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Problem:

If $G$ is a group in which $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a,b \in G$, then $G$ is abelian. If we assume the relation $(ab)^i = a^ib^i$ for just two consecutive integers can we conclude that $G$ is abelian?

Thank all!

user26857
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Minh
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  • The assumption holds for $i=0$ and $i=1$ in any group. Perhaps you intended to assume that it holds for two consecutive positive integers? – Derek Holt Oct 10 '17 at 14:46
  • Yes you are right, so the problem in Herstein's book is trivial. But it is slightly less trivial if you assume that the condition holds for two consecutive positive integers. – Derek Holt Oct 10 '17 at 15:07
  • In my opinion, with this case we can't conclude that $G$ is abelian, because $(a.b)^0 = e$ does not implies $ba = e$ and $a.b = a.b$ does not implies $ab = ba$. – Minh Oct 10 '17 at 15:14
  • Yes exactly. What I am saying is that in every group $G$ and for all $a,b \in G$ we have $(ab)^0 = a^0b^0 = 1$, and $(ab)^1 = ab$. Since this is true in every group, we clearly cannot conclude that $G$ is abelian, . So your question is true but trivial (and the same applies to Herstein's question, so there is no need to feel bad about it). – Derek Holt Oct 10 '17 at 15:21
  • I ask this quention for find an example to prove that isn't true. – Minh Oct 10 '17 at 15:25

3 Answers3

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Two consecutive integers is not enough to conclude that $G$ is abelian.

Let $G$ be any non-abelian group and $e$ the exponent of $G$ (so the least positive integer such that $g^e=1$ for all $g \in G$). Then $(ab)^e=1=a^eb^e$ and $(ab)^{e+1} = ab=a^{e+1}b^{e+1}$ for all $a,b \in G$.

Sebastian Schoennenbeck
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4

Three is enough.

First we have that $$ a^{i+2} b^{i+2} = (ab)^{i+2} = (ab)^{i+1} (ab) = a^{i+1} b^{i+1} ab, $$ that is, $$ a^{i+2} b^{i+2} = a^{i+1} b^{i+1} ab, $$ implying that $$ a b^{i+1} = b^{i+1} a,$$ meaning that $i+1$ powers commute with all elements.

Now, $$ a^{i+2} b^{i+2} = (ab)^{i+2} = (ab)^{i} (ab)^2 = a^i b^i (ab)^2 = a^i b^i (ab)(ab) = a^i b^i ab ab , $$ that is, $$ a^{i+2} b^{i+2} = a^i b^i a b a b, $$ from which it follows that $$ a^2 b^{i+1} = b^i a b a.$$

Reorder the terms on the LHS and conclude $$ b^{i+1} a^2 = b^i a b a, $$ so that $$ b a = ab, $$ as required.

Saaqib Mahmood
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darko
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We have:

$$(ab)^iab=(ab)^{i+1}=a^{i+1}b^{i+1}=aa^ib^ib=a(ab)^ib$$

Right-cancelling with $b^{-1}$, we get:

$$(ab)^ia=a(ab)^i$$

By a similar argument (starting with $ab(ab)^i$), we have that $b$ commutes with $(ab)^i$. This will be true in any group where $g^i=e$ for all $g$. However, this is true in any finite group, Abelian or not, if we set $i=k|G|$ for any integer $k$, for example.

G Tony Jacobs
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  • If we assume $G$ is any group (finite or infinite), then the statement isn't true, it's only true with the finite group with $i$ is an exponent of $G$. – Minh Oct 10 '17 at 14:40
  • Good point. I was clearly thinking of finite groups. – G Tony Jacobs Oct 10 '17 at 14:41