Problem 9 from Herstein's book Suppose that $H$ is a subgroup of $G$ such that whenever $Ha\neq Hb$ then $aH\neq bH$. Prove that $gHg^{-1}\subset H$ for all $g\in G$.
Remark: Honestly, I had some problems. Firstly, after some thoughts I have realized that condition on $H$ is equiavalent to the following: If $aH=bH$ then $Ha=Hb$.
I have found the duplication of that problem in this forum and it is stating that condition on $H$ implies that if $a^{-1}b\in H$ $\Rightarrow$ $ba^{-1}\in H$. I was not able to derive it. Please help how to do it.
Suppose we proved above implication. Taking $x\in gHg^{-1}$ $\Rightarrow$ $x=ghg^{-1}$ for some $h\in H$ $\Rightarrow$ $g^{-1}xg=h\in H$. And taking $a=g$ and $b=xg$ it follows that $ba^{-1}=xgg^{-1}=x\in H$. Thus $gHg^{-1}\subset H$.
Also I can not understand one moment: $a$ and $b$ are certain elements. Is it normal to put $g$ and $xg$ instead of $a$ and $b$, respectively?
Please explain these two questions.

Here is one part:

$aH=bH \implies a^{-1}b\in H $

Indeed, $b=be \in bH=aH \implies b=ah$ for some $h \in H$ and so $a^{-1}b=h\in H $.

$a^{-1}b\in H \implies ba^{-1}\in H$

Indeed, $a^{-1}b=h \in H \implies b^{-1}a=(a^{-1}b)^{-1}=h^{-1} \in H $.
