Problem 22 - Algebra: Pure and Applied

Question

Let \(H\) and \(K\) be subgroups of a group \(G\), with \(|H|=n\) and \(\mid K\\}=m\), where \(\operatorname{gcd}(n, m)=1 .\) Show that \(H \cap K=\\{e\\} .\)

Step-by-Step Solution

Answer

Since the gcd of their orders is 1, the intersection of \(H\) and \(K\) contains only the identity element.

Step 1: Understanding the Problem

We need to show that the intersection of two subgroups, \(H\) and \(K\), within a group \(G\) is solely the identity element \(e\). We know the orders \(|H| = n\) and \(|K| = m\) have no common factors other than 1, i.e., \(\operatorname{gcd}(n, m) = 1\).

Step 2: Applying Lagrange's Theorem

Lagrange's theorem tells us that the order of any subgroup of a finite group divides the order of the group. If \(x\) is an element in both \(H\) and \(K\), then \(x\) must belong to the intersection \(H \cap K\). Therefore, the order of \(x\) must divide both \(n\) and \(m\).

Step 3: Analyzing the Divisibility

Since \(x\)'s order divides both \(n\) and \(m\), it must also divide the greatest common divisor of \(n\) and \(m\), which is 1. The only element with order 1 in a group is the identity element, \(e\). Thus, \(x = e\).

Step 4: Conclusion

Since the only element in \(H \cap K\) is the identity, we conclude that \(H \cap K = \{e\}\). This shows that \(H\) and \(K\) are subgroups whose only common element is the identity element in \(G\).

Key Concepts

Lagrange's Theorem

Lagrange's Theorem is a fundamental result in group theory named after the mathematician Joseph Louis Lagrange. It states that for any finite group \( G \), the order (or the number of elements) of any subgroup \( H \) of \( G \) divides the order of \( G \) itself. This theorem is essential in understanding the structure of groups and their subgroups.
In the context of the exercise, both subgroups \( H \) and \( K \) have orders \( n \) and \( m \) respectively, and both need to satisfy this property. This theorem is crucial when analyzing the intersection of two subgroups since it helps us deduce the possible orders of the elements in \( H \cap K \).

• Understanding that the orders must divide the total elements in \( G \) can determine possible subgroups.
• Applying to intersections, it ensures we predict divisibility conditions like the Greatest Common Divisor insightfully.

Subgroup Intersection

In any group, the intersection of two subgroups \( H \) and \( K \) is defined as the set of elements common to both: \( H \cap K = \{ x \mid x \in H \text{ and } x \in K \} \). Understanding subgroup intersections is a critical aspect of solving many problems in group theory.
For our exercise, the goal is to show that \( H \cap K = \{ e \} \). This means the only thing these subgroups share is the identity element of the group. This scenario arises due to the unique property of prime orders of each subgroup, specified by their orders having no common divisors greater than 1.

• With \( \text{gcd}(n, m) = 1 \), no non-identity element can have an order satisfying both \( n \) and \( m \).
• Thus, the only intersection point, keeping both subgroup properties intact, is the identity element.

Greatest Common Divisor (GCD)

The greatest common divisor (GCD) is the largest positive integer that divides two given integers without leaving a remainder. In mathematics, understanding GCD concepts helps us analyze relationships between different numerical quantities, including group elements' orders.
In the exercise, \( n \) and \( m \) are the orders of the subgroups \( H \) and \( K \) respectively, and we know that \( \operatorname{gcd}(n, m) = 1 \). This fact is crucial because when two numbers are coprime, their only GCD is 1, implying no other number divides both.

• When orders \( n \) and \( m \) are coprime, it ensures the subgroups cannot intersect meaningfully except at their shared identity.
• This is the mathematical foundation for the exercise's conclusion of \( H \cap K = \{ e \} \).

Identity Element

The identity element in a group is a special element that, when combined with any element of the group using the group operation, leaves that element unchanged. For a group \( G \), the identity is usually denoted as \( e \).
In the context of groups and subgroup intersections, the identity element plays a crucial role in showcasing the intersection's triviality when two subgroups are coprime, as seen in our exercise. The only element that is a common solution to the intersection when \( \operatorname{gcd}(n, m) = 1 \) is \( e \).

• The identity element is unique for every group and signifies the definitive closure of the group operation.
• In intersections like \( H \cap K = \{ e \} \), it carries the entire concept of minimal non-trivial intersection.