Prove that SL,(R) is a normal subgroup of GL,(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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Transcribed Image Text:**Problem Statement:**
Prove that \( \text{SL}_n(\mathbb{R}) \) is a normal subgroup of \( \text{GL}_n(\mathbb{R}) \).
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**Explanation:**
1. **Definitions:**
- \( \text{GL}_n(\mathbb{R}) \) is the general linear group consisting of all \( n \times n \) invertible matrices with real entries.
- \( \text{SL}_n(\mathbb{R}) \) is the special linear group of all \( n \times n \) matrices with real entries and determinant equal to 1.
2. **Subgroup and Normal Subgroup:**
- A subgroup \( H \) of a group \( G \) is *normal* if for every element \( g \) in \( G \), the conjugate \( gHg^{-1} = H \).
3. **Proof Outline:**
- Check that \( \text{SL}_n(\mathbb{R}) \) is a subgroup of \( \text{GL}_n(\mathbb{R}) \).
- Show normality by demonstrating that conjugation by elements in \( \text{GL}_n(\mathbb{R}) \) maps \( \text{SL}_n(\mathbb{R}) \) back to itself.
4. **Detailed Explanation:**
- Choose any \( A \in \text{SL}_n(\mathbb{R}) \) and \( B \in \text{GL}_n(\mathbb{R}) \).
- The goal is to show \( BAB^{-1} \in \text{SL}_n(\mathbb{R}) \).
- Calculate the determinant: \(\det(BAB^{-1}) = \det(B) \cdot \det(A) \cdot \det(B^{-1})\).
- Since \( \det(A) = 1 \) (property of elements in \( \text{SL}_n(\mathbb{R}) \)), the equation simplifies to \(\det(BAB^{-1}) = \det(B) \cdot 1 \cdot \det(B)^{-1} = 1\).
- Thus, \( BAB^{-1} \) also has determinant 1, meaning it is in \( \text{SL}_n(\mathbb{
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