Question

Let G be a subgroup of GL₂(ℝ) generated by

\[ \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix} \] Then the order of G is ...........

Hint:

The order of a subgroup generated by two matrices can be found by calculating the distinct products of powers of the matrices and checking their closure under matrix multiplication.

Answer 1

Correct Answer: 5.9 - 6.1

Step 1: Understanding the given group.
We are given a subgroup G of GL₂(ℝ) generated by two matrices:

\[ A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix} \] The order of the subgroup is the number of distinct elements obtained by all products and powers of these matrices.

Step 2: Checking the properties of the matrices.
• A is an involution (A² = I).
• By computing B, B², B³, AB, BA, etc., we find that only 4 distinct matrices appear.

Step 3: Conclusion.
The order of G is 4.