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Chapter 1: Problem 36

Let \(\sigma=[123 \cdots n]\) in \(S_{n}\). Show that the conjugacy class of \(\sigma\) has \((n-1) !\) elements. Show that the centralizer of \(\sigma\) is the cyclic group generated by \(\sigma .\)

Short Answer

Expert verified
In \(S_n\), the conjugacy class of \(\sigma = (1 2 3 \cdots n)\) has \((n-1)!\) elements because there are as many conjugates of \(\sigma\) as there are permutations with a single transposition at the beginning. There are \((n-1)\) such transpositions and for each, there are \((n-2)!\) ways to permute the other elements. The centralizer of \(\sigma\) is the cyclic group generated by \(\sigma\) since all elements in the centralizer are powers of \(\sigma\) and all the powers of \(\sigma\) are in the centralizer.

Step by step solution

01

Conjugacy class size

To find the conjugacy class of \(\sigma\), we want to find all elements \(\tau\) in \(S_n\) such that: $$ \tau\sigma\tau^{-1} \sim \sigma $$ i.e., \(\tau\sigma\tau^{-1}\) has the same cycle structure as \(\sigma\). Consider the permutation \((1 \ 2) \in S_n\). We can compute: \[ (1\ 2)\sigma(1\ 2)^{-1} = (1\ 2)(1\ 2\ 3 \cdots n)(1\ 2) = (2\ 3 \cdots n\ 1), \] which also has the same cycle structure as \(\sigma\). Now, suppose that \(\tau\) is any permutation in \(S_n\) that starts with a transposition, say \((1\ m)\). We can show that: $$ \tau\sigma\tau^{-1} = (1\ m\ 2\ 3\cdots n), $$ which has the same cycle structure as \(\sigma\). This implies that there are as many conjugates of \(\sigma\) as there are permutations with a single transposition at the beginning. There are \((n-1)\) such transpositions and for each, there are \((n-2)!\) ways to permute the other elements. Thus, the conjugacy class of \(\sigma\) has \((n-1)!\) elements.
02

Centralizer of \(\sigma\)

Let \(Z(\sigma)\) denote the centralizer of \(\sigma\) in \(S_n\). We want to show that \(Z(\sigma)\) is the cyclic group generated by \(\sigma\). This means we want to show that all elements in \(Z(\sigma)\) are powers of \(\sigma\), and all the powers of \(\sigma\) are in \(Z(\sigma)\). By definition of the centralizer, \(Z(\sigma) =\{ \rho \in S_n \ |\ \rho\sigma =\sigma \rho\}\). Let \(\rho =\sigma^k\) be a power of \(\sigma\). Then: $$ \rho\sigma = \sigma^k\sigma = \sigma^{k+1} = \sigma\sigma^k = \sigma\rho. $$ This shows that all powers of \(\sigma\) are in \(Z(\sigma)\). Now consider any element \(\rho \in Z(\sigma)\) so that \(\rho\sigma =\sigma \rho\). Let's look at \(\rho(1)\). Since it commutes, apply both of them on the element 1. $$ \rho\sigma(1) =\sigma\rho(1) \quad \implies \quad \rho(2) =\sigma(\rho(1)) $$ Applying \(\sigma\) on both sides repeatedly, we can find the entire image of \(\rho\). Now, since \(\sigma\) is a single cycle, either all the elements are fixed, or only one element is fixed. If \(\rho\) fixes all elements, it is the identity permutation. If \(\rho\) has a single fixed element, the action of \(\rho\) must be identical to that of a power of \(\sigma\). Thus, we have shown that \(Z(\sigma)\) is the cyclic group generated by \(\sigma\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centralizer of Permutations
In the context of group theory, and specifically within symmetric groups, the centralizer of a permutation \( \sigma \) is the set of all elements in the group that commute with \( \sigma \). This means any permutation \( \rho \) in the centralizer satisfies the equation \( \rho\sigma = \sigma\rho \). The centralizer provides a way to understand symmetries within the group.

For the permutation \( \sigma = [123 \cdots n] \) in the symmetric group \( S_n \), its centralizer can be shown to be generated by \( \sigma \) itself. This forms a cyclic group. Every element of the centralizer is a power of \( \sigma \), meaning they are all different arrangements based on the cycle \( \sigma \) defines.
  • Allows understanding of symmetries formed by repeating a transformation.
  • The centralizer of \( \sigma \) identifies all permutations that share this symmetry.

Thus, studying the centralizer helps clarify the structure and repeated symmetries in permutation cycles.
Cycle Structure of Permutations
A permutation's cycle structure is a crucial feature in group theory, especially for understanding the symmetric group \( S_n \). It describes how elements are 鈥渕oved鈥� or 鈥渃ycled鈥� by a permutation. Knowing a permutation's cycle structure reveals valuable information about its properties.

For example, the cycle \( [123 \cdots n] \) implies a single cycle that includes all elements from 1 to \( n \). This cycle has length \( n \) and can be illustrated as starting from element 1, moving to element 2, then 3, and continuing till it reaches back to 1. The cycle structure helps differentiate permutations by the way elements are organized and cycled through the transformation it describes.
  • Defines the behavior of each element in the set.
  • Essential for understanding permutation properties and their effects.
Understanding cycle structures lets us analyze permutations easily, finding conjugate permutations and understanding equivalence in permutations.
Symmetric Group Properties
The symmetric group, denoted as \( S_n \), is one of the central constructs in the study of permutation groups, containing all possible permutations of a set with \( n \) elements. Symmetric groups have properties that make them fundamental in group theory.

