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Chapter 8: Problem 17

Let \(\sigma=24513\) and \(\tau=41352\) be permutations in \(S_{5} .\) Find (a) \(\tau \circ \sigma,(\mathrm{b}) \quad \sigma^{-1}\) Recall that \(\sigma=24513\) and \(\tau=41352\) are short ways of writing $$ \begin{aligned} &\sigma=\left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 4 & 5 & 1 & 3 \end{array}\right) \text { or } \quad \sigma(1)=2, \quad \sigma(2)=4, \quad \sigma(3)=5, \quad \sigma(4)=1, \quad \sigma(5)=3 \\ &\tau=\left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 4 & 1 & 3 & 5 & 2 c \end{array}\right) \quad \text { or } \quad \tau(1)=4, \quad \tau(2)=1, \quad \tau(3)=3, \quad \tau(4)=5, \quad \tau(5)=2 \end{aligned} $$ (a) The effects of \(\sigma\) and then \(\tau\) on \(1,2,3,4,5\) are as follows: $$ 1 \rightarrow 2 \rightarrow 1, \quad 2 \rightarrow 4 \rightarrow 5, \quad 3 \rightarrow 5 \rightarrow 2, \quad 4 \rightarrow 1 \rightarrow 4, \quad 5 \rightarrow 3 \rightarrow 3 $$ [That is, for example, \((\tau \circ \sigma)(1)=\tau(\sigma(1))=\tau(2)=1 .]\) Thus, \(\tau \circ \sigma=15243\). (b) By definition, \(\sigma^{-1}(j)=k\) if and only if \(\sigma(k)=j\). Hence, $$ \sigma^{-1}=\left(\begin{array}{lllll} 2 & 4 & 5 & 1 & 3 \\ 1 & 2 & 3 & 4 & 5 \end{array}\right)=\left(\begin{array}{lllll} 1 & 2 & 3 & 4 & 5 \\ 4 & 1 & 5 & 2 & 3 \end{array}\right) \quad \text { or } \quad \sigma^{-1}=41523 $$

Short Answer

Expert verified
a) \(\tau \circ \sigma = (1 4 5)(2 3)\) b) \(\sigma^{-1} = (1 4 2)(3 5)\)

Step by step solution

01

Write given permutations in cycle notation

To calculate \(\tau \circ \sigma\) and \(\sigma^{-1}\), it would be helpful to represent \(\sigma\) and \(\tau\) in cycle notation. The given permutations are as follows: \(\sigma = 24513\) \(\tau = 41352\) In cycle notation, we have: \(\sigma = (1 2 4 5 3)\) \(\tau = (1 4 3)(2 5)\) Now, we can perform the specified operations.
02

Calculate the composition \(\tau \circ \sigma\)

In order to calculate \(\tau \circ \sigma\), we apply \(\sigma\) first and then the \(\tau\) permutation. 1. Applying the \(\sigma\) permutation: \(\sigma(1) = 2,\, \sigma(2) = 4,\, \sigma(3) = 5,\, \sigma(4) = 1,\, \sigma(5) = 3\) 2. Now, we apply the \(\tau\) permutation to the intermediate result from \(\sigma\): \(\tau(2) = 5,\, \tau(4) = 1,\, \tau(5) = 2,\, \tau(1) = 4,\, \tau(3) = 3\) This gives us the final output for \(\tau \circ \sigma\): \(\tau \circ \sigma = (1 4 5)(2 3)\)
03

Calculate the inverse of the permutation \(\sigma\), \(\sigma^{-1}\)

The inverse of permutation is obtained by reversing the cycles. Given \(\sigma = (1 2 4 5 3)\), we can invert the cycle as follows: \(\sigma^{-1} = (1 4 2)(3 5)\) #Summary# To sum up, we have found: a) The composition of the two permutations: \(\tau \circ \sigma = (1 4 5)(2 3)\) b) The inverse of the permutation \(\sigma\): \(\sigma^{-1} = (1 4 2)(3 5)\)

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Key Concepts

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

Cycle Notation
Understanding the cycle notation is a crucial part of working with permutations, particularly in the symmetric group, denoted as Sn, where n represents the number of elements. In our context, we're dealing with permutations in S5, which consists of all possible permutations of five elements.

Cycle notation is an expressive way to represent a permutation by showing how groups of elements are permuted in cycles. A single cycle consists of a list of numbers in parentheses, indicating that each element is sent to the next element in the list, with the last element going back to the first. For example, the cycle (1 3 2) means that 1 goes to 3, 3 goes to 2, and 2 goes back to 1.

