Solution: We recall that the center of a group $G$ is defined as \[ Z(G) \coloneqq \left\{ g\in G : g g^\prime = g^\prime g \ \text{for all } g^\prime \in G \right\}. \] We have the following important properties for the center of a group.

• A group $ G $ is abelian if and only if $Z(G) = G$.
• If the group is generated by $g_1,g_2,\dots, g_k$, then \[ Z(G) = \left\{ g \in G : g g_j = g_j g\ \text{for } j = 1,2,\dots, k \right\} . \] Since each $g\in G$ can be written as the product of $g_i$'s, we have, if $g_i g_j= g_j g_i$ for all $i,j \in \{ 1,2,\dots, k \} $, then $Z(G) = G$, that is, $G$ is abelian.

Let us find the center of each of these groups. Note that \[ Z \left( G_j \right) = \left\{ A \in GL_2(\mathbb{C} ) : A S = S A \text{ and } A T_j = T_j A ,\ j = 1,2,3 \right\} . \] Let $A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$. Consider the first condition, that is, $AS = SA$. \begin{align*} AS = SA & \implies \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \\[1ex] & \implies \begin{bmatrix} b & -a \\ d & -c \\ \end{bmatrix} = \begin{bmatrix} -c & -d \\ a & b \\ \end{bmatrix} \\[1ex] & \implies b = -c, a = d. \end{align*} Thus, the matrix $A$ reduces to \[ A = \begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix}. \]

Now for each $T_j$, we need the commutative property to be satisfied. Note that, \[ T_1 = \begin{bmatrix} \iota & 0 \\ 0 & \iota \\ \end{bmatrix} = \iota I_2 \implies A T_1 = \iota A = T_1 A,\ \forall\ A. \] In particular, $S T_1 = T_1 S$ and hence, $Z\left( G_1 \right) = G_1$, that is, $G_1$ is abelian.

Now we consider $T_2$. \begin{align*} \textcolor{red}{j = 2} & \implies A T_2 = T_2 A \\ & \implies \begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix} \begin{bmatrix} \iota & 0 \\ 0 & -\iota \\ \end{bmatrix} = \begin{bmatrix} \iota & 0 \\ 0 & -\iota \\ \end{bmatrix} \begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix} \\[1ex] & \implies \begin{bmatrix} a \iota & -b \iota \\ -b \iota & -a \iota \\ \end{bmatrix} = \begin{bmatrix} a \iota & b \iota \\ b \iota & -a \iota \\ \end{bmatrix} \\[1ex] & \implies b = -b \text{ that is},\ b = 0 \\ & \implies A = \begin{bmatrix} a & 0 \\ 0 & a \\ \end{bmatrix} = a I_2. \end{align*} Since $\det S = \det T_2 = 1$ and $A = a I_2 \in \left\langle S, T_2 \right\rangle $ implies $\det \left( a I_2 \right) = 1$, that is $a = \pm 1$. So, the only possibility for $A$ to be either $\pm I_2$. Since, $S^2 = -I_2$, so \[ Z\left\{ G_2 \right\} = \left\{ \pm I_2 \right\} . \]

Finally, \begin{align*} \textcolor{red}{j = 3} & \implies A T_3 = T_3 A \\ & \implies \begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix} \\[1ex] & \implies \begin{bmatrix} b & a \\ a & -b \\ \end{bmatrix} = \begin{bmatrix} -b & a \\ a & b \\ \end{bmatrix} \\[1ex] & \implies b = -b \text{ that is, } b = 0 \\ & \implies A = a I_2. \end{align*} Here again, since $S, T_3 \in GL_2(\mathbb{R} ) \subseteq GL_2(\mathbb{C} )$ and $\det S = 1$ and $\det T_3 = -1$, therefore, $aI_2 \in G_3 = \left\langle S, T_3 \right\rangle \subseteq GL_2(\mathbb{R} ) $. This implies $\det a I_2 = 1$ and hence, $a = \pm 1$. Since $S^2 = -I_2$, so \[ Z\left( G_3 \right) = \left\{ \pm I_2 \right\} . \]

In summary, we got

• $Z\left( G_1 \right) = G_1$, that is, $G_1$ is abelian.
• $Z\left( G_2 \right) = \{ \pm I_2 \} \neq G_2$, so $G_2$ is not abelian.
• $Z\left( G_3 \right) = \{ \pm I_2 \} \neq G_3$, so $G_2$ is not abelian.

Therefore, options (a), (b) and (c) are FALSE and option (d) is TRUE.