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Homework Help: P-groups and orders of elements

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Let p be a prime and let H be a group of order p^n, some n > 0. Prove that for any x not equal to 1 in H, some power of x has order p.

    2. Relevant equations
    --

    3. The attempt at a solution
    I know that by lagrange, for any x in G, if x is not the identity, then x has an order p^r for some r>0. Also I know that there is some element of order p. But I don't know how to show that some power of every p^r = p, even though it seems almost intuitive.

    Thanks very much!
     
  2. jcsd
  3. Apr 11, 2010 #2
    Suppose that x is not the identity and has order [itex]p^2[/itex]. Can you find a power of x that has order p? Now try to solve the general case where x has order [itex]p^r[/itex].

    Petek
     
  4. Apr 11, 2010 #3
    Sorry, I don't even know which power of x would yield order p in the case of [itex]p^2[/itex]! Is it [itex]x^2[/itex]?
     
  5. Apr 11, 2010 #4

    Dick

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    If you know x has order p^r for some r, then x^(p^r)=1. What's (x^(p^(r-1)))^p?
     
  6. Apr 11, 2010 #5
    Using the example of the cyclic group of order 9, I found that (x^(p^(r-1)))^p is 1 as well. Is that correct?
     
  7. Apr 11, 2010 #6

    Dick

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    Probably. But I meant it more generally. In any group, how does (x^(p^(r-1)))^p compare with x^(p^r)?
     
  8. Apr 11, 2010 #7
    (x^(p^(r-1)))^p is equal to x^(p^r)?

    So, is (x^(p^(r-1))) always of order p?
     
    Last edited: Apr 11, 2010
  9. Apr 11, 2010 #8

    Dick

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    Yes, just use the rules of exponents. (x^n)^m=x^(nm).
     
  10. Apr 11, 2010 #9
    Okay, so I have:

    For any element x, |x|=p^r, x^p^r = 1 = (((x^p)^(r-1))^p) by the properties of exponents. Then x^p^(r-1) equals an element of order p.

    Thank you so much!
     
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