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.
 A: 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 $.
A: First of all

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?

$a$ and $b$ are not certain, they arbitrary. This can be deduced from the statement "whenever $Ha\neq Hb$ then $aH\neq bH$". Also (what is more important) the statement is not true if $a,b$ are fixed. For example if $a=b$ then $aH=bH$ and $Ha=Hb$ while $gHg^{-1}\subseteq H$ does not have to hold (indeed, the stament holds for all $g$ if and only if $H$ is normal).
Now for the proof:

Lemma 1. Let $G$ be a group, $H$ a subgroup and $a\in G$. Then $aH=H$ if and only if $a\in H$ which is if and only if $Ha=H$.

Proof.


*

*"$aH=H\ \Rightarrow a\in H$"
Assume that $aH=H$. In particular for any $h\in H$ we have $ah\in H$. In particular for $e\in H$ we have $a=ae\in H$.

*"$a\in H\Rightarrow aH=H$" The inclusion $aH\subseteq H$ follows simply from the fact that $H$ is a subgroup. For the other inclusion, assume that $h\in H$, then $h=aa^{-1}h$ and since $a\in H$, then $a^{-1}\in H$ and therefore $a^{-1}h\in H$. Put $h':=a^{-1}$. Thus $h=ah'$ and so $h\in aH$.


The other two implications are proven analogously and I leave them as an exercise. $\Box$

Lemma 2. Let $G$ be a group, $H$ a subgroup and let $a,b\in H$. Then $aH=bH$ if and only if $b^{-1}aH=H$ and if and only if $H=a^{-1}bH$.

Proof.
"$aH=bH\ \iff b^{-1}aH=H$" The following statements are equivalent:


*

*$aH=bH$

*for all $h\in H$ there is $h'\in H$ such that $ah=bh'$ and for all $k\in H$ there is $k'\in H$ such that $ak'=bk$

*for all $h\in H$ there is $h'\in H$ such that $b^{-1}ah=h'$ and for all $k\in H$ there is $k'\in H$ such that $b^{-1}ak'=k$

*$b^{-1}aH=H$


"$aH=bH\ \iff H=a^{-1}bH$" The following statements are equivalent:


*

*$aH=bH$

*for all $h\in H$ there is $h'\in H$ such that $ah=bh'$ and for all $k\in H$ there is $k'\in H$ such that $ak'=bk$

*for all $h\in H$ there is $h'\in H$ such that $h=a^{-1}bh'$ and for all $k\in H$ there is $k'\in H$ such that $k'=a^{-1}bk$

*$H=a^{-1}bH$


$\Box$
Analogously we prove

Lemma 3. Let $G$ be a group, $H$ a subgroup and let $a,b\in H$. Then $Ha=Hb$ if and only if $H=Hba^{-1}$ and if and only if $Hab^{-1}=H$.

Proof. Just follow the proof of lemma 2. and do appropriate modifications (be careful about the order of multiplication). $\Box$
With these 3 lemmas I'm pretty sure you are able to prove that the following two conditions are equivalent:


*

*$aH=bH\ \Rightarrow\ Ha=Hb$

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

