Proving Properties and Calculating the Commensurator of Subgroups in $\mathrm{GL}_{2}^{+}(\mathbb{R})$

Question

I have a multi-part question involving commensurability and the commensurator of subgroups in a group $G$ and I would love your suggestions/edits/corrections to my workings!

Question: Let $G$ be a group. For two subgroups $\Gamma_{1}, \Gamma_{2}$ recall what it means for $\Gamma_{1}$ and $\Gamma_{2}$ to be commensurable. Further recall the definition of the commensurator $\widetilde{\Gamma_{1}}$ of $\Gamma_{1}$ in $G$. (These are Definitions 5.1.1 and 5.1.2 in the lecture notes.)

a) Show that being commensurable defines an equivalence relation on subgroups of $G$.

b) Let $\Gamma \subseteq G$ be a subgroup. Show that the commensurator, $\widetilde{\Gamma}$, of $\Gamma$ in $G$ is a subgroup of $G$.

c) Let $G=\mathrm{GL}_{2}^{+}(\mathbb{R})$ and let $\Gamma=\mathrm{SL}_{2}(\mathbb{Z})$. Compute the commensurator of $\Gamma$ in $G .\left(\right.$ Note that $\left.\mathrm{GL}_{2}^{+}(\mathbb{R})=\left\{g \in \mathrm{GL}_{2}(\mathbb{R}): \operatorname{det}(g)>0\right\}.\right)$

My Solutions: (a) Let's recall the definitions of commensurable subgroups and commensurators before proceeding.

Two subgroups $\Gamma_1, \Gamma_2$ of a group $G$ are said to be commensurable if the intersection $\Gamma_1 \cap \Gamma_2$ has finite index in both $\Gamma_1$ and $\Gamma_2$. Here, the index of a subgroup $H$ in a group $G$, denoted $[G: H]$, is the number of left cosets of $H$ in $G$. We are asked to show that being commensurable defines an equivalence relation on subgroups of $G$. Recall that an equivalence relation must satisfy three properties: reflexivity, symmetry, and transitivity. a) Reflexivity: A subgroup $\Gamma_1$ is commensurable with itself because $\Gamma_1 \cap \Gamma_1=\Gamma_1$ and $\Gamma_1$ clearly has finite index in itself. b) Symmetry: If $\Gamma_1$ and $\Gamma_2$ are commensurable, then $\Gamma_1 \cap \Gamma_2$ has finite index in both $\Gamma_1$ and $\Gamma_2$. Therefore, $\Gamma_2$ is also commensurable with $\Gamma_1$. c) Transitivity: If $\Gamma_1$ is commensurable with $\Gamma_2$, and $\Gamma_2$ is commensurable with $\Gamma_3$, then $\Gamma_1 \cap \Gamma_2$ has finite index in $\Gamma_1$ and $\Gamma_2$, and $\Gamma_2 \cap \Gamma_3$ has finite index in $\Gamma_2$ and $\Gamma_3$. Therefore, $\Gamma_1 \cap \Gamma_3=\left(\Gamma_1 \cap \Gamma_2\right) \cap$ $\left(\Gamma_2 \cap \Gamma_3\right)$ has finite index in $\Gamma_1$, since the index $\left[\Gamma_1: \Gamma_1 \cap \Gamma_3\right]$ is at most $\left[\Gamma_1: \Gamma_1 \cap \Gamma_2\right] \cdot\left[\Gamma_1 \cap \Gamma_2: \Gamma_1 \cap \Gamma_3\right]$, both of which are finite. Similarly, $\Gamma_1 \cap \Gamma_3$ has finite index in $\Gamma_3$. So, $\Gamma_1$ is commensurable with $\Gamma_3$. Thus, commensurability is an equivalence relation on the set of subgroups of $G$.

The commensurator $\widetilde{\Gamma_1}$ of $\Gamma_1$ in $G$ is defined as the set of all elements $g \in G$ such that $g \Gamma_1 g^{-1}$ is commensurable with $\Gamma_1$.

(b) To prove that the commensurator $\widetilde{\Gamma}$ of a subgroup $\Gamma$ in $G$ is a subgroup of $G$, we need to show that it satisfies the following properties:

1. Identity: The identity element $e \in G$ must be in $\widetilde{\Gamma} .2$. Closure: If $g_1, g_2 \in \widetilde{\Gamma}$, then their product $g_1 g_2$ must also be in $\widetilde{\Gamma}$. 3 . Inverses: For every $g \in \widetilde{\Gamma}$, its inverse $g^{-1}$ must also be in $\widetilde{\Gamma}$. We proceed to demonstrate these three properties:
2. Identity: By definition, $e \Gamma e^{-1}=\Gamma$, and $\Gamma$ is commensurable with itself, so $e \in \widetilde{\Gamma}$.
3. Closure: Suppose $g_1, g_2 \in \widetilde{\Gamma}$. This means $g_1 \Gamma g_1^{-1}$ and $g_2 \Gamma g_2^{-1}$ are commensurable with $\Gamma$. We have $$ \left(g_1 g_2\right) \Gamma\left(g_1 g_2\right)^{-1}=g_1\left(g_2 \Gamma g_2^{-1}\right) g_1^{-1} $$ Since $g_1 \Gamma g_1^{-1}$ is commensurable with $\Gamma$, and conjugation is an automorphism (i.e., it preserves the group structure), it follows that $g_1\left(g_2 \Gamma g_2^{-1}\right) g_1^{-1}$ is commensurable with $\Gamma$. Thus, $g_1 g_2 \in \widetilde{\Gamma}$.
4. Inverses: Suppose $g \in \widetilde{\Gamma}$. Then, $g \Gamma g^{-1}$ is commensurable with $\Gamma$. We want to show that $g^{-1} \Gamma g$ is also commensurable with $\Gamma$. Notice that $g^{-1} \Gamma g=\left(g \Gamma g^{-1}\right)^{-1}$. Since inverses and taking the inverse are both automorphisms, we conclude that $g^{-1} \Gamma g$ is commensurable with $\Gamma$. Therefore, $g^{-1} \in \widetilde{\Gamma}$. So, the commensurator $\widetilde{\Gamma}$ satisfies all properties of a subgroup and is thus a subgroup of $G$.

(c) The group $\mathrm{GL}_2^{+}(\mathbb{R})$ is the group of $2 \times 2$ real matrices with positive determinant, and $\mathrm{SL}_2(\mathbb{Z})$ is the special linear group of $2 \mathrm{x} 2$ integer matrices with determinant 1 .

To compute the commensurator of $\Gamma=\mathrm{SL}_2(\mathbb{Z})$ in $G=\mathrm{GL}_2^{+}(\mathbb{R})$, we want to find all $g \in G$ such that $g \Gamma g^{-1}$ is commensurable with $\Gamma$.

Let $g=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) \in G$, so that $g \Gamma g^{-1}$ consists of all matrices of the form $g \gamma g^{-1}$ for $\gamma \in \Gamma$. We can compute this explicitly as: $$ g\left(\begin{array}{cc} m & n \\ p & q \end{array}\right) g^{-1}=\frac{1}{a d-b c}\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)\left(\begin{array}{cc} m & n \\ p & q \end{array}\right)\left(\begin{array}{cc} d & -b \\ -c & a \end{array}\right) $$ for all $m, n, p, q \in \mathbb{Z}$ with $m q-n p=1$. The resulting group $g \Gamma g^{-1}$ is commensurable with $\Gamma$ if and only if it is a discrete subgroup of $\mathrm{GL}_2^{+}(\mathbb{R})$ (the reason being that $\Gamma$ itself is a discrete subgroup, and two discrete subgroups are commensurable if and only if they have a common discrete subgroup, which must be of finite index in each).

It is a classic result in the theory of modular forms that $g \Gamma g^{-1}$ is a discrete subgroup of $\mathrm{GL}_2^{+}(\mathbb{R})$ if and only if the entries of $g$ are all rational numbers and $\operatorname{det}(g)= \pm 1$. Therefore, the commensurator of $\Gamma$ in $G$ is the group $$ \widetilde{\Gamma}=\left\{g=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right) \in \mathrm{GL}_2^{+}(\mathbb{R}): a, b, c, d \in \mathbb{Q} \text { and } a d-b c= \pm 1\right\} . $$