When a permutation consists of non-overlapping cycles, it's possible for us to write it as a product of these cycles. For a given permutation 蟽 in S5, with cycle notation, you can easily visualize the action of 蟽 on each element. It's especially handy when computing compositions of permutations, as it allows for a straightforward tracing of each element's movement.
Inverse Permutation
The concept of an inverse permutation is quite similar to the idea of inverting a function. If you think of a permutation as a way to shuffle objects, then the inverse permutation is the sequence of moves that would shuffle the objects back to their original places.

In mathematical terms, given a permutation 蟽, its inverse, denoted as 蟽-1, is another permutation such that 蟽-1(蟽(i)) = i for every element i in the set. This essentially means that if permutation 蟽 takes element i to position j, then 蟽-1 must take j back to position i.

In cycle notation, finding the inverse involves reversing the order of the elements within each cycle (but not the order of the cycles themselves). For example, the inverse of the cycle (1 2 4 5 3) is (3 5 4 2 1), which becomes (1 4 2)(3 5) when written with smallest number first and with disjoint cycles.
Permutation in S5
The symmetric group S5 includes all permutations of five elements, which are 1 through 5 in our case. There are exactly 5! (5 factorial), which is 120, different permutations in S5, and they can be quite complex to work with without a systematic approach like cycle notation.

Permutations in S5 can be composed and inverted, as we鈥檝e seen with our example permutations 蟽 and 蟿. The composition of two permutations 蟽 and 蟿, denoted 蟿 鈭� 蟽, follows the rule that for each element, you first apply 蟽 and then 蟿 on the result. This is a bit like following two sets of instructions in order.

Cycle notation really simplifies both the visualization and the calculations of such procedures. It organizes a potentially messy process into a series of individual mappings that can easily be followed or reversed, which is an invaluable tool for anyone studying permutations, algebra, or engaging in tasks that involve sorting or arranging items in a specific order.

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Most popular questions from this chapter

Let \(\mathbf{D}: V \rightarrow V\) be the differential operator; that is, \(\mathbf{D}(f(t))=d f / d t .\) Find \(\operatorname{det}(\mathbf{D})\) if \(V\) is the vector space of functions with the following bases: (a) \(\left\\{1, t, \ldots, t^{5}\right\\},\) (b) \(\left\\{e^{t}, e^{2 t}, e^{3 t}\right\\}\) (c) \(\\{\sin t, \cos t\\}\)

Prove Theorem \(8.8\) (Laplace): Let \(A=\left[a_{i j}\right]\), and let \(A_{i j}\) denote the cofactor of \(a_{i j}\). Then, for any \(i\) or \(j\) $$ |A|=a_{i 1} A_{i 1}+\cdots+a_{i n} A_{i n} \quad \text { and } \quad|A|=a_{1 j} A_{1 j}+\cdots+a_{n j} A_{n j} $$Because \(|A|=\left|A^{T}\right|\), we need only prove one of the expansions, say, the first one in terms of rows of \(A\) Each term in \(|A|\) contains one and only one entry of the ith row \(\left(a_{i 1}, a_{i 2}, \ldots, a_{i n}\right)\) of \(A\). Hence, we can write \(|A|\), in the form $$ |A|=a_{i 1} A_{i 1}^{3}+a_{i 2} A_{12}^{*}+\cdots+a_{i n} A_{i n}^{*} $$ (Note that \(A_{1 j}^{\text {* }}\) is a sum of terms involving no entry of the ith row of \(A\).) Thus, the theorem is proved if we can show that $$ A_{i j}^{4}=A_{i j}=(-1)^{i+j}\left|M_{i j}\right| $$ where \(M_{i j}\) is the matrix obtained by deleting the row and column containing the entry \(a_{i j}\). (Historically, the expression \(A_{i j}^{\text {w }}\) was defined as the cofactor of \(a_{i j}\), and so the theorem reduces to showing that the two definitions of the cofactor are equivalent.) First we consider the case that \(i=n, j=n\). Then the sum of terms in \(|A|\) containing \(a_{n n}\) is $$ a_{m u} A_{n n}^{n}=a_{n n} \sum_{\sigma}(\operatorname{sgn} \sigma) a_{1 \sigma(1)} a_{2 \sigma(2)} \cdots a_{n-1, \sigma(n-1)} $$ where we sum over all permutations \(\sigma \in S_{n}\) for which \(\sigma(n)=n .\) However, this is equivalent (Prove!) to summing over all permutations of \(\\{1, \ldots, n-1\\} .\) Thus, \(A_{m}^{*}=\left|M_{n n}\right|=(-1)^{n+n}\left|M_{n n}\right|\). Now we consider any \(i\) and \(j\). We interchange the ith row with each succeeding row until it is last, and we interchange the \(j\) th column with each succeeding column until it is last. Note that the determinant \(\left|M_{i j}\right|\) is not affected, because the relative positions of the other rows and columns are not affected by these interchanges. However, the "sign" of \(|A|\) and of \(A_{i j}^{\mathrm{~ \cjkstart ? ? ? ?}?\) $$ A_{i j}^{*}=(-1)^{n-i+n-j}\left|M_{i j}\right|=(-1)^{i+j}\left|M_{i j}\right| $$

