# 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]
2 If $g \in K$, then $g^{-1} \in K$ [SHOW MORE]
3 If $g_1,g_2 \in K$, then $g_1g_2 \in K$. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]