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 fact about::commensurable subgroups. Then, $$K$$ is a subgroup of $$G$$.

Facts used

 * 1) uses::Group acts as automorphisms by conjugation
 * 2) uses::Index satisfies transfer inequality
 * 3) uses::Index is multiplicative

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:

Proof for identity element
Let $$e$$ denote the identity element of $$G$$. We have $$eHe^{-1} = H$$, so the intersection $$H \cap eHe^{-1}$$ also equals $$H$$. This has index $$1$$ in both $$H$$ and $$eHe^{-1}$$, which is finite.

Proof for inverses
Additional given: $$g \in K$$. In other words, $$H \cap gHg^{-1}$$ has finite index in both $$H$$ and $$gHg^{-1}$$.

To prove: $$g^{-1} \in K$$, i.e., $$H \cap g^{-1}Hg$$ has finite index in both $$H$$ and $$g^{-1}Hg$$.

Proof

Proof for products
Additional given: $$g_1, g_2 \in K$$

To prove: $$g_1g_2 \in K$$, i.e., $$H \cap (g_1g_2)H(g_1g_2)^{-1}$$ has finite index in both $$H$$ and in $$(g_1g_2)H(g_1g_2)^{-1}$$.

Proof: