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Thread: 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 $\displaystyle N $ is a normal subgroup of $\displaystyle G $ and if every element of $\displaystyle N $ and of $\displaystyle G/N $ has finite order, prove that every element of $\displaystyle G $ has finite order.

    I started off doing this:

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

    $\displaystyle (Ng)^k = Ne $ for some $\displaystyle k \in \mathbb{Z} $
    $\displaystyle \Rightarrow (eg)^k \in (Ng)^k = Ne = N $
    $\displaystyle \Rightarrow (eg)^k \in N $
    $\displaystyle \Rightarrow (e^k)(g^k) \in N $
    $\displaystyle \Rightarrow eg^k \in N $
    $\displaystyle \Rightarrow g^k \in N $
    $\displaystyle \Rightarrow \exists $ $\displaystyle n \in \mathbb{Z} $ such that $\displaystyle (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; Feb 9th 2010 at 04:10 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by crushingyen View Post
    If $\displaystyle N $ is a normal subgroup of $\displaystyle G $ and if every element of $\displaystyle N $ and of $\displaystyle G/N $ has finite order, prove that every element of $\displaystyle G $ has finite order.

    I started off doing this:

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

    $\displaystyle (Ng)^k = Ne $ for some $\displaystyle k \in \mathbb{Z} $
    $\displaystyle \Rightarrow (eg)^k \in (Ng)^k = Ne = N $
    $\displaystyle \Rightarrow (eg)^k \in N $
    $\displaystyle \Rightarrow (e^k)(g^k) \in N $
    $\displaystyle \Rightarrow eg^k \in N $
    $\displaystyle \Rightarrow g^k \in N $
    $\displaystyle \Rightarrow \exists $ $\displaystyle n \in \mathbb{Z} $ such that $\displaystyle (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 $\displaystyle N\lhd G$ and $\displaystyle m=\left[G:N\right]$, then $\displaystyle g^m\in N$ for every $\displaystyle g\in G$

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

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

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