Exercise 3.5.2 (Center of the modular group)

Proof.

(a)

Here $(G,\cdot )$ is a group, and $$C=\{c\in G\mid \forall <!--\; FUNCTION\; APPLICATION\; -->g\in G,\mathit{cg}=\mathit{gc}\}.$$

• For all $g\in G$, $\mathit{eg}=\mathit{ge}=g$, thus $e\in C$, so $G\ne \varnothing$.

• If $c,d\in C$, then for all $g\in G$,

  $$g(\mathit{cd})=(\mathit{gc})d=(\mathit{cg})d=c(\mathit{gd})=c(\mathit{dg})=(\mathit{cd})g.$$

  This shows that $\mathit{cd}\in C$.

• If $c\in C$, then for all $g\in G$, $\mathit{cg}=\mathit{gc}$, thus $c^{-1}\mathit{cg}=c^{-1}\mathit{gc}$, so $g=c^{-1}\mathit{gc}$, and $g{c}^{-1}=c^{-1}\mathit{gc}{c}^{-1}=c^{-1}g$. This proves that $c^{-1}\in C$.

The center $C$ is a subgroup of $G$.

(b)

Now we compute the center $C$ of $\Gamma ={\mathrm{SL}}_2(\mathbb{Z})$.

Let $M=\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill \gamma \hfill & \hfill \delta \hfill \end{array}\right)\in C$. Then $M$ commute with $S$ and $T$, where

$$S=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right)\in \Gamma ,T=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\in \Gamma .$$

(We can prove that $\Gamma =⟨S,T⟩$.)

$$\begin{array}{llll}\hfill \mathit{MS}=\mathit{SM}& ⟺\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill \gamma \hfill & \hfill \delta \hfill \end{array}\right)\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill \gamma \hfill & \hfill \delta \hfill \end{array}\right)& \hfill & \\ \hfill & ⟺\left(\begin{array}{cc}\hfill -\beta \hfill & \hfill \alpha \hfill \\ \hfill -\delta \hfill & \hfill \gamma \hfill \end{array}\right)=\left(\begin{array}{cc}\hfill \gamma \hfill & \hfill \delta \hfill \\ \hfill -\alpha \hfill & \hfill -\beta \hfill \end{array}\right)& \hfill & \\ \hfill & ⟺\{\begin{array}{c}\delta =\alpha \hfill \\ \gamma =-\beta .\hfill \end{array}& \hfill & \end{array}$$ Therefore $$M=\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill -\beta \hfill & \hfill \alpha \hfill \end{array}\right),$$

and

$$\begin{array}{llll}\hfill \mathit{MT}=\mathit{TM}& ⟺\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill -\beta \hfill & \hfill \alpha \hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill -\beta \hfill & \hfill \alpha \hfill \end{array}\right)& \hfill & \\ \hfill & ⟺\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill \alpha +\beta \hfill \\ \hfill -\beta \hfill & \hfill \alpha -\beta \hfill \end{array}\right)=\left(\begin{array}{cc}\hfill \alpha -\beta \hfill & \hfill \alpha +\beta \hfill \\ \hfill -\beta \hfill & \hfill \alpha \hfill \end{array}\right)& \hfill & \\ \hfill & ⟺\beta =0.& \hfill & \end{array}$$

Therefore $M=\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \alpha \hfill \end{array}\right)$. Since $M\in \Gamma$, $\det <!--\; FUNCTION\; APPLICATION\; -->M=1$, thus ${\alpha }^2=1$, so $\alpha =\pm 1$, and $M=\pm I$.

Conversely $\pm I$ commute with every matrix in $\Gamma$.

This shows that

$$C=\{-I,I\}$$

is the center of the modular group.