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Suppose $G$ is a group such that there exists $k\in\mathbb{Z}$ satisfying $(ab)^k = a^k b^k$ and $(ab)^{k+1} = a^{k+1} b^{k+1}$.

Is $G$ abelian?

I've tried to use the same argument in this post, but it doesn't work.

Is there a non-abelian group $G$ with this property?

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Rodrigo Dias
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    What if the exponent of $G$ is $k$? – Angina Seng Aug 09 '17 at 11:47
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    Take $k = |G|$... – lhf Aug 09 '17 at 11:53
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    @lhf The question nowhere says or suggests that $G$ is finite. – Marc van Leeuwen Aug 09 '17 at 12:27
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    Is $k=0$ admissible? – lhf Aug 09 '17 at 12:31
  • Sure $k=0$ is admissible! And it works for all group... Nice! Such a dumb question I've made. – Rodrigo Dias Aug 09 '17 at 12:34
  • While $0\in\Bbb N$ still raises some debate, I don't think anybody really contests $0\in\Bbb Z$. Since we get no information from that case, the answer must obviously be "not necessarily" (and "yes, all non-abelian groups do satisfy the hypothesis"). – Marc van Leeuwen Aug 09 '17 at 12:34
  • Possible duplicate of https://math.stackexchange.com/questions/1199479/non-abelian-group-g-satisfying-a-cdot-bi-ai-cdot-bi-for-two-consecut and https://math.stackexchange.com/questions/103124/an-example-of-a-non-abelian-group-g-where-for-all-a-b-in-g-the-equality. See also https://math.stackexchange.com/questions/467323/non-abelian-groups-where-abni-anibni-for-all-0-leq-i-leq-k-k. – lhf Aug 09 '17 at 12:37

3 Answers3

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In this answer I assume that the $k=0$ isn't allowed.

As given in the comments above, say that $k = \lvert G\rvert$. Then for any $g\in G$, $g^k = e$ (identity in $G$). In particular $(ab)^k = e$ for all $a,b\in G$. In this case

$$ (ab)^k = e = a^kb^k $$ and $$ (ab)^{k+1} = (ab)^kab = eab = ab \\ a^{k+1}b^{k+1} = a^kab^kb = eaeb= ab $$ So in fact, for all finite groups (even the non Abelian ones) $G$ there is indeed a $k$ such that $$ (ab)^k = a^k b^k$$ and $$ (ab)^{k+1} = a^{k+1} b^{k+1} $$

Just for fun, think about the case where there is a $k\neq 0$ not a multiple of $\lvert G\rvert$ such that \dots

Thomas
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  • $G$ isn't necessarily finite though, at least not in the question posed? – SvanN Aug 09 '17 at 12:30
  • Thanks for the answer. I was wondering why this argument doesn't work for the case in the post that I've related.. – Rodrigo Dias Aug 09 '17 at 12:31
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    @S.vanNigtevecht: Right, the question is if you can conclude that $G$ is Abelian. The answer is that you can't conclude this because there are groups that satisfy the equations while not being Abelian. Here is another question: Are there any infinite non-Abelian groups satisfying the equations? – Thomas Aug 09 '17 at 12:35
  • @rldias: Because in that post you assume one extra equation $(ab)^{k+2} = a^{k+2}b^{k+2}$ (the three consecutive integers thing.) – Thomas Aug 09 '17 at 12:36
  • For every group $G$, it is true that $(ab)^0 = a^0 b^0$ and $(ab)^1 = a^1 b^1$ for all $a,b\in G$. Just take $k=0$ and conclude that my question was horrible. – Rodrigo Dias Aug 09 '17 at 12:38
  • @rldias: Yeah, I see that now. I will keep my answer since it might be interesting to you with the condition that $k\neq0$. – Thomas Aug 09 '17 at 12:40
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$k=0$ works for all groups, even infinite nonabelian ones.

lhf
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You can use the following property :$G$ be a group and $k$ be a integer having the property that $(ab)^k=a^kb^k$ and $(ab)^{k+1}=a^{k+1}b^{k+1}$ then
$G$ is abelian iff $(ab)^{k+2}=a^{k+2}b^{k+2}$.

Supriyo Halder
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