Given an example of a non-abelian group $G$ satisfying $(a \cdot b)^i=a^i \cdot b^i, \forall a, b \in G$ for two consecutive integers.
This is question 5 from Herstein Page 35. I have proved that conclusion holds for three positive integers.Thanks.
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Krish
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Taylor Ted
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1If $(ab)^i=a^i b^i$ and $(ab)^{i+1}=a^{i+1}b^{i+1}$, then $ab=b^{-i}ab^{i+1}$, and so $ab^i=b^ia$. Therefore, every $i$th power must be in the center of the group. If you are trying to construct a group with this property, it will be easiest to assume that $a^i=e$ for every $a$. – Aaron Mar 21 '15 at 12:34
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4Take the consecutive integers to be $i=0$ and $i=1$. – Marc van Leeuwen Aug 09 '17 at 12:40
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This is actually possible for any finite nonabelian group. To see this, suppose $G$ is a finite nonabelian group with $|G|=n$. Then the order of any element divides $n$, hence for any element $x\in G$ we have $x^n=e$. Thus for $a,b\in G$ we have $(ab)^n=a^nb^n=e$. Furthermore, $a^{n+1}=a$, $b^{n+1}=b$, $(ab)^{n+1}=ab=a^{n+1}b^{n+1}$.
Matt Samuel
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Order of every element divides order of group ? How ? and how come u have written $x^{n}$=e – Taylor Ted Mar 22 '15 at 04:36