Commensurator of subgroup is subgroup

Statement

Suppose $H$ is a subgroup of a group $G$. Consider the commensurator $K$ of $H$ in $G$, defined as the set of all $g\in G$ such that $H\cap gH{g}^{-1}$ is a subgroup of finite index in both $H$ and $gH{g}^{-1}$, i.e., $H$ and $gH{g}^{-1}$ are Commensurable subgroups (?). Then, $K$ is a subgroup of $G$.

Proof

Given: A group $G$, a subgroup $H$ of $G$. $K$ is the set of all $g\in G$ such that $H\cap gH{g}^{-1}$ has finite index in both $H$ and $gH{g}^{-1}$.

To prove: $K$ is a subgroup of $G$.

Proof:

Step no.	Assertion/construction	Explanation
1	The identity element $e$ of $G$ is in $K$	$eH{e}^{-1}=H$, so the intersection $H\cap eH{e}^{-1}$ also equals $H$. This has index $1$ in both $H$ and $eH{e}^{-1}$, which is finite.
2	If $g\in K$, then $g^{-1}\in K$	Consider the map $\sigma :G\to G$ given by $x\mapsto g^{-1}xg$. This is an inner automorphism of $G$, and hence preserves intersections and the index of subgroups. We have $\sigma (H)=g^{-1}Hg$ and $\sigma (gH{g}^{-1})=H$. Thus, we get $\sigma (H\cap gH{g}^{-1})=g^{-1}Hg\cap H$. Since the index of $H\cap gH{g}^{-1}$ in both $H$ and $gH{g}^{-1}$ is finite, so is the index of $\sigma (H\cap gH{g}^{-1})=g^{-1}Hg$ in both $\sigma (H)=g^{-1}Hg$ and $\sigma (gH{g}^{-1})=H$.
3	If $g_1,g_2\in K$, then $g_1{g}_2\in K$.	PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.