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$gHg^{-1}\subset H$ 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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marked as duplicate by Jack Schmidt, sdcvvc, William, wentaway, t.b. Sep 7 '12 at 14:07

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i have proved by contrapositive of given statement i.e if $aH = bH$ then $ Ha = Hb$. But i want some other point of view. please help in this. – White Dwarf Sep 1 '12 at 15:51
$aH=bH$ iff $a^{-1}b\in H$ – i. m. soloveichik Sep 1 '12 at 16:01
@WhiteDwarf: I am not sure that I understand what it is that you have already proved. – Thomas Sep 1 '12 at 17:08
up vote 2 down vote accepted

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 $gHg^{-1} \subset H$.

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