Ler $G $ be a group and $X $ a non empty subset of $G $ such that for all $g\in G $ we have $XgX=Xg$ . How can we show that $Xg=gX $, for all $g\in G $?
A relation in group theory
3
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group-theory
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0The group is not assumed to be *finite*, right? – 2017-02-28
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0No, $G $ is an arbitrary group. – 2017-02-28
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0Observations: We have $Xg\cdot Xh = XgX\cdot h =Xgh$. so $Q:=\{\,Xg\mid g\in G\,\}$ is a group with $X$ as neutral element and $Xg^{-1}$ as inverse of $Xg$, and $g\mapsto Xg$ is a group epimorphism $\phi\colon G\to Q$. In particular, $XX=X$ and so $x\in X$ implies $x^n\in X$. – 2017-02-28