/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Let \(\langle G,-\rangle\) be a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 40

Let \(\langle G,-\rangle\) be a group. Consider the binary operation * on the set \(G\) defined by $$ a+b=b \cdot a $$ for \(a, b \in G\). Show that \((G, \theta)\) is a group and that \((G, *)\) is actually isomorphte to \(\\{G, \cdot\\} .[\) Hint Consider the \(\operatorname{map} \phi\) with \(\phi(a)=a^{\prime}\) for \(a \in G .1\)

Short Answer

Expert verified
\((G, *)\) is a group, and it's isomorphic to \(\langle G, \cdot \rangle\).

Step by step solution

01

Understand Group Properties

To show that \( (G, *) \) is a group, we need to verify the group properties: closure, associativity, identity, and existence of inverses under the operation \( * \).
02

Verify Closure

Closure means that for all \( a, b \in G \), the result of \( a * b = b \cdot a \) is also in \( G \). In \( \langle G, \cdot \rangle \), \( b \cdot a \) is an element of \( G \) because the group is closed under its operation.
03

Verify Associativity

For associativity, we need \( (a * b) * c = a * (b * c) \) for all \( a, b, c \in G \). This becomes \( (b \cdot a) \cdot c = c \cdot (b \cdot a) \) in the group \( \langle G, \cdot \rangle \), which holds due to associativity of \( \cdot \).
04

Identity Element

The identity element \( e \) for \( * \) is such that for any \( a \in G \), \( e * a = a \) and \( a * e = a \). Using the same element from \( \langle G, \cdot \rangle \), we have \( a \cdot e = a \) and \( e \cdot a = a \), so \( e * a = a \) and \( a * e = a \) are satisfied.
05

Existence of Inverses

For each element \( a \in G \), there exists an inverse \( a^{-1} \) such that \( a * a^{-1} = e \) and \( a^{-1} * a = e \). These are also true in \( \langle G, \cdot \rangle \), confirming the inverses under \( * \).
06

Define the Isomorphism

To show that \( (G, *) \) is isomorphic to \( \langle G, \cdot \rangle \), consider the map \( \phi: G \rightarrow G \) where \( \phi(a) = a' \) such that \( a' = a \). This map must be bijective and homomorphic.
07

Verify Bijectivity

The map \( \phi(a) = a \) is bijective because it is both injective (distinct inputs have distinct outputs) and surjective (every element in \( G \) has a pre-image).
08

Verify Homomorphism

To check for homomorphism, ensure that \( \phi(a * b) = \phi(a) \cdot \phi(b) \). We have \( \phi(a * b) = \phi(b \cdot a) = b \cdot a \), and \( \phi(a) \cdot \phi(b) = a \cdot b \). Since these equal for \( * \) and \( \cdot \), the map preserves group operation.
09

Conclusion of Isomorphism

As \( \phi \) is bijective and a homomorphism, \( (G, *) \) is isomorphic to \( \langle G, \cdot \rangle \) satisfying the problem's conditions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Group Properties
In the realm of group theory, understanding the properties that define a group is crucial. A group must satisfy four key properties: closure, associativity, identity, and the existence of inverses. These properties ensure a well-structured and predictable mathematical set under a given operation.

  • Closure: This property ensures that if you take any two elements from a group and combine them using the group's operation, the result is also a member of the group. If we consider the group \( \langle G, \cdot \rangle \), any operation \( b \cdot a \) results in another element within \( G \).
  • Associativity: This means the way in which elements are grouped in an operation doesn't affect the result. More formally, \((a * b) * c = a * (b * c)\). Applying this to our group operation \((G, *)\), it translates to \((b \cdot a) \cdot c = b \cdot (a \cdot c)\), which holds true since \( \cdot \) is associative.
  • Identity Element: An element \( e \) exists such that for any element \( a \) in \( G \), both \( e \cdot a \) and \( a \cdot e \) return \( a \). This \( e \) is the identity element and works equally in our defined operation *.
  • Existence of Inverses: For every element \( a \) in the group, there is an \( a^{-1} \) such that \( a \cdot a^{-1} = e \) and \( a^{-1} \cdot a = e \). In the new operation * on \( G \), every element still retains its inverse under the multiplication \( \cdot \).
Understanding these properties helps confirm whether a set with a specific operation qualifies as a group.
Isomorphism
In mathematics, two groups are considered isomorphic if there exists a bijective homomorphism between them. This means they share a structure that allows them to be transformed into one another without loss of their inherent properties. To demonstrate that the group \((G, *)\) is isomorphic to \(\langle G, \cdot \rangle \), we can use the map \(\phi\) defined for each element \( a \) as \( \phi(a) = a' \), with \( a'=a \).

