Problem 6
Question
Prove that each of the following is a homomorphism, and describe its kernel. Let \(G\) be the multiplicative group of all \(2 \times 2\) matrices $$ \left(\begin{array}{ll} a & b \\ c & d \end{array}\right) $$ satisfying \(a d-b c \neq 0\). Let \(f: G \rightarrow \mathbb{R}^{*}\) be given by \(f(A)=\) determinant of \(A=a d-b c\)
Step-by-Step Solution
Verified Answer
The function is a homomorphism because it respects determinant multiplication. Its kernel is the set of matrices with determinant 1.
1Step 1: Understand the Homomorphism Definition
A homomorphism between two groups is a function that respects the group operation. For groups \( (G, \cdot) \) and \( (H, *) \), a function \( f: G \rightarrow H \) is a homomorphism if for all \( x, y \in G \), we have:\[ f(x \cdot y) = f(x) * f(y) \]
2Step 2: Formulate Function for Homomorphism Verification
In our case, the function \( f: G \rightarrow \mathbb{R}^{*} \) is given by \( f(A) = \det(A) = ad - bc \). Here, \( G \) is the multiplicative group of \( 2 \times 2 \) matrices, and \( \mathbb{R}^{*} \) is the multiplicative group of non-zero real numbers.
3Step 3: Verify Homomorphism Property
Take two matrices \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) and \( B = \begin{pmatrix} e & f \ g & h \end{pmatrix} \) in \( G \). Their product \( AB \) is:\[ AB = \begin{pmatrix} ae+bg & af+bh \ ce+dg & cf+dh \end{pmatrix} \]The determinant of \( AB \) is \((ae+bg)(cf+dh) - (af+bh)(ce+dg)\). Compute this using expansion and verify it equals \( \det(A) \times \det(B) \):\[ \det(AB) = (ad-bc)(eh-fg) = \det(A) \cdot \det(B) \].Hence, \( f(AB) = \det(AB) = (ad-bc)(eh-fg) = \det(A) \cdot \det(B) = f(A) \cdot f(B) \).
4Step 4: Identify the Kernel of the Homomorphism
The kernel of a homomorphism \( ker(f) \) is the set of elements mapping to the identity element of the target group. Here, \( \mathbb{R}^{*} \) has identity element 1. Therefore, solve:\[ f(A) = \det(A) = ad-bc = 1 \].The kernel consists of all matrices \( A \) in \( G \) with determinant 1, i.e., all \( 2 \times 2 \) matrices with \( ad - bc = 1 \).
Key Concepts
Kernel of HomomorphismMatrix GroupsDeterminantHomomorphism Definition
Kernel of Homomorphism
In the realm of abstract algebra, homomorphisms play a vital role by linking different algebraic structures while respecting their operations. The **kernel** of a homomorphism is an essential concept to understand, as it pinpoints the elements within the domain that get mapped to the identity element of the codomain. This concept becomes clearer with an example: Think of the function \( f: G \to H \), where \( G \) and \( H \) are groups. The kernel \( \ker(f) \) includes all elements \( x \) in \( G \) for which \( f(x) = e_H \), with \( e_H \) being the identity in the group \( H \).
In our exercise, given the group \( G \) of invertible \( 2 \times 2 \) matrices and the homomorphism \( f \) that maps each matrix to its determinant, the kernel comprises matrices for which the determinant equals one. These are often referred to as "special matrices," particularly the special linear group, denoted as \( SL(2, \mathbb{R}) \). Understanding kernels aids in comprehending how structures can be simplified while retaining essential properties, making them indispensable in studies of algebra.
In our exercise, given the group \( G \) of invertible \( 2 \times 2 \) matrices and the homomorphism \( f \) that maps each matrix to its determinant, the kernel comprises matrices for which the determinant equals one. These are often referred to as "special matrices," particularly the special linear group, denoted as \( SL(2, \mathbb{R}) \). Understanding kernels aids in comprehending how structures can be simplified while retaining essential properties, making them indispensable in studies of algebra.
Matrix Groups
Matrix groups represent collections of matrices that possess a group structure via matrix multiplication. When we talk about matrix groups like \(G\), the multiplicative group of all \(2 \times 2\) invertible matrices, we essentially mean that each element (a matrix) follows certain rules:
- There is a unique identity element, which is the identity matrix.
- Every element must have an inverse, which is another matrix that results in the identity when multiplied with the initial matrix.
- Matrix multiplication is associative.
Determinant
Determinants are a fundamental concept in linear algebra that allow us to assess a matrix's properties, such as invertibility. For a \(2 \times 2\) matrix \(\begin{pmatrix} a & b c & d \end{pmatrix}\), the determinant is calculated by:\[ad - bc\]If the determinant is non-zero, the matrix is invertible; otherwise, it's singular and lacks an inverse.
Determinants go beyond mere invertibility checks. They also offer insights like:
Determinants go beyond mere invertibility checks. They also offer insights like:
- The determinant's magnitude can indicate the linear transformation's scaling effect related to area or volume in transformations.
- Sign tells whether a transformation preserves orientation.
Homomorphism Definition
Understanding **homomorphisms** is crucial for appreciating how algebraic structures relate to each other. At its core, a homomorphism is a function that preserves the defining operations of the groups involved. Formally, a function \( f: G \to H \) between groups is a homomorphism if, whenever you take two elements \( x \) and \( y \) in \( G \), the following holds:\[ f(x cdot y) = f(x) \ast f(y) \]This means the operation in \( G \), denoted by \( \cdot \), is seamlessly transferred to the operation in \( H \), denoted by \( \ast \).
In this exercise, our function \( f \), mapping matrices to their determinants, respects the multiplicative structure. Here’s what happens:
In this exercise, our function \( f \), mapping matrices to their determinants, respects the multiplicative structure. Here’s what happens:
- The determinant of a product of matrices equals the product of their determinants.
- This property confirms \( f \) is indeed a homomorphism as \( f(AB) = f(A) \cdot f(B) \).
Other exercises in this chapter
Problem 6
In the following, let \(G\) denote an arbitrary group. Any intersection of normal subgroups of \(G\) is a normal subgroup of \(G\).
View solution Problem 6
Let \(G, H\), and \(K\) be groups. Prove the following: For any group \(G\), the function \(f: G \rightarrow G\) defined by \(f(x)=e\) is a homomorphism.
View solution Problem 7
Let \(G, H\), and \(K\) be groups. Prove the following: For any group \(G,\\{e\\}\) and \(G\) are homomorphic images of \(G\).
View solution Problem 8
Let \(G, H\), and \(K\) be groups. Prove the following: The function \(f: G \rightarrow G\) defined by \(f(x)=x^{2}\) is a homomorphism iff \(G\) is abelian.
View solution