Problem 7 - Applied Discrete Structures

Question

Prove that if \(H, K \leq G,\) and \(H \cup K=G,\) then \(H=G\) or \(K=G\). Hint. Use an indirect argument.

Step-by-Step Solution

Answer

If \(H \cup K = G\), then \(H = G\) or \(K = G\) by contradiction.

Step 1: Understand the Given Information

You are given that \(H\) and \(K\) are subgroups of \(G\), and their union forms the entire group \(G\). This means every element \(g\) in \(G\) is either in \(H\) or in \(K\), or both.

Step 2: Set Up Indirect Argument

Assume indirectly that neither \(H\) nor \(K\) equals \(G\). This means there exist elements \(x\) and \(y\) such that \(x \in G\setminus H\) and \(y \in G\setminus K\). In other words, \(x otin H\) and \(y otin K\).

Step 3: Utilize the Group Properties

Since \(H\) and \(K\) are subgroups, they must be closed under the group operation. Therefore, the product \(xy\) must lie in either \(H\) or \(K\) as \(x, y \in G\) and \(H \cup K = G\).

Step 4: Contradiction Occurrence

If \(xy \in H\), then since subgroups are closed under taking inverses, \(x = (xy)y^{-1} \in H\). But this contradicts \(x otin H\). Similarly, if \(xy \in K\), then \(y = x^{-1}(xy) \in K\), contradicting \(y otin K\).

Step 5: Conclude the Proof

The contradiction proves our assumption wrong, meaning at least one of the original assumptions is incorrect. Therefore, it must be that \(H=G\) or \(K=G\), otherwise, a contradiction arises.

Key Concepts

Subgroups

In group theory, the concept of subgroups is fundamental. A subgroup is essentially a smaller group within a larger group, sharing the same operational rules as the larger group.

If we have a group \( G \), a subset \( H \) of \( G \) is a subgroup if it satisfies three main properties:

• It contains the identity element of \( G \).
• It is closed under the group operation. This means for any two elements \( a, b \) in \( H \), the product \( ab \) is also in \( H \).
• It is closed under taking inverses. For any element \( a \) in \( H \), its inverse \( a^{-1} \) is also in \( H \).

Understanding these properties ensures that any subgroup shares the key structural characteristics of the entire group, hence contributing to its overall behavior.

Indirect Argument

An indirect argument, also known as proof by contradiction, is a powerful method in mathematics. The idea is to assume the opposite of what you want to prove, and then demonstrate that this assumption leads to a contradiction.

For example, in proving that at least one of the subgroups \( H \) or \( K \) equals the group \( G \) given their union equals \( G \), we assume the opposite scenario: neither \( H \) nor \( K \) equals \( G \).

If such an assumption leads to a logical inconsistency, we can conclude that the original statement must hold true. This technique is particularly useful when direct verification of the statement is complex or not straightforward.

Group Properties

To comprehend groups and their subgroups, it's crucial to grasp the properties that define them. Groups are defined by a set paired with an operation that satisfies four key properties:

• Closure: If \( a \) and \( b \) are in the group, then \( ab \) is also in the group.
• Associativity: The equation \( (ab)c = a(bc) \) holds for any elements \( a, b, \, \text{and} \, c \) in the group.
• Identity Element: There is an element \( e \) such that for any element \( a \) in the group, the equation \( ea = ae = a \) holds.
• Inverse Element: For each element \( a \) in the group, there is an element \( b \) such that \( ab = ba = e \), where \( e \) is the identity element.

Subgroups that respect these group properties maintain the inner coherence of the group, ensuring all standard operations and relations are preserved within the subgroup.

Union of Subgroups

A very interesting scenario arises when we consider the union of subgroups. Typically, the union of two subgroups is not itself a subgroup. This is because the set may fail to satisfy the closure and inverse properties.

However, in the problem where \( H \cup K = G \), every element of \( G \) belongs to at least one of the subgroups. Therefore, in this special situation, the union comprises all elements of the group \( G \), which forces one of the subgroups to be equivalent to \( G \).

• When \( H \cup K = G \), without either \( H \) or \( K \) equalling \( G \), some elements and operations would have to simultaneously belong to both subgroups due to the group properties.
• This imposes a structure that aligns with group operations, making it necessary for one of the subgroups to be the entire group \( G \).

This unique characteristic of subgroup unions reveals the intricate balance in mathematical structures, showing how assumptions in mathematics can sometimes lead to unexpected and insightful results.