Skip to main content

Cosets and Lagrange's Theorem

  • Understand the concept of cosets.
  • Learn about left and right cosets.
  • Explore Lagrange's Theorem and its implications.

Definition of Cosets

Given a group \( G \) and a subgroup \( H \) of \( G \), a coset is a form of subset that can be defined as follows:

Left Coset: For any element \( g \in G \), the left coset of \( H \) with respect to \( g \) is the set \( gH = \{ gh \mid h \in H \} \).

Right Coset: For any element \( g \in G \), the right coset of \( H \) with respect to \( g \) is the set \( Hg = \{ hg \mid h \in H \} \).

Left and Right Cosets

Cosets can be classified into two types:

  • Left Cosets: A left coset of \( H \) in \( G \) is a set formed by multiplying \( H \) with a fixed element from the left.
  • Right Cosets: A right coset of \( H \) in \( G \) is a set formed by multiplying \( H \) with a fixed element from the right.

Note that while left cosets and right cosets may differ, they both partition the group \( G \) into disjoint subsets of equal size.

Lagrange's Theorem

Lagrange's Theorem is a fundamental result in group theory that relates the order of a subgroup to the order of the entire group.

Theorem: If \( G \) is a finite group and \( H \) is a subgroup of \( G \), then the order of \( H \) (the number of elements in \( H \)) divides the order of \( G \) (the number of elements in \( G \)).

Implication: The number of distinct left cosets (or right cosets) of \( H \) in \( G \) is equal to the index of \( H \) in \( G \), denoted as \( |G : H| = \frac{|G|}{|H|} \).

Proof of Lagrange's Theorem

Consider the group \( G \) and its subgroup \( H \). The left cosets of \( H \) in \( G \) are \( gH \) for \( g \in G \).

First, we show that the left cosets of \( H \) partition \( G \):

  • Every element \( g \in G \) belongs to some left coset \( gH \).
  • Two left cosets \( g_1H \) and \( g_2H \) are either identical or disjoint.

Next, we show that each left coset \( gH \) has the same number of elements as \( H \). This is because the function \( f_h: H \to gH \) defined by \( f_h(h) = gh \) is a bijection.

Since the left cosets partition \( G \) and each coset has the same number of elements as \( H \), the order of \( G \) is the product of the number of cosets and the order of \( H \):

\( |G| = |H| \times \text{number of distinct left cosets} \)

Thus, the order of \( H \) divides the order of \( G \), completing the proof.

Examples of Cosets and Lagrange's Theorem

  • Example 1: Consider the group \( \mathbb{Z}_6 \) (integers modulo 6) and its subgroup \( \{0, 3\} \). The left cosets of this subgroup in \( \mathbb{Z}_6 \) are \( \{0, 3\} \), \( \{1, 4\} \), and \( \{2, 5\} \).
  • Example 2: For the symmetric group \( S_3 \), consider the subgroup \( H \) generated by the identity permutation and a transposition. Lagrange's Theorem tells us that the order of \( H \) (which is 2) divides the order of \( S_3 \) (which is 6).

Exercises

  • Identify Cosets: Given the group \( \mathbb{Z}_{10} \) (integers modulo 10) and its subgroup \( \{0, 5\} \), identify all distinct left cosets.
  • Apply Lagrange's Theorem: For the group \( \mathbb{Z}_{12} \) and its subgroup \( \{0, 4, 8\} \), use Lagrange's Theorem to determine the number of distinct cosets.
  • Coset Verification: Verify that the left cosets and right cosets of the subgroup \( \{0, 2, 4\} \) in \( \mathbb{Z}_6 \) partition the group into disjoint subsets.

Summary

In this lesson, we explored the concepts of cosets and Lagrange's Theorem, including their definitions, properties, and examples. Understanding these fundamental ideas is crucial as they provide insights into the structure of groups and their subgroups.

This Week's Best Picks from Amazon

Please see more curated items that we picked from Amazon here .

Popular posts from this blog

Gaussian Elimination: A Step-by-Step Guide

Gaussian Elimination: A Step-by-Step Guide Gaussian Elimination is a systematic method for solving systems of linear equations. It works by transforming a given system into an equivalent one in row echelon form using a sequence of row operations. Once in this form, the system can be solved efficiently using back-substitution . What is Gaussian Elimination? Gaussian elimination consists of two main stages: Forward Elimination: Convert the system into an upper triangular form. Back-Substitution: Solve for unknowns starting from the last equation. Definition of a Pivot A pivot is the first nonzero entry in a row when moving from left to right. Pivots are used to eliminate the elements below them, transforming the system into an upper triangular form. Step-by-Step Example Consider the system of equations: \[ \begin{aligned} 2x + 3y - z &= 5 \\ 4x + y...

Singular Value Decomposition

Lesson Objectives ▼ Understand the concept of Singular Value Decomposition (SVD). Decompose a matrix into orthogonal matrices and singular values. Learn how to compute SVD step by step. Explore applications of SVD in data compression and machine learning. Lesson Outline ▼ Definition of SVD Computing SVD Step by Step Applications of SVD Examples Definition of Singular Value Decomposition The **Singular Value Decomposition (SVD)** of an \( m \times n \) matrix \( A \) is a factorization of the form: \[ A = U \Sigma V^T \] where: \( U \) is an \( m \times m \) orthogonal matrix (left singular vectors). \( \Sigma \) is an \( m \times n \) diagonal matrix with **singular values** on the diagonal. \( V \) is an \( n \times n \) orthogonal matrix (right singular vectors). Computing SVD Step by Step To compute \( A = U \Sigma V^T \): Find the eigenvalues and eige...

LU Decomposition

LU Decomposition: A Step-by-Step Guide LU Decomposition, also known as LU Factorization, is a method of decomposing a square matrix into two triangular matrices: a lower triangular matrix L and an upper triangular matrix U . This is useful for solving linear equations, computing determinants, and inverting matrices efficiently. What is LU Decomposition? LU Decomposition expresses a matrix A as: \[ A = LU \] where: L is a lower triangular matrix with ones on the diagonal. U is an upper triangular matrix. Step-by-Step Process Consider the matrix: \[ A = \begin{bmatrix} 2 & 3 & 1 \\ 4 & 7 & 3 \\ 6 & 18 & 5 \end{bmatrix} \] Step 1: Initialize L as an Identity Matrix Start with an identity matrix for \( L \): \[ L = \begin{bmatrix} 1 & 0 & 0 \\ 0 ...