• Support PF! Buy your school textbooks, materials and every day products Here!

Group theory problem

  • Thread starter ehrenfest
  • Start date
  • #1
2,012
1
[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
Staff Emeritus
Science Advisor
Gold Member
14,916
17
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
200
0
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
2,012
1
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.
 

Related Threads for: Group theory problem

  • Last Post
Replies
5
Views
659
  • Last Post
Replies
13
Views
4K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
0
Views
772
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
Top