Problem 1
Question
If \(\alpha=\left(a_{1} \cdots a_{2}\right)\) is a cycle of length \(s\), then
\(\alpha^{s}=\varepsilon, \alpha^{2 \mathrm{~s}}=\varepsilon\), and \(\alpha^{3
s}=\varepsilon .\) Is \(\alpha^{k}=\varepsilon\) for any positive integer \(k
Step-by-Step Solution
Verified Answer
No, \( \alpha^k \neq \varepsilon \) for any positive integer \( k < s \).
1Step 1: Understanding the Given Cycle
A cycle \( \alpha = (a_1 \cdots a_s) \) of length \( s \) is a permutation of \( s \) elements, where each element point to the next one in the cycle, and the last element points back to the first. Applying \( \alpha \) \( s \) times results in each element returning to its original position, hence \( \alpha^s = \varepsilon \), where \( \varepsilon \) is the identity permutation.
2Step 2: Analyzing Powers of the Cycle
Based on the definition, raising \( \alpha \) to any multiple of \( s \) will result in the identity permutation as every element will complete full cycles and return to its initial position. Therefore, \( \alpha^{2s} = \varepsilon \), \( \alpha^{3s} = \varepsilon \), and similarly \( \alpha^{ns} = \varepsilon \) for any positive integer \( n \).
3Step 3: Determining Cycle Powers Less Than Length
For a positive integer \( k < s \), raising \( \alpha \) to the power of \( k \) results in a permutation that has not completed full cycles, meaning not all elements have returned to their starting points. Therefore, \( \alpha^k eq \varepsilon \) for any positive integer \( k < s \), because \( k \) applications of the cycle do not cover all positions.
Key Concepts
Cycle NotationPermutationsIdentity PermutationCycle LengthGroup Theory
Cycle Notation
Cycle notation is a way to express permutations by listing the elements in the order they are arranged in the cycle. Let's say we have a cycle \( \alpha = (a_1 \, a_2 \, \cdots \, a_s) \). This represents that the element \( a_1 \) is sent to \( a_2 \), \( a_2 \) is sent to \( a_3 \), and so on. Finally, \( a_s \) wraps around to map back to \( a_1 \).
Using this notation simplifies examining how elements shift under a permutation. It also helps in calculating the power of cycles, where repeated application is easily visualized with this notation.
Using this notation simplifies examining how elements shift under a permutation. It also helps in calculating the power of cycles, where repeated application is easily visualized with this notation.
Permutations
Permutations are arrangements or orderings of elements in a set. The cycle notation, as explained before, is a form of expressing these permutations. In abstract algebra, permutations are crucial because they help us understand the structure of various mathematical objects, especially in group theory.
Each permutation can be decomposed into one or more cycles. This breakdown allows for a straightforward interpretation of the permutation's behavior, making it approachable to determine how an application of permutations affects a set.
Each permutation can be decomposed into one or more cycles. This breakdown allows for a straightforward interpretation of the permutation's behavior, making it approachable to determine how an application of permutations affects a set.
Identity Permutation
The identity permutation, denoted as \( \varepsilon \), is a specific permutation where all elements remain in their original position.
Think of it as doing nothing to the arrangement—every element maps to itself. Therefore, when a cycle is applied such that all elements return to their original position, we say the cycle results in the identity permutation. This property is essential in observing the repetition of cycles in permutation groups.
Think of it as doing nothing to the arrangement—every element maps to itself. Therefore, when a cycle is applied such that all elements return to their original position, we say the cycle results in the identity permutation. This property is essential in observing the repetition of cycles in permutation groups.
Cycle Length
The cycle length is the total number of elements that a cycle permutes. If a cycle \( \alpha \) has a length \( s \), it means \( s \) elements are in a repeated sequence.
A crucial point here is that applying a cycle to itself as many times as its length produces the identity permutation. That is, \( \alpha^s = \varepsilon \). For any positive integer \( k < s \), repeated applications do not complete the cycle, so the result will not be the identity permutation, as some elements have yet to return to their starting position.
A crucial point here is that applying a cycle to itself as many times as its length produces the identity permutation. That is, \( \alpha^s = \varepsilon \). For any positive integer \( k < s \), repeated applications do not complete the cycle, so the result will not be the identity permutation, as some elements have yet to return to their starting position.
Group Theory
Group theory is a branch of mathematics that studies the algebraic structures known as groups. In the context of permutations, a group consists of a set combined with an operation satisfying a few conditions like closure, associativity, identity, and invertibility.
When considering permutations, each permutation is an element of a symmetric group. This group represents all possible permutations of a set, and the identity permutation acts as the identity element of the group. Understanding the behavior of cycles and their lengths, as in the concept of cycle length, is a pivotal aspect of recognizing the broader structure and function of permutation groups in algebraic systems.
When considering permutations, each permutation is an element of a symmetric group. This group represents all possible permutations of a set, and the identity permutation acts as the identity element of the group. Understanding the behavior of cycles and their lengths, as in the concept of cycle length, is a pivotal aspect of recognizing the broader structure and function of permutation groups in algebraic systems.
Other exercises in this chapter
Problem 1
Prove that the set \(T\) of all the transpositions in \(S_{n}\) generates \(S_{n}\).
View solution Problem 1
Let \(\alpha=\left(a_{1} \cdots a_{s}\right)\) and \(\beta=\left(b_{1} \cdots b_{s}\right)\) be cycles of the same length, and let \(\pi\) be any permutation. I
View solution Problem 1
Determine which of the following permutations is even, and which is odd. (a) \(\pi=\left(\begin{array}{llllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 7 & 4 & 1 & 5 &
View solution Problem 1
Compute each of the following products in \(S_{9}\). (Write your answer as a single permutation.). (a) \((145)(37)(682)\) (b) \((17)(628)(9354)\) (c) \((71825)(
View solution