Problem 19 - Algebra: Pure and Applied

Question

Let \(\phi: G \rightarrow G^{\prime}\) be an onto homomorphism with Kern \(\phi=K,\) and let \(H^{\prime}\) be a subgroup of \(G^{\prime}\). Show that there exists a subgroup \(H\) of \(G\) such that \(K \subseteq H\) and \(H / K \propto H^{\prime}\)

Step-by-Step Solution

Answer

Define \( H = \phi^{-1}(H') \) to ensure \( K \subseteq H \) and \( H / K \cong H' \).

Step 1: Understand the Problem

We need to find a subgroup \( H \) of \( G \) such that the kernel \( K \) of the homomorphism \( \phi \) is contained in \( H \) and the quotient group \( H / K \) is isomorphic to the subgroup \( H' \) of \( G' \). Here, \( \phi \) is onto or surjective, which is crucial for constructing \( H \).

Step 2: Utilize the Surjective Property of \( \phi \)

Since \( \phi \) is an onto homomorphism, for every element \( h' \in H' \), there exists an element \( h \in G \) such that \( \phi(h) = h' \). Define \( H \) as the pre-image of \( H' \) under \( \phi \), i.e., \( H = \phi^{-1}(H') \). This means \( H = \{ g \in G \ | \ \phi(g) \in H' \} \).

Step 3: Show \( K \subseteq H \)

The kernel \( K \) consists of all elements in \( G \) that map to the identity element in \( G' \). Since the identity element is part of any subgroup, \( \phi^{-1}(e') \subseteq \phi^{-1}(H') \), with \( e' \) being the identity in \( G' \). Therefore, \( K \subseteq H \).

Step 4: Establish the Isomorphism \( H / K \cong H' \)

Consider the map \( \psi: H \rightarrow H' \), defined by \( \psi(h) = \phi(h) \). Since \( \phi \) is a homomorphism, \( \psi \) inherits this property. We show \( \psi \) is onto because \( \phi \) is onto. The first isomorphism theorem tells us that \( H / \ker(\psi) \cong H' \). Since \( \ker(\psi) = K \), we have \( H / K \cong H' \).

Step 5: Conclusion

By defining \( H = \phi^{-1}(H') \), we have created a subgroup of \( G \) such that \( K \subseteq H \) and \( H / K \cong H' \). This satisfies the requirements of the problem.

Key Concepts

Subgroups

In group theory, a subgroup is a smaller group that is formed from a larger group, following the same operation as the larger group. To consider a subset of a group as a subgroup, it must satisfy certain conditions.

• Closure: If you take any two elements in the subgroup and perform the group operation, the result must also be in the subgroup.
• Identity: The identity element of the larger group must also be an element of the subgroup.
• Inverses: For every element in the subgroup, its inverse must also be in the subgroup.

Understanding subgroups is crucial because they reveal the internal structure and symmetry of a group. In the context of the exercise, we are looking to identify a subgroup of the original group, ensuring that certain conditions involving a homomorphism and its kernel are met. Subgroups are important because they help us understand more complex groups by studying these simpler, more manageable pieces.

Homomorphism

A homomorphism is a special kind of function that connects two groups, preserving the group operation. If you have two groups, say, \( G \) and \( G' \), a homomorphism \( \phi: G \rightarrow G' \) respects the structure of these groups. This means for any two elements \( a \) and \( b \) in \( G \), \( \phi(ab) = \phi(a)\phi(b) \).
The concept of homomorphism is central to understanding how different groups can relate to each other. In particular, an onto or surjective homomorphism can cover the entire group \( G' \), meaning every element of \( G' \) is the image of at least one element of \( G \). This is critical in our exercise, where \( \phi \) is onto, of \( G \) mapping onto \( H' \), which ultimately helps establish that \( H / K \) is isomorphic to \( H' \). Homomorphisms are essential for developing the concept of kernels and quotient groups, which lead us to isomorphism theorems.

Kernel

The kernel of a homomorphism is a fundamental concept in group theory, representing those elements in the original group that are mapped to the identity element in the target group. For a homomorphism \( \phi: G \rightarrow G' \), the kernel \( \ker(\phi) \) is defined as \( \ker(\phi) = \{ g \in G \ | \ \phi(g) = e' \} \), where \( e' \) is the identity element in \( G' \).
The kernel plays a crucial role in understanding the structure of groups because it forms a normal subgroup of \( G \). This means any element in the kernel, when combined with any element in \( G \), still adheres to the group structure.

• A trivial kernel (only the identity element) means that the homomorphism is injective, forming a one-to-one mapping.
• The kernel helps in constructing quotient groups, which, in turn, assist in establishing isomorphisms between groups.

In our exercise, \( K \), the kernel, is a crucial part in forming the subgroup \( H \), such that \( H / K \) relates to \( H' \) through isomorphism.

Isomorphism Theorem

The Isomorphism Theorem is a key principle in group theory, providing a bridge between different group structures. In essence, the theorem states that if there is a homomorphism \( \phi: G \rightarrow G' \), then the quotient group \( G / \ker(\phi) \) is isomorphic to the image of \( G \) under \( \phi \). This is written as \( G / \ker(\phi) \cong \phi(G) \).
This theorem helps us understand that when we factor out the kernel of a homomorphism from its domain, we are left with a group that is structurally the same as the image of that homomorphism.

• In a simple language, it ensures that even though the groups may appear different (one being a quotient group and the other an image), they possess the same algebraic structure.
• The isomorphism theorem is incredibly useful when dealing with quotient structures and subgroups, granting us a powerful tool to navigate complex group relations.

In the tackled exercise, applying the isomorphism theorem allows us to conclude that \( H / K \) is indeed isomorphic to \( H' \), achieving the solution's final requirement. Understanding this theorem provides a solid foundation for appreciating the beautiful interconnectedness between group theoretical concepts.