Problem 1 - A Book of Abstract Algebra

Question

Let \(G\) and \(H\) be groups. Suppose \(J\) is a normal subgroup of \(G\) and \(K\) is a normal subgroup of \(H\) Show that the function \(f(x, y)=(J x, K y)\) is a homomorphism from \(G \times H\) onto \((G / J) \times(H / K)\)

Step-by-Step Solution

Answer

The function \( f(x, y) = (Jx, Ky) \) is a homomorphism from \( G \times H \) onto \( (G/J) \times (H/K) \).

Step 1: Define the Required Homomorphism

The function in question is defined as \( f : G \times H \to (G/J) \times (H/K) \) given by \( f(x,y) = (Jx, Ky) \). We need to show that \( f \) is a group homomorphism and onto.

Step 2: Verify Group Operation Properties for Homomorphism

For a function to be a homomorphism, it must preserve the group operation. Consider two elements \( (x_1, y_1) \), \( (x_2, y_2) \) in \( G \times H \). We need to show that \( f((x_1, y_1)(x_2, y_2)) = f(x_1, y_1)f(x_2, y_2) \). The left-hand side is \( f(x_1x_2, y_1y_2) = (J(x_1x_2), K(y_1y_2)) \). The right-hand side is \( (Jx_1, Ky_1)(Jx_2, Ky_2) = (Jx_1Jx_2, Ky_1Ky_2) = (J(x_1x_2), K(y_1y_2)) \), showing \( f \) is a homomorphism.

Step 3: Show Surjectivity of the Homomorphism

To show \( f \) is onto, take any element \( (aJ, bK) \) in \( (G/J) \times (H/K) \). We need preimages in \( G \times H \) such that they map to \( (aJ, bK) \). Consider \( (a, b) \) in \( G \times H \), then \( f(a, b) = (Ja, Kb) = (aJ, bK) \), thus \( f \) is surjective.

Step 4: Check the Function is a Well-Defined Map

Since \( J \) and \( K \) are normal subgroups, cosets like \( Jx \) and \( Ky \) are well-defined in the factor groups. This ensures that the map \( f \) does not depend on the specific representatives chosen from \( G \) and \( H \). Since each element maps uniquely, \( f \) is well-defined.

Key Concepts

Normal Subgroup

A normal subgroup is a subgroup that remains invariant under conjugation by any element of the original group. This means that for a subgroup \(J\) of \(G\), \(gJg^{-1} = J\) for all \(g\) in \(G\). This property is crucial because it ensures that the left and right cosets of \(J\) in \(G\) are the same, i.e., \(gJ = Jg\).
This characteristic is essential when creating factor groups, as it simplifies the process and guarantees that operations on the cosets are well-defined. Normal subgroups are significant because they automatically satisfy the requirements needed for the construction of quotient groups, allowing us to form meaningful results like the map \(f(x, y) = (Jx, Ky)\) in the given exercise.

Coset

A coset is a subset formed by multiplying all elements of a subgroup by a fixed element from the group. Specifically, given a subgroup \(H\) of a group \(G\) and an element \(g\) from \(G\), the left coset is represented as \(gH\), and the right coset as \(Hg\).
The concept of cosets is foundational in group theory. If \(H\) is a normal subgroup of \(G\), then the coset \(gH\) is the same as \(Hg\), simplifying operations within the factor group \(G/H\). Cosets play a central role in understanding the structure of the group and are studied using the properties of normal subgroups.

• Cosets partition the group into equal-sized, disjoint subsets.
• They are crucial for defining quotient groups, like \(G/J\) and \(H/K\) in the problem.

Factor Group

A factor group, or quotient group, is constructed by taking the cosets of a normal subgroup. This is expressed as \(G/J\) where \(J\) is a normal subgroup of \(G\).
Factor groups allow us to break down complex groups into simpler components by considering the equivalence classes formed by these cosets. In the context of the exercise, each element of the factor group \((G/J) \times (H/K)\) corresponds to a coset of the normal subgroups \(J\) in \(G\) and \(K\) in \(H\).

• The group operation in a factor group is defined by multiplying representatives from each coset.
• Factor groups simplify and provide insight into the structure of larger groups by examining their smaller, more manageable parts.

Surjectivity

Surjectivity, or being a surjective function, means that every element in the codomain is the image of at least one element from the domain. This property ensures that the function ‘covers’ the entire codomain.
To show that \(f(x, y) = (Jx, Ky)\) is surjective, we need to verify that for every element of \((G/J) \times (H/K)\), there exists a pair in \(G \times H\) mapping to it. This exercise demonstrates this by choosing an element \((a, b)\) from \(G \times H\) such that \(f(a, b) = (Ja, Kb) = (aJ, bK)\), ensuring that any potential coset from the quotient groups is achievable.
Key points about surjectivity include:

• It guarantees that all parts of the target set are represented.
• In the context of homomorphisms, surjectivity confirms that the map is compressing the group structure without leaving any gaps.