Abstract Algebra, I don't understand what my HW question is asking!

  • Thread starter gmn
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gmn
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Abstract Algebra- VERY SIMPLE but I don't understand what my HW question is asking!

Homework Statement




Hi. I am having trouble simply understanding what the question is here:

(6) let w = (1 2 3 4 5 6 7 8 9 10 11 12 13 14). For which integers i is w^i a 14-cycle?



Here is a link to the assignment if you would prefer to read it there, it's #6:

http://math.berkeley.edu/~rdore/113/hw3.pdf [Broken]


Homework Equations



SX ={set of permutations on 1,......,X}

definition of a cycle of length k, or k-cycle:

A permutation o in SX is a cycle of length k if there exist elements
a1; a2; : : : ; ak in X such that
o(a1) = a2
o(a2) = a3
...
o(ak) = a1
and o(x) = x for all other elements x in X. We will write (a1 a2 ....... ak) to
denote the cycle o. Cycles are the building blocks of all permutations.


SO, I'm pretty sure I'm not confused about the following information right:

*so in w, take id = (1 2 3 4 5 6 7 8 9 10 11 12 13 14)
and w(id) = (2 3 4 5 6 7 8 9 10 11 12 13 14 1) [1 goes to 2, 2 goes to 3, etc.]

w^3= w(w(w(id):

w(w(w(1)))= w(w(2))= w(3) = 4
www(2) = ww(3)=w(4)=5
.
.
.
www(14) = ww(1) = w(2) = 3
so w^3=(4 5 6 7 8 9 10 11 12 13 14 1 2 3)
.
.
.

w^i = w(w(w(.........(w(id))))) w composed with w i times.


The Attempt at a Solution



I have

w^2 = w(w(id))= w(2 3 4 5 6 7 8 9 10 11 12 13 14 1) = ( 3 4 5 6 7 8 9 10 11 12 13 14 1 2).

w^3 = ( 4 5 6 7 8 9 10 11 12 13 14 1 2 3).
w^4 = ( 5 6 7 8 9 10 11 12 13 14 1 2 3 4).
.
.
.
w^13 = id = ( 14 1 2 3 4 5 6 7 8 9 10 11 12 13).
w^14 = w = ( 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ).
w^15= w^2
.
.
.
w^i = w ^ imod13

all 14-length cycles. ???? Is that all? I dunno. I'm confused.

What I'm having trouble with is the question of whether or not there are any i's for which w^i is not a 14 cycle? I can't really extract any relevent information with which to make a substantial proof. Maybe I'm not getting something? It just seems to me like it would go on and on in a loop forever, as there does not seem to be any disjoint cycles to be extracted or anything that should change when taking w with itself.
thanks!
 
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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,836
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Hi gmn! :smile:

(try using the X2 tag just above the Reply box :wink:)
w = (1 2 3 4 5 6 7 8 9 10 11 12 13 14). For which integers i is w^i a 14-cycle?[/B]

w^2 = w(w(id))= w(2 3 4 5 6 7 8 9 10 11 12 13 14 1) = ( 3 4 5 6 7 8 9 10 11 12 13 14 1 2).

Noooo :cry:

w sends 1 to 2 and 2 to 3, so w2 sends 1 to 3, 3 to 5, …

so w2 = (1 ? ? ? …)(? ? …) :smile:
 

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