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

If $H$ be a subgroup of a group $G$ such that $Ha \not=Hb$ implies that $aH\not=bH$.Then how can I show that $gHg^{-1}\subset H$ $\forall$ $g\in G$?

I do not think I have made any progress.However, this is what I have done: For any $h\in H$,$(ghg^{-1})(gh^{-1}g^{-1})=e$ where $e$ is the identity element of the group $G$.

And for $h,k\in H$, $(ghg^{-1})(gkg^{-1})=g(hk)g^{-1}$ which is in $gHg^{-1}$ as $hk\in H$. Thus $gHg^{-1}$ is a group.

I do not think I am getting anywhere.

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Hint: by contraposition, the hypothesis on $H$ is equivalent to $aH = bH \Rightarrow Ha = Hb$. This in turn is equivalent to $b^{-1}a \in H \Rightarrow ba^{-1} \in H$. Now pick the right choice of $b$ and $a$ to show that $h \in H \Rightarrow ghg^{-1} \in H$. –  user29743 Aug 13 '12 at 4:15
It looks like I brushed this approach aside.How weird of me .Thanks for the hint. –  user37450 Aug 13 '12 at 5:35

Using countinghaus's hint, let $b^{-1}= g^{-1}$ and $a = gh$. Then $h = b^{-1}a \in H$ means $ba^{-1} = gh^{-1}g^{-1} \in H$, and since $H$ is a subgroup, $(gh^{-1}g^{-1})^{-1} = ghg^{-1} \in H$. It's actually the case that $gHg^{-1} = H$ for every $g$, since if $gHg^{-1} \subset H$, by letting $y = g^{-1}$, we get $yHy^{-1} \subset H \subset y^{-1}Hy =gHg^{-1}$.

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By the assumption, $aH = bH$ implies $Ha = Hb$.

Conversly suppose $Ha = Hb$. Then $a^{-1}H = b^{-1}H$. Hence $Ha^{-1} = Hb^{-1}$ by the assumption. Hence $aH = bH$.

Now suppose $b \in aH$. Then $aH = bH$. Hence $Ha = Hb$ by the assumption. Hence $b \in Ha$. Hence $aH \subset Ha$.

Conversely suppose $b \in Ha$. Then $Ha = Hb$. Hence $aH = bH$ by the above. Hence $b \in aH$. Hence $Ha \subset aH$.

Therefore $Ha = aH$. Hence $aHa^{-1} = H$ as desired.

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Thanks.I looked up your profile.Did you actually work as a worker once? In that case it is amazing. –  user37450 Aug 13 '12 at 5:35