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Quotient groups related problem

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

    Let G be a finite group and N[tex]\triangleleft[/tex]G such that |N| = n, and gcd(n,[G:N]) = 1.
    Proof that if x[tex]^{n} = e[/tex] then x[tex]\in[/tex]N.

    2. Relevant equations

    none.

    3. The attempt at a solution

    I defined |G| = m and and tried to find an integer which divides both n and m/n.
    I went for some X which is not in N, for which X^n = e.
    I defined o(X) = q and then q | n.
    No luck showing that q | m/n.


    Any other ideas?


    Thanks ahead,
    Zvi
     
    Last edited: Jun 1, 2010
  2. jcsd
  3. Jun 1, 2010 #2
    This is not true! Take some subgroup H of the Klein group K of order 2. It is normal as K is abelian. K/H also has order 2, and so gcd(H,[K:H]) = 2. But there is certainly an element of order 2 in K that is not in H.

    Did you mean to write that the gcd is 1?
     
  4. Jun 1, 2010 #3
    Yes I did, sorry for the mistake...
    *Corrected*
     
  5. Jun 1, 2010 #4
    Here's a rough sketch:

    Suppose x is not in N. Let M=<x>. If xn=e then M has order dividing n. |MN|=mn/|M[tex]\cap[/tex]N|, where m is the order of M. MN is a subgroup of G by the second (or third, depending on your numbering!) isomorphism theorem. So mn/|M[tex]\cap[/tex]N| divides |G|.

    Now, |G| = tn, where t=[G:N] is coprime to n. So m/|M[tex]\cap[/tex]N| must divide t. But this is impossible, as m divides n, and t is coprime to n. So x is in N.

    As a disclaimer, it's a while since I did any group theory! So you should check all this carefully.
     
    Last edited: Jun 1, 2010
  6. Jun 1, 2010 #5
    Thanks for the response.

    By stating that M[tex]\cap[/tex]N = <1> did you mean M[tex]\cap[/tex]N = e?
    If so, it seems to me that it is not necessarily true.
    If you meant something else, please elaborate.


    Zvi.
     
  7. Jun 1, 2010 #6
    hmmm..yes, sorry about that. I've modified the argument above. Still no guarantee it is right!

    Yes, when I say <1> I mean e....itis just to distinguish the identity element from the trivial group.
     
  8. Jun 1, 2010 #7
    Sorry... I still see a problem with the conclusion that mn divides |G|...
     
  9. Jun 1, 2010 #8
    That was just an error with my previous editing. Hopefully it is fixed now.

    If you spot another problem, try filling in the details yourself...it's the only way to learn! I have done this very much off the cuff, and intended it to be a "suggested method" rather than a full solution to be copied down.
     
  10. Jun 1, 2010 #9
    Thanks for you help!...
     
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