Problem 1
Question
A property of groups is said to be "preserved under homomorphism" if, whenever a group \(G\) has that property, every homomorphic image of \(G\) does also. In this exercise set, we will survey a few typical properties preserved under homomorphism. If \(f: G \rightarrow H\) is a homomorphism of \(G\) onto \(H\), prove each of the following: If \(G\) is abelian, then \(H\) is abelian.
Step-by-Step Solution
Verified Answer
If \( G \) is abelian and \( f: G \rightarrow H \) is a homomorphism onto \( H \), then \( H \) must be abelian.
1Step 1: Understand Group Homomorphism
A homomorphism between two groups, say \( f: G \to H \), is a function that preserves the group operation. This means for all elements \( a, b \) in \( G \), we have \( f(ab) = f(a)f(b) \). Moreover, since \( f \) is onto, every element \( h \) in \( H \) can be written as \( f(g) \) for some element \( g \) in \( G \).
2Step 2: Use Abelian Property in G
The group \( G \) is given as abelian, which means for all elements \( x, y \) in \( G \), the equation \( xy = yx \) holds true. This is the defining characteristic of abelian groups: their operations are commutative.
3Step 3: Map Elements from G to H
Consider two elements in \( H \), say \( h_1 \) and \( h_2 \). Since \( f \) is onto, there exist \( g_1, g_2 \) in \( G \) such that \( f(g_1) = h_1 \) and \( f(g_2) = h_2 \).
4Step 4: Prove H is Abelian
According to the homomorphism property, \( h_1h_2 = f(g_1)f(g_2) = f(g_1g_2) \). Since \( G \) is abelian, \( g_1g_2 = g_2g_1 \), thus \( f(g_1g_2) = f(g_2g_1) = f(g_2)f(g_1) = h_2h_1 \). Therefore, \( h_1h_2 = h_2h_1 \), proving \( H \) is also abelian.
Key Concepts
Abelian GroupsCommutative PropertyHomomorphic Images
Abelian Groups
Abelian groups are a special class of groups defined by the property that their group operation is commutative. This means within an abelian group, the order in which two elements are combined does not affect the result. If we have a group \( G \) and elements \( x \) and \( y \) in \( G \), this commutative property can be mathematically expressed as \( xy = yx \).
- This is the core quality that distinguishes abelian groups from other types, making them particularly well-behaved and widely studied in algebra.
- Many familiar groups in algebra and other areas of mathematics, such as the integers under addition, are abelian groups.
Commutative Property
The commutative property is a fundamental concept in mathematics, dictating that the result of a binary operation on two elements does not depend on their order. In the context of groups, this means that if you take any two elements \( a \) and \( b \) from a group, the product of these two elements will be the same regardless of the order, i.e., \( ab = ba \).
- This property holds true in abelian groups, which is why they are sometimes referred to as commutative groups.
- Understanding if a group is commutative influences both the structure and type of algebraic problems it can solve.
- Knowing an operation is commutative simplifies many computations, making it easier to work with equations and transformations.
Homomorphic Images
A homomorphism is a special type of function between two groups that respects the group structure, meaning it preserves the ways elements combine. When we apply a homomorphism from one group \( G \) to another group \( H \), the image of this function, which consists of all elements in \( H \) that the function maps elements of \( G \) to, is referred to as a homomorphic image.
- A homomorphic image retains many properties of the original group \( G \), such as being abelian if \( G \) is abelian, as shown in the proof exercise.
- This feature is valuable because it enables the study of potentially more complex groups through their more simplified or reduced images in another group.
- Homomorphic images facilitate a deeper understanding of group properties, helping to discover invariant characteristics under group operations.
Other exercises in this chapter
Problem 1
If \(G\) is any group, let \(H\) and \(K\) be normal subgroups of \(G\) such that \(H \cap K=\\{e\\}\). Prove the following: \(\mathbf{1}\) Let \(h_{1}\) and \(
View solution Problem 1
If \(f: G \rightarrow H\) is a homomorphism, prove each of the following: For each element \(a \in G\), the order of \(f(a)\) is a divisor of the order of \(a\)
View solution Problem 1
In the following, let \(G\) denote an arbitrary group. Find all the normal subgroups (a) of \(S_{3}\) (b) of \(D_{4}\). Prove the following:
View solution Problem 1
Let \(G, H\), and \(K\) be groups. Prove the following: If \(f: G \rightarrow H\) and \(g: H \rightarrow K\) are homomorphisms, then their composite \(g \circ f
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