}ab^{n}b=b^{-1}ab$.

In addition, if $k=1$ then for every $n$ we have an (infinite) group $G=\langle a, b\mid a=b^{-n}ab^{n}\rangle$.

Boris Novikov
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