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Math Help - Lagrange's Theorem & orders

  1. #1
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    Lagrange's Theorem & orders

    My question is: If |G|=15, show that G must have an element of order 3.

    We're currently learning about Lagrange's Theorem, so I know that I need to use it somehow, and I can't use Sylow's or Cauchy's Theorems yet. It makes sense that elements of order 3 and 5 (and 1 and 15) would be in G but I don't know how to use Lagrange to show that.

    Thanks! Any help would be greatly appreciated!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kimberu View Post
    My question is: If |G|=15, show that G must have an element of order 3.

    We're currently learning about Lagrange's Theorem, so I know that I need to use it somehow, and I can't use Sylow's or Cauchy's Theorems yet. It makes sense that elements of order 3 and 5 (and 1 and 15) would be in G but I don't know how to use Lagrange to show that.

    Thanks! Any help would be greatly appreciated!
    Any ideas? So, it cannot have an element of order 3 and it can't have an element of order 15 and so it must have all elements of order 5...except the identity element.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    Any ideas? So, it cannot have an element of order 3 and it can't have an element of order 15 and so it must have all elements of order 5...except the identity element.
    Sorry, I don't quite understand what you mean here. G must have an element of order 3 -- are you saying to try a proof by contradiction?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kimberu View Post
    Sorry, I don't quite understand what you mean here. G must have an element of order 3 -- are you saying to try a proof by contradiction?
    Yes.
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  5. #5
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    Okay, so what I have so far is that
    If G is cyclic, then it must contain some <g> where |g| = 3.
    So, let G be non-cyclic. Then for all G, |g| = 1, 3, or 5. If |g|=1, it is the identity element. Assume |g|=5 for all g except the identity. Then G is cyclic which is a contradiction.

    Or am I way off?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kimberu View Post
    Okay, so what I have so far is that
    If G is cyclic, then it must contain some <g> where |g| = 3.
    So, let G be non-cyclic. Then for all G, |g| = 1, 3, or 5. If |g|=1, it is the identity element. Assume |g|=5 for all g except the identity. Then G is cyclic which is a contradiction.

    Or am I way off?
    How did you arrive at that last part? If G were cyclic then there would be some g\in G such that |g|=15. Thus, the assumption that no element of G is of order 15 means it can't be cyclic.
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  7. #7
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    Oops. Well then, if they're all of order 5, does that mean there are multiple unique subgroups of order 5, which is impossible since 5 is prime and so they must be the same group?
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  8. #8
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    sorry. i just learnt about Lagrange theorem too and im confused about how this question has been proven.

    why is it that we have made an assumption that there is no element of G that is of order 15? and how does that imply that it is not cyclic?

    thanks!
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