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Proving the elements of a group are of finite order

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data

    If N is a normal subgroup of a group G and every element of N and G/N has finite order, prove that every element of G has finite order.

    2. Relevant equations

    Um, a finite group element is given such that g^k = e (identity) for some integer k.
    We also know that the quotient of a group by a group is itself a group, with cosets as elements. If G/N is the quotient space, N is the identity element.

    3. The attempt at a solution

    The elements of N, call them n, have some finite order, so for some integer k we can say that n^k=e (the identity). The elements of the quotient group, G/N are the distinct cosets of N in G, so if they have some finite order m, then (Ng)^m=e=Ne (because N is the identity in the quotient group).

    I was hoping to do something like saying

    (ng)^m = ne ==> (n^m)*(g^m)= ne ==> n^(m-1)*g^m=e ==> somehow g^something =e.


    First of all, I am not sure if this is general enough to prove this for ALL cosets of G/N. Second, can I use little n instead of big N (we usually denote the cosets Ng. I think I can do the little one, though, because we're taking an element of N and operating on the left by some group element g). I don't know what else to do!! :confused:
     
  2. jcsd
  3. Feb 14, 2008 #2
    ooh, wait. Maybe someone can just tell me if this works... I am thinking that maybe

    (Ng)^m = N <==> (ng)^m=n. If that is true, then I can say (ng)^m = n ==> n^(m-1)*g^(m)=e ==> (n^(m-1)*g^(m))^k=e^k=e ==> (n^k)^(m-1)*g^(km))=e ==> g^km = e ==> g has finite order.

    Does that do it? Does that prove it for all g? Was my first assertion even valid?
     
  4. Feb 14, 2008 #3

    Dick

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    Not quite. Try this. G/N has finite order. So for every g there is an m such that (gN)^m=N. (gN)^m=(g^m)N. If (g^m)N=N then g^m is in N. Can you fill in any missing arguments?
     
  5. Feb 14, 2008 #4
    Oh, that's very cool... I think we'd about be done then, because we already are given the fact that the elements of N are finite as well... So what mechanism is it that allows us to take the m inside the (gN) and just apply it to g? That's the part I am having trouble seeing.
     
  6. Feb 14, 2008 #5

    Dick

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    An element of (gN)^m can be written as g*n1*g*n2*...*g*nm where the n's are in N, right? Since N is normal gN=Ng. So every g*n with n in N can also be written as n'*g where n' is a possibly different element of N. This lets you collect all of the g's together with the rest of the product being a product of n's. The same reason g1*N*g2*N=g1*g2*N.
     
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