/*! 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 45 Prove that the intersection of t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 45

Prove that the intersection of two subgroups of a group \(G\) is also a subgroup of \(G\).

Short Answer

Expert verified
Question: Prove that the intersection of two subgroups of a group is also a subgroup. Answer: To prove that the intersection of two subgroups is also a subgroup, we can use the Subgroup Test. First, we showed that the intersection of the two subgroups is nonempty, as it contains the identity element of the parent group. Next, we demonstrated that the intersection satisfies the closure property, meaning that the product of any two elements in the intersection is also an element of the intersection. Finally, we showed that the intersection satisfies the inverse property, meaning that if an element is in the intersection, so is its inverse. Since the intersection meets all these conditions, it is a subgroup of the parent group.

Step by step solution

01

Proving the intersection is nonempty

Let's first prove that the intersection \(H_1 \cap H_2\) is nonempty. Since \(H_1\) and \(H_2\) are subgroups of \(G\), they both contain the identity element \(e\) of \(G\). Thus, \(e \in H_1 \cap H_2\), making the intersection nonempty.
02

Closure property

We must now show that \(H_1 \cap H_2\) satisfies the closure property. Let \(a, b \in H_1 \cap H_2\). This means that \(a, b \in H_1\) as well as \(a, b \in H_2\). Since both \(H_1\) and \(H_2\) are subgroups, they are closed under the group operation. This means that \(ab \in H_1\) and \(ab \in H_2\). Therefore, \(ab \in H_1 \cap H_2\), and the intersection is closed under the group operation.
03

Inverse property

Lastly, we must demonstrate that the intersection satisfies the inverse property. Let \(a \in H_1 \cap H_2\). This means that \(a \in H_1\) and \(a \in H_2\). As \(H_1\) and \(H_2\) are subgroups, they contain the inverse of \(a\), denoted as \(a^{-1}\). Thus, \(a^{-1} \in H_1\) and \(a^{-1} \in H_2\). Therefore, \(a^{-1} \in H_1 \cap H_2\), and the intersection satisfies the inverse property. Since \(H_1 \cap H_2\) is nonempty and satisfies both the closure and inverse properties, it is a subgroup of \(G\), as required.

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.

Subgroups
In group theory, a subgroup is essentially a smaller group within a larger group. Think of it like a small circle inside a big circle, where each small circle is fully governed by the same rules as the big circle. For a subset of a group to be a subgroup, it must satisfy some vital criteria:
  • The identity element of the larger group must be part of this subgroup.
  • It must fulfill the closure property, meaning if you take any two elements from the subgroup and perform the group operation, the result should also be in the subgroup.
  • Finally, it should satisfy the inverse property, meaning every element in the subgroup should have an inverse within it.
Subgroups are crucial in understanding the structure of groups, as they help to break down complex groups into simpler ones.
Intersection of Subgroups
The intersection of two subgroups is the element set that they both share. It's a fascinating concept because it still preserves the structure of a subgroup. If you have two subgroups, say, \( H_1 \) and \( H_2 \), their intersection \( H_1 \cap H_2 \) will always form a subgroup. This is important because it allows for combining properties of different subgroups and studying them together.
Why does this work? It's pretty straightforward:
  • Every subgroup contains the identity element, so their intersection will naturally include it.
  • Both are closed under the group operation, meaning any operation results in another element of the subgroup, making the intersection closed.
  • Each subgroup contains the inverse of its elements, so the intersection does too. Therefore, it fulfills the inverse property.
Because of these points, the intersection maintains all the necessary criteria to be a subgroup, which is a fundamental concept in group theory.
Closure Property
The closure property is like a safety net for a subgroup. It ensures that whenever you take two elements from the subgroup and apply the group operation, the resulting element still belongs to the same subgroup. This is essential in maintaining the integrity of a subgroup.
To understand it better, think of it as making sure any possible action or operation done within the subgroup doesn't take you outside of it. For example, in the context of numbers:
  • If you pick any two even numbers and add them, you'll always get another even number. This property satisfies closure within even numbers under addition.
  • Similarly, if \( a \) and \( b \) are elements of subgroup \( H \), then \( ab \) being part of \( H \) is the closure property, conforming to the group's operation.
Closure ensures operations stay 'in-bounds,' preventing any leap out of the subgroup's confines.
Inverse Property
The inverse property is another pillar that supports the existence of subgroups. It declares that for every element in a subgroup, there exists another element within the same subgroup that is its inverse. In simpler terms, when an element combines with its inverse through the group operation, the result is the identity element of the group.
This is crucial because:
  • If you have an element \( a \) in a subgroup, its inverse \( a^{-1} \) must also be present in the same subgroup.
  • To visualise, if you're multiplying numbers, the inverse is like the reciprocal of a number鈥攖ogether, they yield the number one, considered the 'identity' in multiplication.
  • The presence of inverses in subgroups ensures every element can 'undo' itself, reinforcing the underlying symmetry and structure of the group.
This property allows subgroups to mirror the consistent functionality of the whole group, maintaining their identity.

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

Find all the subgroups of the symmetry group of an equilateral triangle.

Show that addition and multiplication \(\bmod n\) are well defined operations. That is, show that the operations do not depend on the choice of the representative from the equivalence classes mod \(n\).

Given the groups \(\mathbb{R}^{*}\) and \(\mathbb{Z}\), let \(G=\mathbb{R}^{*} \times \mathbb{Z}\). Define a binary operation o on \(G\) by \((a, m) \circ(b, n)=(a b, m+n)\). Show that \(G\) is a group under this operation.

Describe the symmetries of a square and prove that the set of symmetries is a group. Give a Cayley table for the symmetries. How many ways can the vertices of a square be permuted? Is each permutation necessarily a symmetry of the square? The symmetry group of the square is denoted by \(D_{4}\).

Show that if \(G\) is a finite group of even order, then there is an \(a \in G\) such that \(a\) is not the identity and \(a^{2}=e\).
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

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.