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.
Best Answer
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 $g\in G$.