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Math Help - Finite group with element of prime order

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    Finite group with element of prime order

    Let G be a finite group with more than one element. Show that G has an element of prime order.
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    MHF Contributor red_dog's Avatar
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    Let a\in G, \ ord(a)=n.
    If n is prime, we've done.
    Else, let p be a prime divisor of n.
    Then a^n=a^{pm}=(a^m)^p=e.
    Let b=a^m\in G\Rightarrow b^p=e.
    Let 0<q<p. Then b^q=(a^m)^q=a^{mq}.
    But mq<mp=n\Rightarrow a^{mq}\neq e.
    So ord(b)=p.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by red_dog View Post
    Let a\in G, \ ord(a)=n.
    If n is prime, we've done.
    Else, let p be a prime divisor of n.
    Then a^n=a^{pm}=(a^m)^p=e.
    Let b=a^m\in G\Rightarrow b^p=e.
    Let 0<q<p. Then b^q=(a^m)^q=a^{mq}.
    But mq<mp=n\Rightarrow a^{mq}\neq e.
    So ord(b)=p.
    If I am reading this right, then, the only case left to show is to prove that no group can exist where the order of all of its elements is 1. (I admit this is practically trivial, but it wasn't addressed in the above proof.)

    -Dan
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    Quote Originally Posted by topsquark View Post
    If I am reading this right, then, the only case left to show is to prove that no group can exist where the order of all of its elements is 1. (I admit this is practically trivial, but it wasn't addressed in the above proof.)

    -Dan
    The only element of order 1 is the identity. He said "with more than 1 element (identity)".
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