$H$ is normal whenever $Ha\not = Hb \implies aH\not =bH$ 
Topics in Algebra- Hernstein pg-47,Q.9
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 $gHg^{-1}\subset H$ $\forall$ $g\in G$?

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.
 A: Using @user29743 hint, let   $a = bh$.  Then $h = b^{-1}a \in H$ means $ba^{-1} = bh^{-1}b^{-1} \in H$, and since $H$ is a subgroup, $(bh^{-1}b^{-1})^{-1} = bhb^{-1} \in H$. Replacing $b$ as $g$ we have: $ 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 $yHy^{-1} \subset H \subset y^{-1}Hy =gHg^{-1}$.
A: 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.
A: The existing proofs are actually unnecessarily complicated. There is no need for non-constructive (e.g. by contrapositive/contradiction) proofs, nor any equivalences, nor so many inversions...
For every $g∈G$ and $h∈H$, we have $ghH = gH$ and so $Hgh = Hg$, and hence $ghg^{-1} ∈ Hghg^{-1}$ $= Hgg^{-1} = H$. Therefore $G ◁ H$.
Tada done in one single line...
A: $Ha\neq Hb  \Rightarrow aH \neq bH$ is equivalent to  $aH=bH \Rightarrow Ha=Hb$. This means $b^{−1}a \in  H$ implies $ba^{−1} \in H$. Now let  h∈ H and g ∈ G. Take $b^{-1}= g^{-1}$ and $a = gh$. Then $h = g^{-1}gh = b^{−1}a \in H$. Then  $ba^{−1}\in H$ (given). Since $  (ab^{-1})^{-1} = ba^{−1}$, $(ab^{-1})^{-1} \in  H$, for $H$ is a subgroup. Then $(ghg^{-1})^{-1} = gh^{-1}g^{-1}\in H \Rightarrow (gh^{-1}g^{-1})^{-1} \in H \Rightarrow ghg^{-1} \in H, ~\forall ~g \in G$ and $\forall ~ h \in H$. Therefore $gHg^{-1} \subseteq H$, $\forall ~ g \in G$.
