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 $G$.

Facts used

1. Group acts as automorphisms by conjugation
2. Index satisfies transfer inequality
3. Index is multiplicative

Proof

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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

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Consider the map $\sigma :G\to G$ given by $\sigma (x):=g^{-1}xg$. This is an inner automorphism of $G$	Fact (1)	--	--
2	$\sigma$ preserves intersections and index of subgroups	Follows from definition of automorphism		Step (1)
3	$\sigma (H)=g^{-1}Hg$ and $\sigma (gHg^{-1})=H$.			Step (1)
4	$\sigma (H\cap gHg^{-1})=g^{-1}Hg\cap H$.			Steps (2), (3)
5	$H\cap gHg^{-1}$ has finite index in both $H$ and $gHg^{-1}$.		$g\in K$, the commensurator of $H$.
6	$\sigma (H\cap gHg^{-1})=g^{-1}Hg\cap H$ has finite index in both $\!\sigma (H)=g^{-1}Hg$ and $\!\sigma (gHg^{-1})=H$.			Steps (2), (3), (4), (5)	[SHOW MORE] We apply $\sigma$ to the statement of Step (5), which is permissible because automorphisms preserve index of subgroups. We use Steps (3) and (4) to compute what the images of the three subgroups under $\sigma$.
7	$g^{-1}$ is in $K$.		definition of $K$	Step (6)	Step-definition direct.

Proof for products

Additional given: $g_{1},g_{2}\in K$

To prove: $g_{1}g_{2}\in K$, i.e., $H\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ has finite index in both $H$ and in $(g_{1}g_{2})H(g_{1}g_{2})^{-1}$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Consider the map $\tau :G\to G$ given by $\tau (x)=g_{1}xg_{1}^{-1}$. Then, $\tau$ is an inner automorphism and in particular an automorphism of $G$.	Fact (1)
2	$\tau$ preserves intersections and index of subgroups	follows from definition of automorphism		Step (1)
3	$\tau (H)=g_{1}Hg_{1}^{-1}$ and $\tau (g_{2}Hg_{2}^{-1})=g_{1}g_{2}Hg_{2}^{-1}g_{1}^{-1}=(g_{1}g_{2})H(g_{1}g_{2})^{-1}$.			Step (1)
4	$\tau (H\cap g_{2}Hg_{2}^{-1})=g_{1}Hg_{1}^{-1}\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$.			Steps (2), (3)
5	$H\cap g_{2}Hg_{2}^{-1}$ has finite index in both $\!H$ and $\!g_{2}Hg_{2}^{-1}$.		$g_{2}\in K$, definition of $K$		Given-direct
6	$g_{1}Hg_{1}^{-1}\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ has finite index in (i) $\!g_{1}Hg_{1}^{-1}$ and (ii) $\!(g_{1}g_{2})H(g_{1}g_{2})^{-1}$.			Steps (2), (3), (4), (5)	[SHOW MORE] We apply $\tau$ to both sides of Step (5), permissible by Step (2), and then simplify using Steps (3) and (4).
7	$H\cap (g_{1}Hg_{1}^{-1}\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1})$ has finite index in (i) $H\cap g_{1}Hg_{1}^{-1}$ and (ii) $H\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$.	Fact (2)		Step (6)	[SHOW MORE] Intersect all terms of Step (6) with $H$ and apply Fact (2).
8	$H\cap g_{1}Hg_{1}^{-1}$ has finite index in (i) $\!H$ and in (ii) $g_{1}Hg_{1}^{-1}$		$g_{1}\in K$, definition of $K$		Given-direct
9	$H\cap (g_{1}Hg_{1}^{-1}\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1})$ has finite index in $\!H$.	Fact (3)		Steps (7)(i), (8)(i)	[SHOW MORE] Combine Steps (7)(i) and (8)(i) with Fact (3) for the chain $H\cap (g_{1}Hg_{1}^{-1}\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1})\leq H\cap g_{1}Hg_{1}^{-1}\leq H$
10	$H\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ has finite index in $\!H$	Fact (3)		Step (9)	[SHOW MORE] Follows from Step (9), when we note that $H\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ is an intermediate subgroup between $H\cap (g_{1}Hg_{1}^{-1}\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1})$ and $H$.
11	$(H\cap g_{1}Hg_{1}^{-1})\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ has finite index in (i) $H\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ and (ii) $g_{1}Hg_{1}^{-1}\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$	Fact (2)		Step (8)	[SHOW MORE] Intersect all terms of Step (8) with $(g_{1}g_{2})H(g_{1}g_{2})^{-1}$.
12	$(H\cap g_{1}Hg_{1}^{-1})\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ has finite index in $\!(g_{1}g_{2})H(g_{1}g_{2})^{-1}$	Fact (3)		Steps (6)(ii), (11)(ii)	[SHOW MORE] Combine Steps (11)(ii) and (6)(ii) with Fact (4) for the chain $(H\cap g_{1}Hg_{1}^{-1})\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}\leq g_{1}Hg_{1}^{-1}\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}\leq (g_{1}g_{2})H(g_{1}g_{2})^{-1}$.
13	$H\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ has finite index in $\!(g_{1}g_{2})H(g_{1}g_{2})^{-1}$	Fact (3)		Step (12)	[SHOW MORE] Follows from Step (12), when we note that $H\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ is an intermediate subgroup between $(H\cap g_{1}Hg_{1}^{-1})\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ and $(g_{1}g_{2})H(g_{1}g_{2})^{-1}$.
14	$g_{1}g_{2}\in K$, i.e., $H\cap (g_{1}g_{2})H(g_{1}g_{2})^{-1}$ has finite index in both $H$ and $\!(g_{1}g_{2})H(g_{1}g_{2})^{-1}$.			Steps (10), (13)	Step-combination direct.