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

    When I pursued this line of reasoning, I was trying to show that n|ord(a), and thus 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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    Quote Originally Posted by Tachyon View Post
    This is what I have so far:

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

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

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

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