# 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

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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]
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: $g_1g_2 \in K$, 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:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Consider the map $\tau:G \to G$ given by $\tau(x) = g_1xg_1^{-1}$. 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 $\tau(H) = g_1Hg_1^{-1}$ and $\tau(g_2Hg_2^{-1}) = g_1g_2Hg_2^{-1}g_1^{-1} = (g_1g_2)H(g_1g_2)^{-1}$. Step (1)
4 $\tau(H \cap g_2Hg_2^{-1}) = g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1}$. Steps (2), (3)
5 $H \cap g_2Hg_2^{-1}$ has finite index in both $\! H$ and $\! g_2Hg_2^{-1}$. $g_2 \in K$, definition of $K$ Given-direct
6 $g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1}$ has finite index in (i) $\! g_1Hg_1^{-1}$ and (ii) $\! (g_1g_2)H(g_1g_2)^{-1}$ Steps (2), (3), (4), (5) [SHOW MORE]
7 $H \cap (g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1})$ has finite index in (i) $H \cap g_1Hg_1^{-1}$ and (ii) $H \cap (g_1g_2)H(g_1g_2)^{-1}$. Fact (2) Step (6) [SHOW MORE]
8 $H \cap g_1Hg_1^{-1}$ has finite index in (i) $\!H$ and in (ii) $g_1Hg_1^{-1}$ $g_1 \in K$, definition of $K$ Given-direct
9 $H \cap (g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1})$ has finite index in $\! H$. Fact (3) Steps (7)(i), (8)(i) [SHOW MORE]
10 $H \cap (g_1g_2)H(g_1g_2)^{-1}$ has finite index in $\! H$ Fact (3) Step (9) [SHOW MORE]
11 $(H \cap g_1Hg_1^{-1}) \cap (g_1g_2)H(g_1g_2)^{-1}$ has finite index in (i) $H \cap (g_1g_2)H(g_1g_2)^{-1}$ and (ii) $g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1}$ Fact (2) Step (8) [SHOW MORE]
12 $(H \cap g_1Hg_1^{-1}) \cap (g_1g_2)H(g_1g_2)^{-1}$ has finite index in $\! (g_1g_2)H(g_1g_2)^{-1}$ Fact (3) Steps (6)(ii), (11)(ii) [SHOW MORE]
13 $H \cap (g_1g_2)H(g_1g_2)^{-1}$ has finite index in $\! (g_1g_2)H(g_1g_2)^{-1}$ Fact (3) Step (12) [SHOW MORE]
14 $g_1g_2 \in K$, i.e., $H \cap (g_1g_2)H(g_1g_2)^{-1}$ has finite index in both $H$ and $\! (g_1g_2)H(g_1g_2)^{-1}$. Steps (10), (13) Step-combination direct.