/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Let \(G=S_{4}\). (a) For \(\alph... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 1

Let \(G=S_{4}\). (a) For \(\alpha=\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1\end{array}\right)\), find the subgroup \(H=\langle\alpha\rangle\). (b) Determine the left cosets of \(H\) in \(G\).

Short Answer

Expert verified
The subgroup \(H=\langle\alpha\rangle\) generated by \(\alpha\) is \{e, (1 2 3 4), (1 3 4 2), (1 4 2 3)\}, and there are 6 left cosets of \(H\) in \(G\) as \(|G|=24\) and \(|H|=4\).

Step by step solution

01

Understanding the permutation \(\alpha\)

The permutation \(\alpha\) is given in two-row notation. The top row represents the elements being permuted, and the bottom row represents the targets of these elements under the permutation. So \(\alpha\) consists of the cycle (1 2 3 4)
02

Finding the subgroup generated by \(\alpha\)

To find the subgroup \(H=\langle\alpha\rangle\), generate all the elements by repeatedly applying the operation of \(\alpha\). Start with the identity permutation \(e\), then apply \(\alpha\) to get (1 2 3 4), apply \(\alpha\) again to get (1 3 4 2), apply \(\alpha\) once more to get (1 4 2 3), and apply \(\alpha\) one more time to return to the identity. Thus, \(H=\langle\alpha\rangle=\{e, \alpha, \alpha^2, \alpha^3\}\
03

Generating the left cosets of \(H\) in \(G\)

Generating a left coset of \(H\) in \(G\) involves operating each element in \(G\) on \(H\). As \(G=S_4\) has 24 elements (4!), ideally this procedure should be repeated 24 times. However, noticing that \(|H|=4\), a property of cosets ensures that all left cosets have the same order as \(H\). So, there only can be 24/4=6 different left cosets of \(H\) in \(G\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Why is the set \(\mathbf{Z}\) not a group under subtraction?

Express each of the following elements of \(S_{7}\) as a product of disjoint cycles. $$ \begin{aligned} \alpha &=\left(\begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 4 & 6 & 7 & 1 & 5 & 3 \end{array}\right) \\ \beta &=\left(\begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 3 & 6 & 5 & 2 & 1 & 7 & 4 \end{array}\right) \\ \gamma &=\left(\begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 3 & 1 & 7 & 5 & 4 & 6 \end{array}\right) \\ \delta &=\left(\begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 4 & 2 & 7 & 1 & 3 & 6 & 5 \end{array}\right) \end{aligned} $$

Let \(E: \mathbf{Z}_{2}^{3} \rightarrow \mathbf{Z}_{2}^{9}\) be the encoding function for the \((9,3)\) triple repetition code. a) If \(D: \mathbf{Z}_{2}^{9} \rightarrow \mathbf{Z}_{2}^{3}\) is the corresponding decoding function, apply \(D\) to decode the received words (i) 111101100 ; (ii) 000100011 ; (iii) 010011111 . b) Find three different received words \(r\) for which \(D(r)=\) \(000 .\) c) For each \(w \in \mathbf{Z}_{2}^{3}\), what is \(\left|D^{-1}(w)\right| ?\)

Let \(H\) and \(K\) be subgroups of a group \(G\), where \(e\) is the identity of \(G\). a) Prove that if \(|H|=10\) and \(|K|=21\), then \(H \cap K=\\{e\\}\). b) If \(|H|=m\) and \(|K|=n\), with \(\operatorname{gcd}(m, n)=1\), prove that \(H \cap K=\\{e\\}\)

The \((5 m, m)\) five-times repetition code has encoding function \(E: \mathbf{Z}_{2}^{m} \rightarrow \mathbf{Z}_{2}^{5 m}\), where \(E(w)=w w w w w .\) Decoding with \(D: \mathbf{Z}_{2}^{5 m} \rightarrow \mathbf{Z}_{2}^{m}\) is accomplished by the majority rule. (Here we are able to correct single and double errors made in transmission.) a) With \(p=0.05\), what is the probability for the transmission and correct decoding of the signal \(0 ?\) b) Answer part (a) for the message 110 in place of the signal \(0 .\) c) For \(m=2\), decode the received word $$ r=0111001001 $$ d) If \(m=2\), find three received words \(r\) where \(D(r)=00\). e) For \(m=2\) and \(D: \mathbf{Z}_{2}^{10} \rightarrow \mathbf{Z}_{2}^{2}\), what is \(\left|D^{-1}(w)\right|\) for each \(w \in \mathbf{Z}_{2}^{2} ?\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.