Proving Non-Simplicity in Finite Groups with Two-Element Conjugacy Class

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In summary, if a finite group G has a conjugacy class with exactly two elements, then G cannot be simple. This is because if G has a subgroup of order |G|/2, it must be normal and therefore G is not simple. This can be shown by looking at the centralizers of each element in the conjugacy class and considering left and right cosets.
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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.
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  • #2
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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