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Math Help - In a finite group, the number of elements of order d is divisible by phi(d)

  1. #1
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    In a finite group, the number of elements of order d is divisible by phi(d)

    Proof: Suppose that d(1) , d(2) , ...., d(k) are all the elements of order d.

    Look at the subgroups <d(1) >, <d(2) >, ... <d(k) >

    (i) Each of these is cyclic and so each has phi(d) elements of order d.

    (ii) Suppose that some element x of order d is in two of these subgroups

    x in <d(i)> intersection <d(j)>. Then <d(i)> = <x> = <d(j)>

    (iii) We count the number of these subgroups.
    no two can share any elements of order d.
    The number of elements of order d will be:
    (the number of distinct subgroups) * phi(d).

    Can somebody explain me in part (ii) why <d(i)> = <x> = <d(j)> ?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    If x \in <d_i>\cap<d_j>, then <x> \subset <d_i>\cap<d_j>. But can this containment be proper, if we're given that |<x>|=d?
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  3. #3
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    Thank you, I got it. if they are finite set and one is contained in the other, then they must be equal set. Now it makes sense.
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