Problem 4 - A Book of Abstract Algebra

Question

Let \(G, M\), and \(N\) be groups, let \(f: G \rightarrow M\) be a homomorphism from \(G\) onto \(M\), and let \(h: G \rightarrow N\) be a homomorphism from \(G\) onto \(N\). Show that \(\phi(x)=(f(x), h(x))\) is a homomorphism from \(G\) onto \(M \times N\).

Step-by-Step Solution

Answer

\( \phi(x) = (f(x), h(x)) \) is an onto homomorphism.

Step 1: Understand the Mapping

We are given two homomorphisms: \( f: G \rightarrow M \) and \( h: G \rightarrow N \). We need to show that the function \( \phi: G \rightarrow M \times N \) defined by \( \phi(x) = (f(x), h(x)) \) is a homomorphism from \( G \) onto the product group \( M \times N \).

Step 2: Verify Function \(\phi\) is Well-defined

For every \( x \in G \), \( \phi(x) = (f(x), h(x)) \) is an element of \( M \times N \) because both \( f(x) \in M \) and \( h(x) \in N \). Therefore, \( \phi \) is well-defined.

Step 3: Check Homomorphism Property

A function \( \phi \) is a homomorphism if \( \phi(xy) = \phi(x) \phi(y) \) for all \( x, y \in G \). Calculate:\[ \phi(xy) = (f(xy), h(xy)) \]Since \( f \) and \( h \) are homomorphisms:\[ f(xy) = f(x)f(y) \quad \text{and} \quad h(xy) = h(x)h(y) \]Thus, \[ \phi(xy) = (f(x)f(y), h(x)h(y)) = (f(x), h(x))(f(y), h(y)) = \phi(x) \phi(y) \]Therefore, \( \phi \) preserves the group operation, confirming it is a homomorphism.

Step 4: Show \(\phi\) is Onto

To demonstrate \( \phi \) is surjective (onto), note that both \( f \) and \( h \) are onto. Let \( (m, n) \in M \times N \). Since \( f \) is onto, there exists \( x_1 \in G \) such that \( f(x_1) = m \). Similarly, since \( h \) is onto, there exists \( x_2 \in G \) such that \( h(x_2) = n \). Choose \( x \in G \) such that \( f(x) = m \) and \( h(x) = n \). Then, \( \phi(x) = (m, n) \), proving \( \phi \) is onto.

Key Concepts

Group Theory

Group Theory is a fascinating branch of mathematics dealing with algebraic structures known as groups. A group is a set with a binary operation that satisfies four key properties: closure, associativity, identity, and invertibility. In group theory, understanding mappings like homomorphisms between groups is crucial. A homomorphism is a function between two groups that preserves the group operation. This means if you have two elements, say \(a\) and \(b\), in your group \(G\), the homomorphism \(\phi\) from \(G\) to another group \(H\) satisfies \(\phi(ab) = \phi(a)\phi(b)\).

Different types of homomorphisms include isomorphisms (a bijective homomorphism), endomorphisms (a homomorphism from a group onto itself), and automorphisms (a bijective endomorphism).

• A homomorphism maintains the operation structure of the group.
• It's used to understand the relationship between different algebraic structures.
• They simplify complex structures into more manageable forms.

Grasping these concepts helps in understanding mappings like \(\phi(x)=(f(x), h(x))\) in the exercise.

Surjective Function

A surjective function, also known as an "onto" function, is a type of mapping where every element in the codomain has a pre-image in the domain. This implies that for every element \(y\) in the set \(Y\), there is at least one element \(x\) in set \(X\) such that \(f(x) = y\). In the context of group homomorphisms, checking if a homomorphism is surjective is crucial because it ensures the target group is fully 'covered' or 'reached'.

For the homomorphisms \(f: G \rightarrow M\) and \(h: G \rightarrow N\) given in the exercise, both being surjective means every element in groups \(M\) and \(N\) is the image of some element in \(G\). This property is key for proving that the combined mapping \(\phi(x) = (f(x), h(x))\) is also a surjective function.

• Surjective functions ensure the range and codomain are the same set.
• The exercise demonstrates this by finding elements \(x_1\) and \(x_2\) in \(G\) that map onto \(m\) and \(n\).
• Surjectivity is an important property in proving the existence of solutions in algebraic equations.

Well-Defined Mapping

A function is well-defined when it provides a unique output for each input from its domain. For a mapping \(\phi: G \rightarrow M \times N\) to be well-defined, each \(x\) in \(G\) must map to exactly one element \((f(x), h(x))\) in \(M \times N\). This ensures that there are no ambiguities or contradictions regarding the outputs of the function.

In the original exercise, the well-defined nature of \(\phi\) can be verified by noting that since both \(f\) and \(h\) are functions (and by definition, functions are well-defined), \(\phi(x) = (f(x), h(x))\) must assign exactly one pair in \(M \times N\) for each \(x\).

• Each input has a consistent, predictable output, without hierarchy or dependability issues.
• Well-defined mapping is crucial to ensure the integrity and reliability of a function.
• Breaking the mapping into components helps establish its well-defined status.

Product Groups

In group theory, a product group \(M \times N\) is formed by taking two groups \(M\) and \(N\) and defining operations on ordered pairs created from elements of these groups. The operation on the product group involves performing the group operation from \(M\) on the first elements of the pairs and \(N\) on the second elements.

In the original exercise, the function \(\phi(x) = (f(x), h(x))\) maps the group \(G\) into a product group \(M \times N\). This idea is significant because it captures the behavior of two homomorphisms at once into a single structure, enabling more complex interactions and allowing for multi-faceted analyses.

• Product group operation is defined for pairs \((a,b)\) and \((c,d)\) as \((a \cdot_M c, b \cdot_N d)\).
• They illustrate how different algebraic structures can combine to provide a new, larger group structure.
• Product groups are turning intricate relations between multiple structures into simpler, cohesive units.