Problem 1 - A Book of Abstract Algebra

Question

Let \(E\) designate the group of all the even integers, with respect to addition. Prove that \(\mathbb{Z} \cong E\).

Step-by-Step Solution

Answer

The function \(\phi(n) = 2n\) shows \(\mathbb{Z} \cong E\).

Step 1: 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.

Step 2: 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.

Step 3: 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: \(\phi(a+b) = 2(a+b) = 2a + 2b = \phi(a) + \phi(b)\).

Step 4: 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.

Step 5: 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.

Step 6: 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.

Key Concepts

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.

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:

• \( \phi(a + b) = 2(a + b) \) should equal \( \phi(a) + \phi(b) = 2a + 2b \).

By simple distribution, these expressions are identical. Therefore, the function \( \phi \) maintains the structure of the groups it maps between.
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:

• If \( \phi(a) = \phi(b) \), then \( 2a = 2b \).
• Dividing both sides by 2, we get \( a = b \).

This ensures that there is precisely one domain element for each co-domain element, which confirms the injectiveness of \( \phi \).
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 \( \phi(k) = m \):

• An even integer \( m \) can be represented as \( m = 2k \) for some integer \( k \).
• Thus, \( \phi(k) = 2k = m \).

This shows that the function \( \phi \) maps every integer to every even integer.
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.