Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Proposition 14.4.6. The order of a cycle is always equal to the cycle's length. PROOF. To prove this, we essentially have to prove two things: (A) If o is a cycle of length k, then ok = id;

Proposition 14.4.6. The order of a cycle is always equal to the cycle's length. PROOF. To prove this, we essentially have to prove two things: (A) If o is a cycle of length k, then ok = id;

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question

Please do Exercise 14.4.7 and please show step by step and explain

(ii) rE A. We'll deal with these two cases separately (as we did in Exercise 14.4.4). <2>_ It follows that o (x) = o(0(x)) = (i) In this case, o(r) o<3>) = <4>. We can use the same argument to show that o*(r) = < number <7>. %3D < 5> , and that o (1) = < 6>_ for any natural %3D (ii) In this case, then r = a; for some integer j,1 <j< <8> . It follows from the definition of cycle that o(r) = o(a;) = a mod (j+1,k). Furthermore, o (r) = 0(a mod (j+1,k) = < 9> . Similarly it follows that o*(r) = a mod (j+ <10>k) =a <11>_=1. Cases (i) and (ii) establish that Vr e X, < 12 ; ok = > = x. It follows that < 13 > The proof of (B) is also structured as an exercise.
Transcribed Image Text:(ii) rE A. We'll deal with these two cases separately (as we did in Exercise 14.4.4). <2>_ It follows that o (x) = o(0(x)) = (i) In this case, o(r) o<3>) = <4>. We can use the same argument to show that o*(r) = < number <7>. %3D < 5> , and that o (1) = < 6>_ for any natural %3D (ii) In this case, then r = a; for some integer j,1 <j< <8> . It follows from the definition of cycle that o(r) = o(a;) = a mod (j+1,k). Furthermore, o (r) = 0(a mod (j+1,k) = < 9> . Similarly it follows that o*(r) = a mod (j+ <10>k) =a <11>_=1. Cases (i) and (ii) establish that Vr e X, < 12 ; ok = > = x. It follows that < 13 > The proof of (B) is also structured as an exercise.
Proposition 14.4.6. The order of a cycle is always equal to the cycle's length. PROOF. To prove this, we essentially have to prove two things: (A) If o is a cycle of length k, then ok = id; (B) If o is a cycle of length k, then oi # id Vj:1<j< k. The proof for (A) follows the same lines as our investigations in Exer- cise 14.4.4. In that exercise, we considered separately the elements of X that are moved by the cycle, and those elements that are not moved by the cycle. Exercise 14.4.7. Prove part (A) by filling in the blanks. Let o e Sx be an arbitrary cycle of length k. Then o can be written as (a1 az ... ak), for some set of elements a1, a2,…..az in X. In order to show that ok = id, it is sufficient to show that ok(x) = <l> Væ € X. Let A be the set {a1, a2, ... ag}. Now for any r € X, there are two possibilities: (i) r E X\A;
Transcribed Image Text:Proposition 14.4.6. The order of a cycle is always equal to the cycle's length. PROOF. To prove this, we essentially have to prove two things: (A) If o is a cycle of length k, then ok = id; (B) If o is a cycle of length k, then oi # id Vj:1<j< k. The proof for (A) follows the same lines as our investigations in Exer- cise 14.4.4. In that exercise, we considered separately the elements of X that are moved by the cycle, and those elements that are not moved by the cycle. Exercise 14.4.7. Prove part (A) by filling in the blanks. Let o e Sx be an arbitrary cycle of length k. Then o can be written as (a1 az ... ak), for some set of elements a1, a2,…..az in X. In order to show that ok = id, it is sufficient to show that ok(x) = <l> Væ € X. Let A be the set {a1, a2, ... ag}. Now for any r € X, there are two possibilities: (i) r E X\A;
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS