I took an exam today. If I remember correctly question was like this: let G be a group. if it has "a" element which has exact two conjugates, then G cant be simple.
I answered: let that two conjugates be $g_1a{g_1^{-1}} $and ${g_2}a{g_2^{-1}}$ and H= {a}. then N(H) ={g:$gag^{-1}\in G$}={$g_1,g_2$} so $|H|=1 and|N(H)|=2$
since $[N(H):H]=2$ then $H \triangleleft N(H)$... $H<N(H)<G $ so G has a normal group, therefor G cant be simple.
was my answer correct?
The set of conjugates of a does not have to be equal to N({a}) in size. It might happen that x,y belong to N({a}) ($x,y$ distinct) and $xax^-1$=$yay^-1$
The proof that G is not simple is:
We consider G/N({a}). Note that |G/N(H)|=2
Proof: Consider the function that sends the coset xN({a}) to xax^-1 , this function is a bijection.
If N(H) were equal to G then |G/N(H)|=1. Thus N({a}) is a proper normal subgroup of G