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Homework Help: Prove in U(n) for n finite: x^5 = e has a number of solutions that is multiple of 4

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    In a finite group, show that the number of non-identity elements that satisfy the equation:

    x^5 = e = identity element of multiplication mod n = 1

    is a multiple of 4.


    (Also need to show: if the stipulation that the group be finite is omitted, what can you say about the number of non-identity elements that satisfy x^5 = e ... but if I get the part of the question that's not in parenthesis, I should be able to figure it out...but I'm mentioning it just in case there's a hint in the asking of the question).


    2. Relevant equations

    axioms of groups: associativity, existance of inverse-elements, existance and uniqueness of identity, and we are in the (Abelian) group U(n), which is multiplication mod n. i think those are the ingrediants needed for a group.


    3. The attempt at a solution

    my professor suggested "look for a pattern", so i used Microsoft EXCEL to multiply mod n and find groups for U(n), for the cases n = 4, 5, 6, ... 14.

    [tex]U(4) = \left\{ {1,3} \right\}[/tex]
    [tex]U(5) = \left\{ {1,2,3,4} \right\}[/tex]
    [tex]U(6) = \left\{ {1,5} \right\}[/tex]
    [tex]U(7) = \left\{ {1,2,3,4,5,6} \right\}[/tex]
    [tex]U(8) = \left\{ {1,3,5,7} \right\}[/tex]
    [tex]U(9) = \left\{ {1,2,4,5,7,8} \right\}[/tex]
    [tex]U(10) = \left\{ {1,3,7,9} \right\}[/tex]
    [tex]U(11) = \left\{ {1,2,3,4,5,6,7,8,9,10} \right\}[/tex]
    [tex]U(12) = \left\{ {1,5,7,11} \right\}[/tex]
    [tex]U(13) = \left\{ {1,2,3,4,5,6,7,8,9,10,11,12} \right\}[/tex]
    [tex]U(14) = \left\{ {1,3,5,9,11,13} \right\}[/tex]

    Then: of these elements, find only those guys that satisfy x^5 = e = 1. The elements of U(11): 3, 4, 5, and 9, four elements, satisfy x^5 = e. Everything else, only 1, the identity element (not a non-identity element! ha ha) satisfied x^5 = e.

    Too small a set of instances (one!) to find a pattern. Is there some sort of closed form expression for 3, 4, 5, and 9 "in" U(11) that raising these guys to the 5th power that I could show to be mandatorily divisible by 4 that I am missing? Any suggestions on how to approach this? Am I going down a blind alley by just looking for a pattern? :-|
     
  2. jcsd
  3. Sep 18, 2010 #2
    Re: prove in U(n) for n finite: x^5 = e has a number of solutions that is multiple of

    I think I just "thought" of the instance where the group becomes infinite. if we were to stay in the integers, I think you can NOT have any solutions x^5 = e (if group is infinite, how can we speak of "mod"?). However, if we consider complex numbers, perhaps the solutions to x^5 = e must be the four (out of five) roots that are not equal to exp(i*2*pi) = the identity element.
     
  4. Sep 19, 2010 #3
    Re: prove in U(n) for n finite: x^5 = e has a number of solutions that is multiple of

    UPDATE: I have a nice proof that if x satisfies x^5 = e, then so too must x^2 and x^4.

    Suppose there exists nonidentity element “x” that satisfies x^5 mod n = e. then:

    [tex]\exists x:{x^5}\bmod n = e[/tex] (1)

    That means:

    [tex]n|({x^5} - e)[/tex] (2)

    turn, n will divide any integer multiple of the right hand side of [I.15]. Such an integer-multiple might be:

    [tex]({x^5} - e)({x^5} + e) = {x^{10}} - {e^2} = {x^{10}} - e[/tex] (3)

    Thus, if (1) is true and we thus have that “x”, then “x2” must also satisfy (1), generating our next “x”. By this same method, an “x4” must also exist, too: three out of the four elements that need to exist.

    Does x3 also satisfy (1)? It must, somehow. Perhaps we could reach a contradiction if it didn't, and complete the proof?
     
  5. Sep 19, 2010 #4
    Re: prove in U(n) for n finite: x^5 = e has a number of solutions that is multiple of

    ..at least i THINK i have a proof....critique this proof if you see problems! i have no experience with "pure-math" proofs!
     
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