  • Bijectivity: A function is bijective if it is both injective and surjective. In this case, the map \( \phi(a) = a \) is straightforwardly bijective since each element maps uniquely to itself, ensuring distinct inputs produce distinct outputs (injective) and each element in \( G \) can be reached by some input (surjective).
  • Homomorphism: A homomorphism is a function between two groups that respects the group operation. Specifically, this means \( \phi(a * b) = \phi(a) \cdot \phi(b) \). For the given operations, since \( \phi(a * b) = \phi(b \cdot a) = b \cdot a \) and \( \phi(a) \cdot \phi(b) = a \cdot b \), the operation is preserved, establishing the homomorphism.
Through both bijectivity and homomorphism, we establish that the groups are isomorphic, indicating their structural similarity.
Binary Operations
A binary operation is a rule for combining two elements of a set to produce another element of the same set. In our context, the operation \(*\) is defined on \(G\) such that for any elements \(a, b \in G\), \(a * b = b \cdot a\). This operation must fulfill specific properties to maintain the group's structure.

  • Understanding the Operation: Here, \(a * b = b \cdot a\). Notice the change in order; it's essentially the regular multiplicative operation in the group \(\langle G, \cdot \rangle\) but flipped. Understanding how this reversal impacts operation is pivotal for manipulating the group's elements.
  • Associating with Group Properties: Our operation \(*\) is shown to fulfill all critical group properties when analyzed under conclusions about closure, associativity, identity, and inverses already known in \(\langle G, \cdot \rangle\). This ensures \((G, *)\) is indeed a group.
  • Importance of Binary Operations: Binary operations define the behavior and structure of groups, fundamentally shaping how we perform and interpret operations with group elements. In this problem, defining an unusual operation \(*\) while maintaining group characteristics offers insights into the flexibility and creativity in group theory.
The nature of binary operations like \(*\) helps illustrate the structural flexibility in defining group actions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(n\) be a posltive integer and let \(n \mathbb{Z}=[n m \mid m \in \mathbb{Z}\\}\). a. Show that \(\langle n \mathbb{Z},+\rangle\) is a group. b. Show that \(\langle n \mathbb{Z},+\rangle \simeq\langle\mathbb{Z},+)\).

In Exercises 11 through 18, determine whether the given set of matrices under the specified operation, matrix addition or multiplication, is a group. Recall that a diagonal matrix is a square matrix whose only nonzero entries lie on the main diagonal, from the upper left to the lower right corner. An upper- triangular matrix is a square, matrix with only zero entries below the main diagonal. Associated with each \(n \times n\) matrix \(A\) is a number called the determinant of \(A\), denoted by \(\operatorname{det}(A)\). If \(A\) and \(B\) are both \(n \times n\) matrices, then \(\operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B)\), Also, \(\operatorname{det}\left(I_{n}\right)=1\) and \(A\) is invertible if and only if \(\operatorname{det}(A) \neq 0\). All \(n \times n\) diagonal matrices with all diagonal entries 1 or \(-1\) under matrix multiplication.

This exercise shows that there are two nonisomorphic group structures on a set of 4 elements. Let the set be \((e, a, b, c)\), with \(e\) the identity clement for the group operation. A group table would then have to start in the manner shown in Table \(4.22\). The square indicated by the question mark cannot be filled in with \(a\). It must be filled in either with the identity clement e or with an clement different from both \(e\) and \(a\). In this latter case, it is noloss of generality to assume that this element is \(b\). If this square is filled in with \(e\), the table can then be completed in two ways to give a group. Fud these two tables. (You need not check the assoclative law.) If this square is filled in with \(b\), then the table can only be completed in one way to give a group. Find this table, (Again, you need not check the associative law,) Of the three tables you now have, two give isomorphic groups. Determine which two tables these are, and give the one-to-one onto renaming function which is an isornorphism. a. Are all groups of 4 clements commutative? b. Which table gives a group isomorphic to the group \(U_{4}\), so that we know the binary operation defined by the table is associative? C. Show that the group given by one of the other tables ls structurally the same as the group in Exercise 14 for one particular value of \(n\), so that we know that the operation defined by that table is assoclative also.

In Exercises 11 through 18, determine whether the given set of matrices under the specified operation, matrix addition or multiplication, is a group. Recall that a diagonal matrix is a square matrix whose only nonzero entries lie on the main diagonal, from the upper left to the lower right corner. An upper- triangular matrix is a square, matrix with only zero entries below the main diagonal. Associated with each \(n \times n\) matrix \(A\) is a number called the determinant of \(A\), denoted by \(\operatorname{det}(A)\). If \(A\) and \(B\) are both \(n \times n\) matrices, then \(\operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B)\), Also, \(\operatorname{det}\left(I_{n}\right)=1\) and \(A\) is invertible if and only if \(\operatorname{det}(A) \neq 0\). All \(n \times n\) matrices with determinant either 1 or \(-1\) under matrix multiplication.

All \(n \times n\) upper-triangular matrices with determinant 1 under matrix multiplication.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.