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
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**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{
Transcribed Image Text:**Problem Statement:** Prove that \( \text{SL}_n(\mathbb{R}) \) is a normal subgroup of \( \text{GL}_n(\mathbb{R}) \). --- **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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