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 $H \cap gHg^{-1}$ is a subgroup of finite index in both $H$ and $gHg^{-1}$, i.e., $H$ and $gHg^{-1}$ 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 $H \cap gHg^{-1}$ has finite index in both $H$ and $gHg^{-1}$.

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

Proof:

Step no.	Assertion/construction	Explanation
1	The identity element $e$ of $G$ is in $K$	[SHOW MORE] $eHe^{-1} = H$, so the intersection $H \cap eHe^{-1}$ also equals $H$. This has index $1$ in both $H$ and $eHe^{-1}$, which is finite.
2	If $g \in K$, then $g^{-1} \in K$	[SHOW MORE] 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 $\sigma(gHg^{-1}) = H$. Thus, we get $\sigma(H \cap gHg^{-1}) = g^{-1}Hg \cap H$. Since the index of $H \cap gHg^{-1}$ in both $H$ and $gHg^{-1}$ is finite, so is the index of $\sigma(H \cap gHg^{-1}) = g^{-1}Hg \cap H$ in both $\sigma(H) = g^{-1}Hg$ and $\sigma(gHg^{-1}) = H$.
3	If $g_1,g_2 \in K$, then $g_1g_2 \in K$.	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.