Prove Odd Length Cycle u2 is Also a Cycle

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Homework Help Overview

The discussion revolves around proving that for any cycle of odd length, denoted as u, the square of that cycle, u2, is also a cycle. The subject area pertains to graph theory and permutations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore examples of odd-length cycles and their squares, with one participant attempting to generalize their observations. Another participant questions the reasoning behind the assertion that u2 is a cycle when u is of odd length.

Discussion Status

The discussion includes attempts to formulate a proof based on the properties of cycles and their permutations. Some participants express uncertainty about their reasoning, while others provide affirmations of understanding regarding the explanations given.

Contextual Notes

There is mention of potential fatigue affecting the clarity of reasoning, and a focus on the specific characteristics of cycles of odd length is noted. The discussion does not reach a consensus but explores various interpretations and approaches.

Metahominid
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Homework Statement


Show that for any cycle of odd length, u, then u2 is a cycle as well.


Homework Equations





The Attempt at a Solution


I looked at examples of it and it does work them but I am unsure of how to generalize it.
 
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There is really only one three cycle, (1,2,3). All of the others just relabel the elements. Ditto for other lengths. Are you sure you can't figure out why if the length of the cycle is odd then a^2 is a cycle?
 
I think I was just really tired and didn't remember what I had looked at before.

but my proof was if u is a cycle of length k.
Then
u(ai) = ai+1 1\leqi\leqk-1
u(ak) = a1

then if v = u2
v(ai) = ai+2 1\leqi\leqk-2
v(ak-1) = a1
v(ak) = a2

so v(a1) = a3 which then we permute, and each goes on to an odd a, until k-2 which goes to ak, this goes to a2 which then covers the even a's, and at ak-1 goes back to a1 making another cycle.
 
Metahominid said:
I think I was just really tired and didn't remember what I had looked at before.

but my proof was if u is a cycle of length k.
Then
u(ai) = ai+1 1\leqi\leqk-1
u(ak) = a1

then if v = u2
v(ai) = ai+2 1\leqi\leqk-2
v(ak-1) = a1
v(ak) = a2

so v(a1) = a3 which then we permute, and each goes on to an odd a, until k-2 which goes to ak, this goes to a2 which then covers the even a's, and at ak-1 goes back to a1 making another cycle.

That's a good enough explanation for me.
 
okay, awesome. Thanks
 

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