Chapter 9: Problem 1
Let \(E\) designate the group of all the even integers, with respect to addition. Prove that \(\mathbb{Z} \cong E\).
Short Answer
Expert verified
The function \(\phi(n) = 2n\) shows \(\mathbb{Z} \cong E\).
Step by step solution
01
Understand the Problem
We need to prove that the group of integers \(\mathbb{Z}\) is isomorphic to the group of even integers \(E\). This means there exists a bijective homomorphism between these two groups.
02
Define a Function
To construct the isomorphism, we define a function \(\phi: \mathbb{Z} \to E\) by \(\phi(n) = 2n\). This function maps each integer to an even integer.
03
Prove Homomorphism Property
To show that \(\phi\) is a homomorphism, we must show \(\phi(a + b) = \phi(a) + \phi(b)\) for all \(a, b \in \mathbb{Z}\). Calculate: \(\phi(a+b) = 2(a+b) = 2a + 2b = \phi(a) + \phi(b)\).
04
Prove Injectivity
For \(\phi\) to be injective, we require that if \(\phi(a) = \phi(b)\), then \(a = b\). Given that \(2a = 2b\), dividing both sides by 2 yields \(a = b\). Thus, \(\phi\) is injective.
05
Prove Surjectivity
To prove \(\phi\) is surjective, for any even integer \(m\) in \(E\), we need an integer \(n\) such that \(\phi(n) = m\). Given \(m\) is even, \(m = 2k\) for some integer \(k\). Thus, \(\phi(k) = 2k = m\), showing surjectivity.
06
Conclude Isomorphism
Since \(\phi\) is a bijective homomorphism, \(\mathbb{Z}\) is isomorphic to \(E\). We have shown that \(\phi\) is a homomorphism and both injective and surjective.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Even Integers Group
The group of even integers, often denoted as \( E \), forms a special set in algebraic structures. In mathematics, a group is a set equipped with an operation that satisfies four fundamental properties: closure, associativity, there exists an identity element, and each element has an inverse. For the even integers, the operation is addition. The set of all even integers is \( \{ \ldots, -4, -2, 0, 2, 4, \ldots \} \). Each integer is divisible by 2, ensuring it is even.
- **Closure:** Adding any two even numbers always yields another even number. For example, \( 2 + 4 = 6 \), which is also even.
- **Associativity:** We know addition is associative. Thus, for any \( x, y, z \in E \), \((x + y) + z = x + (y + z)\).
- **Identity Element:** The identity is 0. Any even integer plus 0 is itself, i.e., \( x + 0 = x \).
- **Inverse Element:** For each even integer, its inverse under addition is its negative. So, for \( x = 2n \), its inverse is \( -x = -2n \).
The group of even integers is typically explored due to its straightforward properties, serving as an essential example when discussing more complex group theories.
- **Closure:** Adding any two even numbers always yields another even number. For example, \( 2 + 4 = 6 \), which is also even.
- **Associativity:** We know addition is associative. Thus, for any \( x, y, z \in E \), \((x + y) + z = x + (y + z)\).
- **Identity Element:** The identity is 0. Any even integer plus 0 is itself, i.e., \( x + 0 = x \).
- **Inverse Element:** For each even integer, its inverse under addition is its negative. So, for \( x = 2n \), its inverse is \( -x = -2n \).
The group of even integers is typically explored due to its straightforward properties, serving as an essential example when discussing more complex group theories.
Homomorphism
A homomorphism in algebra is a map between two algebraic structures (such as groups) that preserves the operations defined on them. For the groups involved in our exercise, we have integers \( \mathbb{Z} \) and even integers \( E \). Our task was to verify that a function could connect these groups in a way that their group operations (addition) align.
We defined a function \( \phi: \mathbb{Z} \to E \) by \( \phi(n) = 2n \). This function carries integers into even integers. To ensure \( \phi \) is a homomorphism, the operation between elements should be preserved:
Homomorphisms are crucial in group theory because they allow the study of groups by examining the simpler properties of known structures.
We defined a function \( \phi: \mathbb{Z} \to E \) by \( \phi(n) = 2n \). This function carries integers into even integers. To ensure \( \phi \) is a homomorphism, the operation between elements should be preserved:
- \( \phi(a + b) = 2(a + b) \) should equal \( \phi(a) + \phi(b) = 2a + 2b \).
Homomorphisms are crucial in group theory because they allow the study of groups by examining the simpler properties of known structures.
Injective Function
An injective function, also known as a one-to-one function, is a type of mapping where each element in the domain maps to a unique element in the co-domain. This is essential to prove that an isomorphism exists between the groups \( \mathbb{Z} \) and \( E \).
For our function \( \phi(n) = 2n \), we needed to ensure that if \( \phi(a) = \phi(b) \), then \( a = b \). This condition checks that no two distinct integers get mapped to the same even integer. Let’s verify:
Understanding injective functions helps in concluding that two algebraic structures behave the same way; one can map directly back and forth without loss of information.
For our function \( \phi(n) = 2n \), we needed to ensure that if \( \phi(a) = \phi(b) \), then \( a = b \). This condition checks that no two distinct integers get mapped to the same even integer. Let’s verify:
- If \( \phi(a) = \phi(b) \), then \( 2a = 2b \).
- Dividing both sides by 2, we get \( a = b \).
Understanding injective functions helps in concluding that two algebraic structures behave the same way; one can map directly back and forth without loss of information.
Surjective Function
A surjective function, often called an onto function, is one where every element in the co-domain has a pre-image in the domain. This means \( \phi: \mathbb{Z} \to E \) should cover every element in \( E \) with an element from \( \mathbb{Z} \).
To prove surjectivity for \( \phi(n) = 2n \), consider any even integer \( m \) in \( E \). It must be demonstrable that there’s an integer \( k \) such that \( \phi(k) = m \):
By establishing that our function is surjective, we ensure comprehensive mapping coverage of the target set, aligning with the group structures we explore in isomorphism.
To prove surjectivity for \( \phi(n) = 2n \), consider any even integer \( m \) in \( E \). It must be demonstrable that there’s an integer \( k \) such that \( \phi(k) = m \):
- An even integer \( m \) can be represented as \( m = 2k \) for some integer \( k \).
- Thus, \( \phi(k) = 2k = m \).
By establishing that our function is surjective, we ensure comprehensive mapping coverage of the target set, aligning with the group structures we explore in isomorphism.