Proving Non-Simplicity

  • Thread starter Obraz35
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Obraz35
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Homework Statement


Let G be a finite group. Prove that if some conjugacy class has exactly two elements then G can't be simple


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The Attempt at a Solution


I originally proved this accidentally assuming that G is abelian, which it isn't. So say x and y are conjugate then we have conjugacy classes {e} and {x,y}. I know that the size of a conjugacy class has to divide the order of a group so |G| is even and must be greater than 2. I'm having trouble showing the existence of a normal subgroup though.
 

Answers and Replies

  • #2
Dick
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If G has a conjugacy class with two elements then it has a subgroup of order |G|/2. Look at the centralizer of each element of the conjugacy class. If G has a subgroup H of order |G|/2, H must be normal. Look at left and right cosets.
 

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