Let \(A\) be an \(n\)-square matrix. Prove \(|k A|=k^{n}|A|\).

Evaluate each of the following determinants: (a) \(\left|\begin{array}{rrrrr}1 & 2 & -1 & 3 & 1 \\ 2 & -1 & 1 & -2 & 3 \\\ 3 & 1 & 0 & 2 & -1 \\ 5 & 1 & 2 & -3 & 4 \\ -2 & 3 & -1 & 1 & -2\end{array}\right|\) (b) \(\left|\begin{array}{lllll}1 & 3 & 5 & 7 & 9 \\ 2 & 4 & 2 & 4 & 2 \\ 0 & 0 & 1 & 2 & 3 \\ 0 & 0 & 5 & 6 & 2 \\ 0 & 0 & 2 & 3 & 1\end{array}\right|\) (c) \(\left|\begin{array}{lllll}1 & 2 & 3 & 4 & 5 \\ 5 & 4 & 3 & 2 & 1 \\ 0 & 0 & 6 & 5 & 1 \\ 0 & 0 & 0 & 7 & 4 \\ 0 & 0 & 0 & 2 & 3\end{array}\right|\)

Prove Theorem 8.12: Suppose \(M\) is an upper (lower) triangular block matrix with diagonal blocks \(A_{1}, A_{2}, \ldots, A_{n} .\) Then $$ \operatorname{det}(M)=\operatorname{det}\left(A_{1}\right) \operatorname{det}\left(A_{2}\right) \cdots \operatorname{det}\left(A_{n}\right) $$ We need only prove the theorem for \(n=2\)-that is, when \(M\) is a square matrix of the form \(M=\left[\begin{array}{cc}A & C \\ 0 & B\end{array}\right]\). The proof of the general theorem follows easily by induction. Suppose \(A=\left[a_{i j}\right]\) is \(r\)-square, \(B=\left[b_{i j}\right]\) is \(s\)-square, and \(M=\left[m_{i j}\right]\) is \(n\)-square, where \(n=r+s\). By definition, $$ \operatorname{det}(M)=\sum_{\sigma \in S_{\Sigma}}(\operatorname{sgn} \sigma) m_{1 \sigma(1)} m_{2 \sigma(2)} \cdots m_{n \sigma(n)} $$ If \(i>r\) and \(j \leq r\), then \(m_{i j}=0\). Thus, we need only consider those permutations \(\sigma\) such that \(\sigma\\{r+1, r+2, \ldots, r+s\\}=\\{r+1, r+2, \ldots, r+s\\} \quad\) and \(\quad \sigma\\{1,2, \ldots, r\\}=\\{1,2, \ldots, r\\}\) Let \(\sigma_{1}(k)=\sigma(k)\) for \(k \leq r\), and let \(\sigma_{2}(k)=\sigma(r+k)-r\) for \(k \leq s\). Then $$ (\operatorname{sgn} \sigma) m_{1 \sigma(1)} m_{2 \sigma(2)} \cdots m_{n \sigma(n)}=\left(\operatorname{sgn} \sigma_{1}\right) a_{1 \sigma_{1}(1)} a_{2 \sigma_{1}(2)} \cdots a_{r \sigma_{1}(r)}\left(\operatorname{sgn} \sigma_{2}\right) b_{1 \sigma_{2}(1)} b_{2 \sigma_{2}(2)} \cdots b_{s \sigma_{2}(s)} $$ which implies \(\operatorname{det}(M)=\operatorname{det}(A) \operatorname{det}(B)\)
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