Question

Let \(GL_2(\mathbb{C})\) denote the group of \(2 \times 2\) invertible complex matrices with usual matrix multiplication. For \(S, T \in GL_2(\mathbb{C})\), \(\langle S, T \rangle\) denotes the subgroup generated by \(S\) and \(T\). Let
\[ S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \in GL_2(\mathbb{C}) \] and \(G_1, G_2, G_3\) be three subgroups of \(GL_2(\mathbb{C})\) given by
\[ G_1 = \langle S, T_1 \rangle, \text{ where } T_1 = \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix}, \]
\[ G_2 = \langle S, T_2 \rangle, \text{ where } T_2 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, \]
\[ G_3 = \langle S, T_3 \rangle, \text{ where } T_3 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. \]
Let \(Z(G_i)\) denote the center of \(G_i\) for \(i = 1, 2, 3\).
Which of the following statements is correct?

• A. \(G_1\) is isomorphic to \(G_3\)
• B. \(Z(G_1)\) is isomorphic to \(Z(G_2)\)
• C. \(Z(G_3) = \left\{ \begin{pmatrix} 1 & 0
  0 & 1 \end{pmatrix} \right\}\)
• D. \(Z(G_2)\) is isomorphic to \(Z(G_3)\)

Hint:

Recognizing standard groups from their generators and relations is a key skill in group theory. The quaternion group \(Q_8\) is generated by \(i, j\) with \(i^2=j^2=k^2=ijk=-1\). The dihedral group \(D_n\) is generated by a rotation \(r\) and a reflection \(s\) with \(r^n=s^2=1, srs=r^{-1}\). Knowing their properties, especially their centers, is very useful.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem involves analyzing the structure of groups generated by given matrices. We need to determine the properties of these groups, such as their order, whether they are abelian, and their centers. The center \(Z(G)\) of a group G is the set of elements that commute with every element of G.
Step 3: Detailed Explanation:
First, let's analyze the matrix \(S\).
\(S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\). \(S^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I\), \(S^3 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\), \(S^4 = I\).
The order of S is 4. The subgroup \(\langle S \rangle\) is cyclic of order 4.
Analysis of \(G_1\):
\(T_1 = iI\). \(T_1\) is a scalar matrix, so it commutes with any matrix, including S. \( (iI)^2 = -I\), \((iI)^3 = -iI\), \((iI)^4 = I\). The order of \(T_1\) is 4.
Since S and \(T_1\) commute, \(G_1\) is an abelian group. The elements are of the form \(S^k T_1^j\).
\(S^2 = -I\), \(T_1^2 = -I\), so \(S^2 = T_1^2\).
The group is \(\{I, S, S^2, S^3, T_1, ST_1, S^2T_1, S^3T_1\}\). Order is 8. \(G_1\) is abelian, so its center is the entire group: \(Z(G_1) = G_1\).
Analysis of \(G_2\):
\(T_2 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}\). \(T_2^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I = S^2\). \(T_2^4=I\), so order of \(T_2\) is 4.
Check commutation:
\(ST_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}\),
\(T_2S = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} = -ST_2\).
They anticommute. The group is the quaternion group \(Q_8\). Its center is \(Z(Q_8) = \{I, -I\} = \{I, S^2\}\).
Analysis of \(G_3\):
\(T_3 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). \(T_3^2 = I\), so order of \(T_3\) is 2.
Check commutation:
\(ST_3 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\),
\(T_3S = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = -ST_3\).
They anticommute. The group \(G_3\) is the dihedral group of order 8, \(D_4\). Its center is \(Z(D_4) = \{I, -I\} = \{I, S^2\}\).
Evaluate the options:
(A) \(G_1\) is isomorphic to \(G_3\): \(G_1\) is abelian, \(G_3\) is not. FALSE.
(B) \(Z(G_1)\) is isomorphic to \(Z(G_2)\): \(Z(G_1)\) has order 8, \(Z(G_2)\) has order 2. FALSE.
(C) \(Z(G_3) = \{I\}\): We found \(Z(G_3)=\{I, -I\}\). FALSE.
(D) \(Z(G_2)\) is isomorphic to \(Z(G_3)\): Both are \(\{I, -I\}\), cyclic of order 2. TRUE.
Step 4: Final Answer:
The correct statement is \(Z(G_2)\) is isomorphic to \(Z(G_3)\).
Step 5: Why This is Correct:
By analyzing the generators and their relations, \(G_2\) is the quaternion group \(Q_8\) and \(G_3\) is the dihedral group \(D_4\). Both centers are \(\{I, -I\}\), cyclic of order 2. Hence they are isomorphic.