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Math Help - abstract algebra problem

  1. #1
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    abstract algebra problem

    Let G be a finite group. Let a and b be in G such that a != b.
    Suppose |a|=|b|= d so |<a>|=|<b>|, and b is not in <a>.
    ...
    Moreover suppose that c is in G such that |c|= d and c is in both <a> and <b>.
    How does this implies that <a> = <c> = <b>

    I clearly understand that <c> is contained in both <a> and <b>. But what argument shows that both <a> and <b> are contained in <c>

    This is just part of a proof that will later yield the contradiction that b is in <a>. This proof is titled the number of elements of order d in a finite group.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Hint : if S,T are finite sets with the same cardinality, then an injection S \to T is also a surjection. Now consider the inclusion <c> \to <a>...
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  3. #3
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    Sure, and clearly |<c>|=|<a>|= d since |c|=|a|=d. But how does this shows that everything that is in <a> is also in <c>?
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