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By analogy with the abelian groups, are there groups $G$ such that: $$\forall x,y\in G:x\bar ky=y\bar kx$$ for some nontrivial $\bar k\in G$?

citadel
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    By setting $x=e$ you can see that we would require $\bar{k} \in Z(G)$. But then we would have $x\bar{k}y = y\bar{k}x \iff \bar{k}xy = \bar{k}yx \iff xy = yx$ for all $x,y \in G$. So in fact $G$ must be abelian. – Dylan Jul 13 '23 at 13:40
  • $(\forall x,y)(\exists \bar{k})$ or $(\exists \bar{k})(\forall x,y)$? – Jeremy Rickard Jul 13 '23 at 14:23
  • The second one, @JeremyRickard – citadel Jul 13 '23 at 14:36

1 Answers1

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If $z \in G$, non-trivial and $xzy=yzx$ for all $x,y \in G$, then (take $y=1$) $z \in Z(G)$ and hence $xy=yx$ for all $x,y \in G$, that is $G$ is abelian...

Nicky Hekster
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