Problem 1 - A Book of Abstract Algebra

Question

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 \(h_{2}\) be any two elements of \(H\), and \(k_{1}\) and \(k_{2}\) any two elements of \(K\). $$ h_{1} k_{1}=h_{2} k_{2} \quad \text { implies } \quad h_{1}=h_{2} \quad \text { and } \quad k_{1}=k_{2} $$ (HINT: If \(h_{1} k_{1}=h_{2} k_{2}\), then \(h_{2}^{-1} h_{1} \in H \cap K\) and \(k_{2} k_{1}^{-1} \in H \cap K\). Explain why.)

Step-by-Step Solution

Answer

If \( h_1 k_1 = h_2 k_2 \), then \( h_1 = h_2 \) and \( k_1 = k_2 \). This uses the fact that \( H \cap K = \{e\} \).

Step 1: Understand the Hypothesis

We have a group \( G \) with two normal subgroups \( H \) and \( K \) such that their intersection is the identity element, i.e., \( H \cap K = \{e\} \). This means that the only element common to both \( H \) and \( K \) is the identity element \( e \).

Step 2: Analyze the Given Equation

We are given that \( h_1 k_1 = h_2 k_2 \) for elements \( h_1, h_2 \in H \) and \( k_1, k_2 \in K \). We aim to prove that this equality implies \( h_1 = h_2 \) and \( k_1 = k_2 \).

Step 3: Rearrange the Equation

Rearrange the equation \( h_1 k_1 = h_2 k_2 \) to obtain \( h_1 k_1 k_2^{-1} = h_2 \). Multiply both sides by \( h_2^{-1} \) from the left to obtain \( h_2^{-1} h_1 = k_2 k_1^{-1} \).

Step 4: Utilize Subgroup Properties

Note that \( h_2^{-1} h_1 \in H \) since \( H \) is a subgroup and \( k_2 k_1^{-1} \in K \) since \( K \) is a subgroup. Thus, \( h_2^{-1} h_1 \in H \cap K \) and \( k_2 k_1^{-1} \in H \cap K \).

Step 5: Apply the Intersection Property

The intersection \( H \cap K = \{e\} \) implies that the only element in both \( H \) and \( K \) is the identity \( e \). Therefore, \( h_2^{-1} h_1 = e \) and \( k_2 k_1^{-1} = e \).

Step 6: Conclude with Element Equality

Since \( h_2^{-1} h_1 = e \), it follows that \( h_1 = h_2 \). Similarly, \( k_2 k_1^{-1} = e \) implies \( k_1 = k_2 \). Therefore, the initial equation \( h_1 k_1 = h_2 k_2 \) indeed implies \( h_1 = h_2 \) and \( k_1 = k_2 \).

Key Concepts

Group Theory

Group theory is a mathematical discipline with a wide range of applications in mathematics and beyond. Essentially, a group is a set combined with an operation that meets four key properties: closure, associativity, identity, and invertibility. This framework helps us understand symmetrical and other algebraical structures more deeply.
In our exercise, the group in question, denoted as \( G \), comes with two significant subgroups, \( H \) and \( K \). A subgroup orders a subset of a group into a new group using the same operation as the parent group. The properties of these subgroups deeply affect how elements inside them interact, especially in situations like normal subgroups, where the subgroup is invariant under conjugation by elements of the larger group.
Normal subgroups play a pivotal role in group theory as they allow us to construct quotient groups, which can simplify complex group operations into more manageable terms. In our scenario, both \( H \) and \( K \) are normal subgroups, emphasizing their robust algebraic structure and making it easier to analyze equations that involve elements from both.

Identity Element

The identity element is a fundamental concept in group theory. Within any group, the identity element is the unique element that, when combined with any other element of the group under the group operation, returns the other element unchanged.
For example, in the group of integers under addition, the identity element is 0 because adding 0 to any integer doesn’t change the integer. In our exercise, the intersection of subgroups \( H \) and \( K \) resulted in the identity element \( e \).
This means that the only element mutual to both subgroups is this identity element. Recognizing that \( h_2^{-1} h_1 \) must equal the identity in \( H \), and similarly, \( k_2 k_1^{-1} \) equals the identity in \( K \), solidifies that the elements must individually be equal, helping unravel the equality conditions in our equation.

Intersection of Subgroups

The concept of the intersection of subgroups refers to the set of elements that are found in both subgroups simultaneously. It essentially highlights shared characteristics between subgroups.
Our exercise introduces \( H \cap K = \{e\} \), meaning the only element shared between the subgroups \( H \) and \( K \) is the identity element \( e \). This intersection is crucial for the solution since it tells us that any element belonging to both subgroups must be this identity element.
The implication here is that both \( h_2^{-1} h_1 \) and \( k_2 k_1^{-1} \), obtained from rearranging our equation involving subgroup elements, have to lie within this intersection. This property simplifies the resolution of equality of elements in group operations significantly.

Element Equality

Element equality is about determining when two elements of a group are, in fact, the same. In group theory, showing that elements are equal can often break down complex problems into solvable pieces.
In our problem, we started with an equation \( h_1 k_1 = h_2 k_2 \), involving elements from the subgroups \( H \) and \( K \). To prove that \( h_1 = h_2 \) and \( k_1 = k_2 \), we had to rearrange the equation in such a way that revealed each element's position within the intersection \( H \cap K = \{e\} \).
By demonstrating that \( h_2^{-1} h_1 \) is the identity, we established \( h_1 = h_2 \). Similarly, showing that \( k_2 k_1^{-1} \) is the identity established \( k_1 = k_2 \). This clear path to element equality echoes the innate order and logic that group properties provide, making such proofs possible.