# If $aH=bH$ implies $Ha=Hb$ then $aHa^{-1}=H$ for all $a\in G$.

The Question:

If $aH=bH$ forces $Ha=Hb$ in a group G, show that $aHa^{-1}=H$ for all $a \in G$.

My Attempt:

I understand that $H$ must be a normal subgroup and that normal subgroups are closed under conjugation, but I don't know how to explicitly show this.

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Choose $h\in H$ and $g\in G$. Then $gH = gh^{-1}H$ trivially, so $Hg = Hgh^{-1}$ by the hypothesis. So by the definition of these sets being equal there is some $h'\in H$ so that $g = h'gh^{-1}$, which is equivalent to $ghg^{-1} = h' \in H$.
And this exactly means that $gHg^{-1} = H$ for every $g\in G$.
We are given that $aH=bH$ implies that $Ha=Hb$, however we can equivalently write that $b^{-1}a\in H$ implies that $ba^{-1}\in H$. (why?) Let $h\in H$, take any $a\in G$, and let $b=ah$. Then $b^{-1}a = h^{-1} \in H$, so by hypothesis $ba^{-1}=aha^{-1}\in H$. Thus $aha^{-1}\in H$ for any $a\in G$.