3. You have already proved that GL(2, R) = {[ª la, b, c, d e R and ad – bc ± 0} forms a group under matrix multiplication. Prove that the set H = {, ) ne z} is a cyclic subgroup of GL(2, R).
3. You have already proved that GL(2, R) = {[ª la, b, c, d e R and ad – bc ± 0} forms a group under matrix multiplication. Prove that the set H = {, ) ne z} is a cyclic subgroup of GL(2, R).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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plz answer Q3 and Q4 with details
![**3.** You have already proved that
\[ GL(2, R) = \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \middle| a, b, c, d \in R \text{ and } ad - bc \neq 0 \right\} \]
forms a group under matrix multiplication. Prove that the set
\[ H = \left\{ \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \middle| n \in \mathbb{Z} \right\} \]
is a cyclic subgroup of \( GL(2, R) \).
**4.** Find an isomorphism from the multiplicative group \( G = \{ 1, -1, i, -i \} \) to the multiplicative group
\[ G' = \left\{ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} \right\}. \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fba322d66-282b-4f13-8778-b2c554594cf2%2F43f93900-b78b-4700-a904-f2b81b5036d8%2Fab4jv1_processed.png&w=3840&q=75)
Transcribed Image Text:**3.** You have already proved that
\[ GL(2, R) = \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \middle| a, b, c, d \in R \text{ and } ad - bc \neq 0 \right\} \]
forms a group under matrix multiplication. Prove that the set
\[ H = \left\{ \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \middle| n \in \mathbb{Z} \right\} \]
is a cyclic subgroup of \( GL(2, R) \).
**4.** Find an isomorphism from the multiplicative group \( G = \{ 1, -1, i, -i \} \) to the multiplicative group
\[ G' = \left\{ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} \right\}. \]
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