Group theory problem

  • Thread starter ehrenfest
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  • #1
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[SOLVED] group theory problem

Homework Statement


A cyclic group of order 15 has an element x such that the set {x^3,x^5,x^9} has exactly two elements. The number of elements in the set {x^{13n} : n is a positive integer} is

a)3
b)5


Homework Equations





The Attempt at a Solution


From the given information, we know that x^6 = 1 or x^4 = 1. In the first case, either answer is possible. In the second case, only answer a is possible. Anyway, do we know enough to decide which case this is or is there a different way to do the problem?
 

Answers and Replies

  • #2
Hurkyl
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From the given information, we know that x^6 = 1 or x^4 = 1.
True. But why stop there? (I'm assuming that you have a valid justification for this assertion)

In the first case, either answer is possible. In the second case, only answer a is possible.
I can't guess at your reasoning for either conclusion; would you show your work?
 
  • #3
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if x^4 = 1, you can conclude what x is on what you know about the size of the group. the answer then follows.
 
  • #4
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Oh. I see. By Lagrange's Theorem, x^4=1 implies x=1 which implies there is exactly one element in the set. x^6 implies x^3=3 again by Lagrange and the fact that there is more than one element in the set. The answer a) is immediate since gcd(13,3)=1. Thanks.
 

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