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Possible Duplicate:
$gHg^{-1}\subset H$ whenever $Ha\not = Hb$ implies $aH\not =bH$

Suppose that $H$ is a subgroup of $G$ such that whenever $H\circ a\neq H\circ b $ then $ a\circ H\neq b\circ H$.
Prove that $ g\circ H\circ g^{-1} \subset H$, $\forall g \in G$.

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    @WhiteDwarf: I am not sure that I understand what it is that you have already proved.2012-09-01

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Let $g \in G$. Suppose $x \in gH$. Then $gH = xH$. By the assumption, $Hg = Hx$. Hence $x \in Hg$. Hence $gH \subset Hg$. Hence $gHg^{-1} \subset H$.