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
was my answer correct?