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Abstract Algebra- VERY SIMPLE but I don't understand what my HW question is asking!
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
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.
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 relevant 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!
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
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 relevant 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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