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Math Help - Prove that every element of G has finite order

  1. #1
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    Prove that every element of G has finite order

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

    I started off doing this:

    Let  g \in G . By showing that  g^n = e ,  n \in \mathbb{Z} ,  |G| = n <\infty .

     (Ng)^k = Ne for some  k \in \mathbb{Z}
     \Rightarrow (eg)^k \in (Ng)^k = Ne = N
     \Rightarrow (eg)^k \in N
     \Rightarrow (e^k)(g^k) \in N
     \Rightarrow eg^k \in N
     \Rightarrow g^k \in N
     \Rightarrow \exists  n \in \mathbb{Z} such that  (g^k)^n = e

    I don't know how to finish off this problem after point. Any help would be greatly appreciated.
    Last edited by crushingyen; February 9th 2010 at 05:10 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by crushingyen View Post
    If  N is a normal subgroup of  G and if every element of  N and of  G/N has finite order, prove that every element of  G has finite order.

    I started off doing this:

    Let  g \in G . By showing that  g^n = e ,  n \in \mathbb{Z}

     (Ng)^k = Ne for some  k \in \mathbb{Z}
     \Rightarrow (eg)^k \in (Ng)^k = Ne = N
     \Rightarrow (eg)^k \in N
     \Rightarrow (e^k)(g^k) \in N
     \Rightarrow eg^k \in N
     \Rightarrow g^k \in N
     \Rightarrow \exists  n \in \mathbb{Z} such that  (g^k)^n = e

    I don't know how to finish off this problem after point. Any help would be greatly appreciated.
    I think this is what you did...

    Lemma: Let N\lhd G and m=\left[G:N\right], then g^m\in N for every g\in G

    Proof: Since N\lhd G we know that G/N is a group of order m and so given any element K in G/N we have that, by basic group theory, that K^m=e=N. In particular, \left(Ng\right)^m=Ng^m=N from where it follows that g^m\in N. \blacksquare.

    Is that what you wre getting at??
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  3. #3
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    Sorry, I edited the my post up above.

    Pretty much, I'm trying to show that the group G is finite, not the lemma.
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  4. #4
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    Never mind, I found out how to do it.

    Since  g^k has finite order,  |g^k| = n for some  n \in \mathbb{Z}

     \Rightarrow (g^k)^n = e
     \Rightarrow g^{kn} = e
     \Rightarrow |g| = kn < \infty
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