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Homework Help: Proof similiar to Lagrange's?

  1. Apr 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose a finite group has exactly n elements of order p where p is prime. Prove that either n=0 or p divides n + 1.


    2. Relevant equations

    My professor says that this proof is similiar to the proof of Lagrange's Theorem, in our Abstract Algebra book (Gallian).

    3. The attempt at a solution

    I am so lost with this question. It is a "special problem" we've been given to work on all semester. I have tried letting H represent a subgroup of the finite group, and using the elements of order p in the original group to form left cosets of H in the group. Not sure I really understand that. Not sure where that is getting me. I am not used to feeling so lost when tackling a problem.

    Please...can anyone steer me in the right direction?

    Thank you!
     
  2. jcsd
  3. Apr 18, 2010 #2

    Dick

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    Start dealing with this by assuming there is an element x of order p. How many more elements of order p does that force the group to have? Assuming that those are all of the elements of order p, then what's the relation between n and p. Now generalize.
     
  4. Apr 19, 2010 #3
    thank you! I will work with that awhile...and let you know how it goes!
     
  5. Apr 27, 2010 #4
    I am sorry, I am still stuck. If x has order p, then x inverse must also have order p, and must also be in the group. then that is 2 elements...unless x is it's own inverse. So I think I am totally missing something here? HELP!!!
    Thank you.....
     
  6. Apr 27, 2010 #5

    Dick

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    x and x inverse aren't all. x generates a subgroup containing p elements. Try writing down a simple group of prime order like Z5. How many elements have order 5?
     
  7. Apr 27, 2010 #6
    What is [tex]\left(x^q\right)^p[/tex] for some integer q? Is there a positive integer r smaller than p for which [tex]\left(x^q\right)^r=e[/tex] might be true for some positive integer q less than p? If x is of order p, how many distinct elements that are powers of x exist? How many of those are of order p? For how many elements y that are powers of x, is x also a power of y, and vice versa? Then, if another element z that is not a power of x is of order p, how many distinct elements that are powers of z exist, and how many of those are of order p? It is seen that every subgroup of order p has only the element e in common, and the element e has order 1. This gives a general formula for the possible numbers of elements with order p.
     
  8. Apr 27, 2010 #7
    If x is of order p, there are p distinct elements of powers of x....right? <x> will have order p also. one of those will be e, so p-1 of them can have order p.
    Z5 has 4 elements with order 5.

    thinking.....


    sorry, I am usually not so dense..
     
  9. Apr 27, 2010 #8
    Yes, now consider the case of multiple subgroups of order p, such as [tex]Z_5\times Z_5[/tex]:

    e A A A A
    B C D E F
    B E C F D
    B D F C E
    B F E D C
     
  10. Apr 27, 2010 #9
    I'm sorry...I am still getting no where....

    I do appreciate everyone's help, though. I will report if I make any progress. Going to talk to the professor tomorrow....this is so discouraging...
     
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