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Thread: [SOLVED] Prove that G has an element of order n

  1. #1
    Newbie Tachyon's Avatar
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    [SOLVED] Prove that G has an element of order n

    If N is a finite normal subgroup of a group G and if G/N contains an element of order n, prove that G contains an element of order n.
    This is what I have so far:

    Let $\displaystyle Na \in G/N$ have order n for some $\displaystyle a \in G$. Then $\displaystyle (Na)^n = Na^n = e$, so $\displaystyle a^n \in N$. Since N is finite, let $\displaystyle ord(N) = m$. Then $\displaystyle (a^n)^m = e$, so $\displaystyle ord(a)|nm$.

    When I pursued this line of reasoning, I was trying to show that $\displaystyle n|ord(a)$, and thus $\displaystyle n = ord(a)$. However, I'm stuck here. Have I made a mistake somewhere, or am I missing something?

    Thanks for any insights.
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  2. #2
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    Quote Originally Posted by Tachyon View Post
    This is what I have so far:

    Let $\displaystyle Na \in G/N$ have order n for some $\displaystyle a \in G$. Then $\displaystyle (Na)^n = Na^n = e$, so $\displaystyle a^n \in N$. Since N is finite, let $\displaystyle ord(N) = m$. Then $\displaystyle (a^n)^m = e$, so $\displaystyle ord(a)|nm$.

    When I pursued this line of reasoning, I was trying to show that $\displaystyle n|ord(a)$, and thus $\displaystyle n = ord(a)$. However, I'm stuck here. Have I made a mistake somewhere, or am I missing something?

    Thanks for any insights.
    $\displaystyle o(a^n) < \infty$ because $\displaystyle a^n \in N$ and $\displaystyle |N| < \infty.$ let $\displaystyle o(a^n)=k.$ then $\displaystyle (a^k)^n=e$ and thus $\displaystyle o(a^k) < \infty.$ let $\displaystyle o(a^k)=m.$ the claim is that $\displaystyle m=n.$ clearly $\displaystyle m \mid n.$ now $\displaystyle (Na)^{km}=N$ and hence $\displaystyle n \mid km.$

    let $\displaystyle \frac{km}{n}=\ell.$ then $\displaystyle a^{\ell n}=e.$ thus $\displaystyle k \mid \ell$ and therefore $\displaystyle n \mid m. $
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    Newbie Tachyon's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    now $\displaystyle (Na)^{km}=N$ and hence $\displaystyle n \mid km.$
    I understand that $\displaystyle (Na)^{km} = N$, but could you clarify why $\displaystyle n \mid km$ necessarily follows? If $\displaystyle o(Na) = n$, wouldn't we need $\displaystyle (Na)^{km} = e$ in order to have $\displaystyle n \mid km$?
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    Quote Originally Posted by Tachyon View Post
    I understand that $\displaystyle (Na)^{km} = N$, but could you clarify why $\displaystyle n \mid km$ necessarily follows? If $\displaystyle o(Na) = n$, wouldn't we need $\displaystyle (Na)^{km} = e$ in order to have $\displaystyle n \mid km$?
    the identity element of $\displaystyle G/N$ is the coset $\displaystyle N.$ so $\displaystyle (Na)^{km}=N=1_{G/N}.$
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  5. #5
    Newbie Tachyon's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    the identity element of $\displaystyle G/N$ is the coset $\displaystyle N.$ so $\displaystyle (Na)^{km}=N=1_{G/N}.$
    D'oh. That's what I deserve for my notation confusing $\displaystyle 1_{G/N}$ and $\displaystyle 1_G$. Thanks for your help; I understand now.
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