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Does every group whose order is a power of .

  1. Oct 12, 2011 #1
    Does every group whose order is a power of .....

    1. The problem statement, all variables and given/known data

    Does every group whose order is a power of a prime p contain an element of order p?

    2. Relevant equations

    I'm trying to get away with only using Lagrange's Theorem and a corollary thereof.

    3. The attempt at a solution

    Seems like I'm almost there. How do I seal the deal? Thanks in advance, brahs.


    screen-capture-1-36.png
     
  2. jcsd
  3. Oct 12, 2011 #2

    Dick

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    Re: Does every group whose order is a power of .....

    You could use Cauchy's theorem, but that's overkill for this problem. Sure, finish your approach. If m is the order of some element g, then the group generated by g is a subgroup of order p^k for some k. And the subgroup generated by g is CYCLIC. Use that!
     
  4. Oct 12, 2011 #3
    Re: Does every group whose order is a power of .....


    The subgroup generated by g is <g>={..., -g2, -g, 1, g, g2, ...}, which is the smallest subgroup of G that contains g, and has order pk for some integer k≥0. Since pk is not infinite, the elements of <g> are not distinct, so .......... am I getting somewhere?
     
  5. Oct 12, 2011 #4

    Dick

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    Re: Does every group whose order is a power of .....

    I think we are using multiplicative notation here. The distinct elements of the subgroup generated by g are {e,g,g^2,...g^(p^k-1)} since g^(p^k)=e. Now get somewhere with that. Isn't there an element of order p in there?
     
    Last edited: Oct 12, 2011
  6. Oct 12, 2011 #5
    Re: Does every group whose order is a power of .....

    *pssst* .... What's e?
     
  7. Oct 12, 2011 #6
    Re: Does every group whose order is a power of .....

    e = indentity element of group G
     
  8. Oct 12, 2011 #7
    Re: Does every group whose order is a power of .....

    Got it.

    Can you explain once more where we got {1, g, g2, ..., gpk-1}.

    My book says "Let <x> be the cyclic group of a group G generated by an element x, and let S denote the set of integers k such that xk=1.

    [...]

    Suppose that S is not the trivial subgroup. Then S = Zn for some positive integer n. The powers 1,x,x2,...,xn-1 are the distinct elements of the subgroup <x>, and the order of <x> is n."


    But I'm a little confused by it.
     
  9. Oct 13, 2011 #8

    Dick

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    Re: Does every group whose order is a power of .....

    The subgroup generated by g is just the collection of all distinct powers of g. It's a subgroup since (g^a)(g^b)=g^(a+b). If the order of g is 5 then that's {1,g,g^2,g^3,g^4}. No use listing g^5 since that's 1 or g^6 since g^6=g^5*g=g. Now your case is order g equal to p^k.This is simple enough that I wouldn't spend a lot of time confusing yourself about it.
     
  10. Oct 13, 2011 #9
    Re: Does every group whose order is a power of .....

    This right?

    screen-capture-2-24.png
     
  11. Oct 13, 2011 #10

    Dick

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    Re: Does every group whose order is a power of .....

    Why do you think g^p has order p???? And that's not generally correct. Proofs usually involve giving reasons. What's (g^p)^p? Why would that be 1?
     
  12. Oct 13, 2011 #11
    Re: Does every group whose order is a power of .....

    Never mind. So I'm looking for an element in {1, g, g2, ..., gpk-1} that has order p, i.e., the cyclic subgroup generated by that element this element has p elements. So if I let x be that element, the cyclic subgroup generated by x is {1, x, x2, ..., xp-1}. Am I right?
     
  13. Oct 13, 2011 #12

    Dick

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    Re: Does every group whose order is a power of .....

    Yes. You want x^p=1. Which nonidentity element of {1, g, g2, ..., gpk-1} might have that property? Think about exponents ok? (g^a)^p=g^(ap) right? THINK!
     
    Last edited: Oct 13, 2011
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