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Math Help - Simple Group order 168

  1. #1
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    Simple Group order 168

    Hello,

    I need help on proving this problem.

    How many elements of order 7 are there in a simple group of order 168?

    Thanks a lot!
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  2. #2
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    Quote Originally Posted by roporte View Post
    How many elements of order 7 are there in a simple group of order 168?
    By Sylow's 3rd theorem the number of Sylow's 7-subgroups of a group of order 168 is either 1 or 8. It cannot be 1 because it would imply such a Sylow 7-subgroup must be normal contradicting the fact the group is simple. Therefore there are 8 Sylow 7-subgroups. Given P_1,P_2 to be two Sylow-7 subgroups with P_1\not =P_2 then it implies P_1 \cap P_2 = \{ e \} (because P_1\cap P_2 is a subgroup of P_1 and my Lagrange's theorem |P_1\cap P_2| divides |P_1| - a prime). This means the non-trivial elements in P_1,P_2,...,P_8 are all distinct. And furthermore each one generates the group i.e. have order 7. There are a total of 6\cdot 8 = 48 of such elements. And conversely any element of order seven gives rise to a Sylow 7-subgroup and must be included among those elements. Therefore there are 48 elements of order 7.
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