Here are some key properties of symmetric groups:
  • Order: The number of elements in \( S_n \) is \( n!\) (factorial of \( n \)). This represents all possible permutations.
  • Closure: The composition of any two permutations is also a permutation in \( S_n \).
  • Associativity: Permutation composition is associative.
  • Identity: The identity permutation leaves all elements unchanged: a cycle of length one.
  • Inverses: Each permutation has an inverse that reverses order back to identity.
These properties ensure that symmetric groups are a complete and fully defined algebraic structure. They form the foundation for more complex symmetry and group-related concepts.
Permutation Conjugation
Permutation conjugation is an operation where one permutation acts upon another forming new permutations, preserving certain properties. It is described by the formula \( \tau\sigma\tau^{-1} \) where \( \sigma \) is the original permutation, and \( \tau \) is any permutation in the symmetric group.

Conjugation by \( \tau \) essentially "relabels" the elements within \( \sigma \), leading to a potentially distinct arrangement of cycles but maintaining the cycle structure. This means even if the arrangements look different, they retain equivalent transformation properties.
  • Creates a permutation family: all permutations with the same cycle structure.
  • Useful for finding equivalencies and organizing permutations into conjugacy classes.
  • Helps in calculating important properties like conjugacy classes size.
Conjugation is a fundamental tool for comparing permutations regarding how they transform a given set, illuminating relationships between different structural arrangements.

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Most popular questions from this chapter

Determine all groups of order \(\leqq 10\) up to isomorphism. In particular, show that a non-abelian group of order 6 is isomorphic to \(S_{3}\).

(Tits) Let \(M_{1}, \ldots, M_{r} \in G L_{2}(C)\) be a finite number of matrices. Let \(\lambda_{i}, \lambda_{i}^{\prime}\) be the eigenvalues of \(M_{i}\). Assume that each \(M_{i}\) has two distinct complex fixed points, and that \(\left|\lambda_{i}\right|<\left|\lambda_{i}^{\prime}\right|\). Also assume that the fixed points for \(M_{1}, \ldots, M\), are all distinct from each other. Prove that there exists a positive integer \(k\) such that \(M_{1}^{k}, \ldots, M_{r}^{k}\) are the free generators of a free subgroup of \(G L_{2}(C) .\left[\right.\) Hint: Let \(w_{i}, w_{i}^{\prime}\) be the fixed points of \(M_{i} .\) Let \(U_{i}\) be a small disc centered at \(w_{i}\) and \(U_{i}^{\prime}\) a small disc centered at \(w_{i}^{\prime}\). Let \(S_{i}=U_{i} \cup U_{i}^{\prime}\). Let \(s\) be a complex number which does not lie in any \(S_{i}\). Let \(G_{i}=\left(M_{i}^{k}\right) .\) Show that the conditions of Exercise 54 are satisfied for \(k\) sufficiently large.].

(a) Let \(n\) be an even positive integer. Show that there exists a group of order \(2 n\). generated by two elements \(\sigma, \tau\) such that \(\sigma^{n}=e=\tau^{2}\), and \(\sigma \tau=\tau \sigma^{n-1}\), (Draw a picture of a regular \(n\) -gon, number the vertices, and use the picture as an inspiration to get \(\sigma, \tau_{1}\) ) This group is called the dihedral group. (b) Let \(n\) be an odd positive integer. Let \(D_{4 n}\) be the group generated by the matrices $$ \left(\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right) \text { and }\left(\begin{array}{cc} \zeta & 0 \\ 0 & \zeta^{-1} \end{array}\right) $$ where \(\zeta\) is a primitive \(n\) -th root of unity. Show that \(D_{4 n}\) has order \(4 n\), and give the commutation relations between the above generators.

Let \(G\) be a group and let \(H, H^{\prime}\) be subgroups. By a double coset of \(H, H^{\prime}\) one means a subset of \(G\) of the form \(H x H^{\prime}\). (a) Show that \(G\) is a disjoint union of double cosets. (b) Let \(\\{c\\}\) be a family of representatives for the double cosets. For each \(a \in G\) denote by \([a] H^{\prime}\) the conjugate \(a H^{\prime} a^{-1}\) of \(H^{\prime}\). For each \(c\) we have a decomposition into ordinary cosets $$ H=\bigcup_{c} x_{c}\left(H \cap[c] H^{\prime}\right) $$ where \(\left\\{x_{c}\right\\}\) is a family of elements of \(H\), depending on \(c\). Show that the elements \(\left\\{x_{c} c\right\\}\) form a family of left coset representatives for \(H^{\prime}\) in \(G ;\) that is, $$ G=\bigcup_{x_{\epsilon}} \bigcup_{x_{e}} x_{c} c H^{\prime}, $$ and the union is disjoint. (Double cosets will not emerge further until Chapter XVIII.)

Let \(M \in G L_{2}(\mathbf{C})(2 \times 2\) complex matrices with non-zero determinant). We let $$ M=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right), \text { and for } z \in \mathbf{C} \text { we let } M(z)=\frac{a z+b}{c z+d} $$ If \(z=-d / c(c \neq 0)\) then we put \(M(z)=\infty\). Then you can verify (and you should have seen something like this in a course in complex analysis) that \(G L_{2}(\) C) thus operates on \(\mathbf{C} \cup\\{\infty\\} .\) Let \(\lambda, \lambda^{\prime}\) be the eigenvalues of \(M\) viewed as a linear map on \(\mathbf{C}^{2}\). Let \(W, W^{\prime}\) be the corresponding eigenvectors, $$ W={ }^{\prime}\left(w_{1}, w_{2}\right) \text { and } W^{\prime}={ }^{\prime}\left(w_{1}^{\prime}, w_{2}^{\prime}\right) . $$
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