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Element of a group composed a times = e -> a = e

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    EDIT: Necessary and sufficient condition for a^m = e -> a=e for a in a finite abelian group of order n, and I think the answer is (m,n)=1, but it is hard for me to show a^m =e -> a=e requires (m,n) =1.


    2. Relevant equations



    3. The attempt at a solution
    I know that if GCD(m,n) =1 then if a^m = e, then o(a) | m, but by Lagrange, o(a) | n -> 1 = GCD(m,n) >= o(a) -> o(a) =1 -> a=e.

    However, if a^m=e -> a=e, I thought the following: suppose d=GCD(m,n). Then, d= bm + cn (b,c integers) suppose a^m = e -> a^bm = e -> a^(bm + cn) = e -> a^d =e, then I thought so now we need a^d =e -> a=e, so o(a) can't divide d, but o(a) divides m and n, since a^m =e, whence o(a) | bm +cn -> o(a)|d

    (but if o(a)|m, then a^m =e doesn't imply a= e) so I don't know if it is right to say o(a)|m, so I am stuck here and any help would be appreciated.

    Thanks

    EDIT: Necessary and sufficient condition for a^m = e -> a=e for a in a finite abelian group of order n, and I think the answer is (m,n)=1, but it is hard for me to show a^m =e -> a=e requires (m,n) =1.
     
    Last edited: Nov 3, 2011
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  3. Nov 3, 2011 #2

    mathwonk

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    no elements of finite order except e. for example the group of non zero real numbers, under multiplication.
     
  4. Nov 3, 2011 #3

    Deveno

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    necessary and sufficient condition on...what? a? m? the group G?
     
  5. Nov 3, 2011 #4

    mathwonk

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    good point. my answer was for m>1. if m = 1, we could say any group.
     
  6. Nov 3, 2011 #5
    Thanks so much for your responses, I forgot to say, sorry, a necessary and sufficient condition on m and the order of the abelian group n. Sorry for leaving out so much, just spent so much time on it, I didn't even ralize I left the details out. Again:

    Necessary and sufficient condition for a^m = e -> a=e for a in a finite abelian group of order n, and I think the answer is (m,n)=1, but it is hard for me to show a^m =e -> a=e requires (m,n) =1.
     
  7. Nov 3, 2011 #6

    micromass

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    By contraposition. Given [itex](m,n)\neq 1[/itex], try to find an element [itex]a\neq e[/itex] such that [itex]a^m=e[/itex].
     
  8. Nov 3, 2011 #7
    I've looked at the fact that GCD(m,n) is a linear combination of them, and said something like: Let d=(m,n) then d=bm + cn, b,c integers, then
    a^(d)= a^(bm +cn) = a^bm -> o(a)|d-bm, but we already knew this by Lagrange.

    a^(d-1)=/ e since d>1, but a^m*(d-1) = a^m*(bm +cn -1) = a^m*(bm-1), but I dont know how to show this is identity.


    EDIT just thought d|m, d|n -> m=q'd, n=qd -> a^(n/d) = a^q, if I had some way to show o(a)|\ q, then I would have a^q =/ e, whence a^(q*m) = a^(n/(m/q`))*m = a^(n*q') = e, but I need to have that o(a) doesn't divide q, I know that if d>1, then n>q, but what if q=k*o(a) for some integer k??
     
  9. Nov 3, 2011 #8

    micromass

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    Do you know Cauchy's theorem?? Or sylows theorems??? Look them up and try to use them in your reasoning.
     
  10. Nov 3, 2011 #9
    Thanks, but

    I do actually know both, but Cauchy's thm. only applies to finite groups whose orders are divisible by a prime, and Sylow's only to a finite group of order p^n *m, where p doesn't divide m, so given n arbitrary, I don't think the proof should require these, (I wanted to do a general case) but does my edit on the last post loook right?
     
  11. Nov 3, 2011 #10

    micromass

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    This is every group. (except the trivial group). Every number n>1 is divisible by a prime.

    This is again every group. (except the trivial group). Every number can be written in those ways.

    What's bothering me about it is that you never actually construct a. Given n and m you must FIND an [itex]a\neq a[/itex] such that [itex]a^m=e[/itex]. So, how would you make a?
     
  12. Nov 3, 2011 #11
    Thanks, yeah, by the thm. n>1 -> n is prime or product of primes, Cauchy's Thm. can be applied, but still I am not quite sure I understand what you mean by find. If we don't know what the group is, what I thought find meant was in terms of some arbitrary non identity element a??
     
  13. Nov 3, 2011 #12

    micromass

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    The element a is obviously not arbitrary. Given n and m, you must show that there EXISTS an [itex]a\neq e[/itex] such that [itex]a^m=e[/itex].

    Indeed, the group is general, so you don't really have much options in constructing your a. Cauchy's theorem will be helpful as it states the existence of elements with certain properties...
     
  14. Nov 3, 2011 #13
    Thanks,
    Ok, sorry for my silly mistakes,

    By Cauchy, we know there exists an x in G with o(x) = p, where p is some prime factor of n. If p>d, p cannot divide d, so (p,d) =1, but d|n, p|n , (p,d) =1 -> dp|n -> x^(dp) = e. Still working, but is this the right track?
     
  15. Nov 3, 2011 #14

    micromass

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    Yes, but which p did you take??
     
  16. Nov 3, 2011 #15
    As written, I guess I took an arbitrary prime divisor of n, and am checking if >d or <d, but (p,d) =1 -> x^d =/ e, but x^md = x^lp
    d=am + bn = am + bp/q
    x^d*m = x^am^2
     
  17. Nov 3, 2011 #16
    Am I supposed to take the least prime divisor exists by law of well-ordering on Z+, and then arrive at contradiction on minimality?
     
  18. Nov 3, 2011 #17

    micromass

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    Remember that you assumed [itex]gcd(m,n)\neq 1[/itex] Do something with that.
     
  19. Nov 3, 2011 #18
    dp|n -> dp can't divide m, since if it did, it would be (m,n),
    but x^d = x^(am+bn) = x^am=/e since (p,d) =1.
     
  20. Nov 3, 2011 #19
    d|n, so let n=q*d. Let x be the above element, then if p>q, p|/ q -> x^q =/e, but x^m*q = x^(q'*d) *q =x^(q'*n_ = e, a contradiction. So, p must divide q. But if p|q ....??
     
  21. Nov 3, 2011 #20
    If p|q, p<q, whence x^q =/ e, since p is the smallest integer which satisfies x^ y = e !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

    Is this right?
     
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