Problem 1 - A Book of Abstract Algebra

Question

Let \(G\) be a group; let \(H\) and \(K\) be subgroups of \(G\), with \(H\) a normal subgroup of \(G\). Prove the following : $$ H \cap K \text { is a normal subgroup of } K \text {. } $$

Step-by-Step Solution

Answer

Since \( H \) is normal in \( G \), \( H \cap K \) is normal in \( K \).

Step 1: Define Normal Subgroup in K

A subgroup \( N \) of a group \( K \) is normal if for every element \( k \in K \) and every element \( n \in N \), the element \( k n k^{-1} \) is also in \( N \). We need to show that this holds for \( H \cap K \) as a subgroup of \( K \).

Step 2: Establish Membership in H \cap K

For any element \( h \in H \cap K \), we know \( h \) is in both \( H \) and \( K \) since \( H \cap K = \{ x \in G : x \in H \text{ and } x \in K \} \).

Step 3: Use Normality of H in G

Since \( H \) is normal in \( G \), for any \( g \in G \) and \( h \in H \), \( g h g^{-1} \in H \). Specifically, for any \( k \in K \subseteq G \), we have \( k h k^{-1} \in H \).

Step 4: Stay in Subgroup K

Since \( k \in K \) and \( h \in K \), it follows that their product \( k h k^{-1} \) must also be in \( K \) because subgroups are closed under group operations and inverses. Hence, \( k h k^{-1} \in K \).

Step 5: Conclusion of Normality for H \cap K in K

Combining steps 3 and 4, for any \( k \in K \) and \( h \in H \cap K \), \( k h k^{-1} \) is in both \( H \) and \( K \). Therefore, \( k h k^{-1} \in H \cap K \), showing \( H \cap K \) is normal in \( K \).

Key Concepts

Group Theory

Group theory is an essential area of mathematics that studies the algebraic structures known as groups. A group consists of a set of elements together with an operation that combines two elements to form a third while satisfying four fundamental properties.

These properties are:

• Closure: The group operation is closed, meaning the operation on any two elements results in another element within the same set.
• Associativity: The operation is associative, which implies that changing the grouping of operations does not change the result.
• Identity Element: There exists an element (usually denoted as "e") in the group that acts as an identity, so for any element "g," the equation \(e \, * \, g = g \) holds true.
• Inverse Element: For every element "g" in the group, there is another element "g\(^{-1}\)" such that \(g \, * \, g^{-1} = e\).

Understanding these properties is crucial as they apply to various mathematical and applied science fields, providing a foundation for more advanced concepts such as subgroups and normal subgroups.

Subgroup

A subgroup is a subset of a group that itself forms a group under the operation defined on the larger group. For a subset "H" to qualify as a subgroup of a group "G," three specific conditions must be satisfied:

• Non-empty: The subset must not be empty; usually, it contains at least the identity element "e" of the group "G."
• Closure: If you take any two elements from "H," their product under the group's operation should also be in "H."
• Inverses: For every element in "H," its inverse must also be in "H."

Subgroups are important because they preserve the group structure of the larger group while existing as smaller, manageable units of analysis within the larger framework of group theory.

Intersection of Subgroups

When considering two subgroups, "H" and "K," within a group "G," their intersection, denoted as \(H \cap K\), is composed of all elements that belong to both subgroups simultaneously.

This intersection itself forms a subgroup, according to subgroup criteria. This is because:

• Non-empty: Since both "H" and "K" cannot be empty by subgroup definition, their intersection contains at least the identity element.
• Closure: Given any two elements in \(H \cap K\), their product under group operations is in both subgroups because both are closed under group operations. Therefore, their intersection is also closed.
• Inverses: For any element in \(H \cap K\), its inverse exists in both "H" and "K". Therefore, the inverse is also in \(H \cap K\).

This intersection concept is foundational in understanding how subgroups relate to each other and helps in proving more complex properties like the normality condition.

Normality Condition

In group theory, a subgroup is called normal if it satisfies a specific condition regarding the conjugation with elements of the larger group. Specifically, a subgroup "N" of a group "G" is said to be "normal" if for every element \(g \in G\) and every element \(n \in N\), the element \(g n g^{-1}\) is also in \(N\).

This property ensures that the subgroup is invariant under conjugation by any group element, which is crucial for forming quotient groups.

In the context of the problem, we focus on demonstrating that the intersection \(H \cap K\) is normal in "K," given that "H" is normal in "G." By analyzing elements of this intersection when conjugated by elements from "K," we observe that

• Since \(H\) is normal in \(G\), for every \(k \in K\), conjugation \(k h k^{-1} \in H\)
• The operation already ensures closure in \(K\), so \(k h k^{-1} \in K\)
• As a result, \(k h k^{-1} \in H \cap K\), demonstrating normality

This conclusion is essential since normal subgroups enable the creation of new groups (quotient groups), contributing significantly to the structure theory in group theory.