Question

Consider the group G = {A ∈ M₂ (ℝ): AA^T = I₂} with respect to matrix multiplication. Let
Z(G) = {A ∈ G : AB = BA, for all B ∈ G}.
Then, the cardinality of Z(G) is

• A. 1
• B. 2
• C. 4
• D. infinite

Answer 1

The Correct Option is B

To determine the cardinality of the center \( Z(G) \) of the group \( G \), we begin by understanding the elements of \( G \). The group \( G \) consists of \( 2 \times 2 \) orthogonal matrices \( A \), such that \( AA^T = I_2 \), where \( I_2 \) is the identity matrix.

The center \( Z(G) \) of the group \( G \) is defined as the set of all matrices \( A \in G \) that commute with every matrix \( B \in G \). Mathematically, this is expressed as:

\(Z(G) = \{ A \in G \, : \, AB = BA \, \text{for all} \, B \in G \}\)

 

For \( A \in Z(G) \), since \( A \) commutes with every matrix in \( G \), it must possess a special form. We know that:

1. Orthogonal matrices must satisfy \( AA^T = I_2 \), meaning \( A^{-1} = A^T \).
2. For \( A \) to be in the center, it must satisfy \( AB = BA \) for all \( B \in G \). 

A notable property of the group of \( 2 \times 2 \) orthogonal matrices is that the subset of scalar matrices of the form:

\[ A = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \]

 

where \( a = \pm 1 \), commutes with any matrix in \( G \), because the multiplication of scalar matrices with any other matrix results in the same matrix scaled by \( a \).

Therefore, the matrices:

\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) and \( \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \)

 

both belong to \( Z(G) \). These are the only such \( 2 \times 2 \) orthogonal matrices satisfying \( AA^T = I_2 \) that commute with every element in \( G \).

Hence, the cardinality of \( Z(G) \) is 2, corresponding to these two distinct matrices.