Problem 14
Question
Let \(X_{1}\) and \(X_{2}\) be \(G\) -sets for the same group \(G,\) and assume \(X_{1} \cap \mathrm{X}_{2}=\varnothing\). Show how \(X_{1} \cup X_{2}\) can be made into a \(G\) -set in a natural way.
Step-by-Step Solution
Verified Answer
The union \(X_1 \cup X_2\) becomes a G-set by extending the group actions on \(X_1\) and \(X_2\).
1Step 1: Understand the concept of a G-set
A G-set is a set on which a group G acts. This means there is a function \(G \times X \to X\) that follows certain properties: for every element \(x \in X\), \(e \cdot x = x\) (where \(e\) is the identity element of \(G\)), and for every \(g, h \in G\) and \(x \in X\), \((gh) \cdot x = g \cdot (h \cdot x)\).
2Step 2: Consider the separate actions on \\(X_1\) and \\(X_2\\)
Given that \(X_1\) and \(X_2\) are both \(G\)-sets, there are actions \(G \times X_1 \to X_1\) and \(G \times X_2 \to X_2\) that satisfy the G-set properties. We denote these actions by \(g \cdot x_1\) and \(g \cdot x_2\) respectively, where \(g \in G\).
3Step 3: Define the action on the union set \\(X_1 \cup X_2\\)
Since \(X_1 \cap X_2 = \varnothing\), we can define an action on \(X_1 \cup X_2\) by specifying: for \(x \in X_1\), \(g \cdot x\) is already defined as in \(X_1\), and for \(x \in X_2\), \(g \cdot x\) is defined as in \(X_2\). Thus, \(g \cdot x\) is naturally extended to act on each subset in the union.
4Step 4: Verify the group action properties on \\(X_1 \cup X_2\\)
Check that the defined action satisfies the properties: for any \(x \in X_1 \cup X_2\), \(e \cdot x = x\) since it holds in each individual set. For \(g, h \in G\) and \(x \in X_1 \cup X_2\), \((gh) \cdot x = g \cdot (h \cdot x)\) is true because it follows from the G-action on \(X_1\) or \(X_2\). Thus, \(X_1 \cup X_2\) is a G-set.
Key Concepts
Understanding G-setsGroup Theory BasicsConnections with Set Theory
Understanding G-sets
Imagine a G-set as a playground for a group \( G \), where each piece of equipment (element of the set) is altered by the children (elements of the group \( G \)) in a consistent and structured manner. Here, the group \( G \) acts on the set \( X \), meaning there is a function \( G \times X \to X \) that determines how each group element transforms the set's elements. To qualify as a G-set, two key properties must be observed:
- Identity Action: For each element \( x \in X \), the identity element \( e \) of the group must satisfy \( e \cdot x = x \), meaning that doing nothing doesn't change anything.
- Compatibility with Group Operations: For any elements \( g \) and \( h \) in \( G \), and any element \( x \in X \), applying one action after the other should be the same as applying their combined action: \( (gh) \cdot x = g \cdot (h \cdot x) \).
Group Theory Basics
At the heart of group theory is the study of sets with operations that meet specific criteria, creating a rich field of symmetrical patterns. A group \( G \) is based on four primary characteristics:
- Closure: If \( a \) and \( b \) are in \( G \), so is their operation result \( a \cdot b \).
- Associativity: The rule \( (a \cdot b) \cdot c = a \cdot (b \cdot c) \) must always hold, ensuring operations are grouped consistently.
- Identity Element: There exists an element \( e \) in \( G \); for every \( a \) in \( G \), \( e \cdot a = a \cdot e = a \).
- Inverses: For each element \( a \) in \( G \), an inverse \( b \) must exist such that \( a \cdot b = b \cdot a = e \).
Connections with Set Theory
Set theory provides the foundational language required for discussing many mathematical concepts, including G-sets and group actions. A set is essentially a collection of distinct objects, known as elements, and these setups provide necessary frameworks for organizing and analyzing data.
- Empty Set: Symbolized as \( \emptyset \), it contains no elements, much like the intersection \( X_1 \cap X_2 = \emptyset \) from the problem, signifying no common elements between sets.
- Union and Intersection: Union \( X_1 \cup X_2 \) combines elements from both sets, while intersection \( X_1 \cap X_2 \) identifies shared elements.
- Subset: \( X_1 \subseteq X \) signifies all elements in \( X_1 \) are within \( X \).
Other exercises in this chapter
Problem 12
In Exercises 10 through 12 let \(G\) be a group let \(P(G)\) be the set of all subsets of \(G\), and consider the map \(G \times P(G) \rightarrow P(G)\) sending
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In Exercises 12 through 14 for the indicated permutation \(\sigma \in S_{5},\) find \((\) a) the number of conjugates of a in \(S_{5},\) (b) the centralizer of
View solution Problem 15
Show that no group of order \(p q\) where \(p\) and \(q\) are distinct primes is simple.
View solution Problem 15
In Exercises 15 through 18 let \(H\) be a subgroup of a group \(G\) and let \(X\) be the set \(\\{x H \mid x \in G\\}\) of all left cosets of \(H\) in \(G\). Le
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