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Group of order 35

  1. Jul 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Does a group of order 35 contain an element of order 5? of order 7?

    *I must prove all my answers.

    2. Relevant equations



    3. The attempt at a solution
    I'm not really sure how to prove this and could really use some hints. Here's what I've got so far:

    Pf. Let [itex]G[/itex] be a group of order 35, and let [itex]x \in G[/itex] such that [itex]x \neq e[/itex], where [itex]e[/itex] is the identity element of [itex]G[/itex]. It follows from Lagrange's Thoerem that [itex]x[/itex] will be of order 5, 7, or 35. If [itex]x[/itex] is of order 35, then [itex]G[/itex] is cyclic and thus has elements of order 5 and 7.

    So at this point, I want to suppose that [itex]x[/itex] doesn't have order 35. But since [itex]G[/itex] has 35 elements, the only other possibilities for the orders of the other nontrivial elements are either 5 or 7. Is this reasoning okay? Being optimistic, I will proceed and claim that each element [itex]x \in G[/itex] is of order 5. Then each [itex]x[/itex] generates an additional three unique elements: [itex]x, \, \, x^2, \, \, x^3[/itex]. But this would imply that the order of [itex]G[/itex] would be either 33 or 37 since elements of order 5 yield sets of four unique elements [itex]\{ x , \, \, x^2, \, \, x^3, \, \, x^4\}[/itex], a contradiction to the fact that [itex]|G| = 35[/itex]. Thus [itex]G[/itex] contains at least one element of order 7.
     
    Last edited: Jul 10, 2011
  2. jcsd
  3. Jul 10, 2011 #2

    HallsofIvy

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    You have said, correctly, that G must have a member of order 5, 7, or 35. You have noted, correctly, that if it has a member of order 35, it must be cyclic and so have members of order 5 and 7.

    Now, it G does not have a member or order 35, then it must have a member of order 5 or 7. Yes, assume, first, that there is a member of order 5. Can you then prove it must also have a member of order 7? Lastly assume there is a member of order 7. Can you then prove it must also have a member of order 5?
     
  4. Jul 10, 2011 #3
    Yes, I believe I can. Is the part where I assumed all members of [itex]G[/itex] have an order 5 then showed that [itex]G[/itex] must contain element of order 7 correct? If so, I can play the same game again and suppose all members of [itex]G[/itex] have an order of 7 and show that this would imply the order of [itex]G[/itex] must be either 31 or 37; again a contradiction to the fact that [itex]|G|=35[/itex].
     
  5. Jul 10, 2011 #4

    Char. Limit

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    Gold Member

    Are you allowed to use Sylow's theorems? If so, they might be useful here.
     
  6. Jul 10, 2011 #5
    No, I am not allowed to use Sylow's theorems.
     
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