1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook