Notation: If $G$ is a countable group and $H$ is a subgroup, for $g\in G$, let $|\mathcal{O}\_{gH}|$ be the size of the $H$-orbit of $gH$ in the $H$-set $G/H$. 


Does there exist a countable group $G$ and a subgroup $H$ with $[G\colon H]=\infty$ such that:


1. There is a $g\in G$ with $|\mathcal{O} \_{gH}|\neq |\mathcal{O} \_{g^{-1}H}|$,
2. $|\mathcal{O} \_{gH}|=\infty$ if and only if $|\mathcal{O} \_{g^{-1}H}|=\infty$, and
3. There is a constant $M>0$ such that $\displaystyle \frac{|\mathcal{O} \_{gH}|}{|\mathcal{O} \_{g^{-1}H}|} \leq M$ for all $g\in G$ with $|\mathcal{O} \_{gH}|,|\mathcal{O} \_{g^{-1}H}|\neq \infty$.


For instance, 2 and 3 above would be satisfied if:


* If $|\mathcal{O} \_{gH}|\neq |\mathcal{O} \_{g^{-1}H}|$, then $0< |\mathcal{O} \_{gH}|,|\mathcal{O} \_{g^{-1}H}|\leq M$ for some $M>1$ independent of $g$.



Does such an infinite index subgroup exist?
I believe (1 and 2) and (3) are mutually exclusive. Here is a proof:


First, the commensurator 
$$
Comm\_G(H) = \{g\in G : |\mathcal{O} \_{gH}|, |\mathcal{O} \_{g^{-1}H}|<\infty\}
$$ 
is a group. We will show:



> 
> Lemma: $\varphi\colon Comm\_G(H)\to \mathbb{Q}\_{>0}$ by 
>  $g\mapsto \displaystyle\frac{[H\colon H\cap gHg^{-1}]}{[H\colon H\cap g^{-1}Hg]}$
>  is a homomorphism.
> 
> 
> 


From the lemma, if we assume there is a $g\in Comm\_G(H)$ such that $\varphi(g)=x>1$ (criteria 1 and 2), then the order of $g$ must be infinite, since $x>1$ implies $x^n>1$ for all $n\geq 1$. Since the order of $g$ is infinite, criterion 3 cannot hold since eventually $\varphi(g^n)=\varphi(g)^n=x^n>M$ for any $M>0$.


Proof of the lemma:


We must show $\varphi(g\_1g\_2)=\varphi(g\_1)\varphi(g\_2)$. Define the following constants:


* For $i=1,2$, $a\_i = [H\colon H\cap g\_iHg\_i^{-1}]$ and $b\_i = [H\colon H\cap g\_i^{-1}Hg\_i]=[g\_iHg\_i^{-1}\colon H\cap g\_iHg\_i^{-1}]$
* $a=[H\colon H\cap (g\_1g\_2)H(g\_1g\_2)^{-1}]$ and $b=[(g\_1g\_2)H(g\_1g\_2)^{-1}\colon H\cap (g\_1g\_2)H(g\_1g\_2)^{-1}]$


Note that since $x\mapsto g\_1xg\_1^{-1}$ is an automorphism of $G$, we have:


* $a\_2=[g\_1Hg\_1^{-1}\colon g\_1Hg\_1^{-1}\cap (g\_1g\_2)H(g\_1g\_2)^{-1}]$ and $b\_2=[(g\_1g\_2)H(g\_1g\_2)^{-1}\colon g\_1Hg\_1^{-1}\cap (g\_1g\_2)H(g\_1g\_2)^{-1}]$


Now look at the subgroup $K=H\cap g\_1 Hg\_1^{-1}\cap (g\_1g\_2)H(g\_1g\_2)^{-1}$, and define


* $a\_1'=[H\cap (g\_1g\_2)H(g\_1g\_2)^{-1}\colon K]$
* $a\_2'=[H\cap g\_1Hg\_1^{-1}\colon K]$
* $b\_1'=[g\_1Hg\_1^{-1}\cap (g\_1g\_2)H(g\_1g\_2)^{-1}\colon K]$


which are all finite, since if we have a quadrilateral of groups $L\_1\cap L\_2\subset L\_1,L\_2\subset G$, we must have $[L\_1\colon L\_1\cap L\_2] \leq [G\colon L\_2]$. Now since index is multiplicative, we have


* $a a\_1'=a\_1a\_2'$
* $ba\_1' = b\_2 b\_1'$
* $a\_2b\_1'=b\_1a\_2'$


Solving for $a$ and $b$, we get
$$
\frac{a}{b} = \frac{a\_1a\_2'}{a\_1'}\frac{a\_1'}{b\_2b\_1'}=\frac{a\_1a\_2'}{b\_2b\_1'}.
$$
Now note that $\displaystyle \frac{a\_2'}{b\_1'}=\frac{a\_2}{b\_1}$, so we have
$$
\varphi(g\_1g\_2)=\frac{a}{b} = \frac{a\_1}{b\_2}\frac{a\_2}{b\_1}= \frac{a\_1}{b\_1}\frac{a\_2}{b\_2}=\varphi(g\_1)\varphi(g\_2).
$$


