If a group $G$ contains an element a having exactly two conjugates, then $G$ has a proper normal subgroup $N \ne \{e\}$
So my take on this is as follows: If we take $C_G(S)$ of S. This is a subgroup of G. If $C_G(S)=G$, then S has no conjugate but itself, so therefore $C_G(S)$ is a proper subgroup. If we suppose $C_G(S)=\{e\}$,then in order for there to be exactly two conjugates of S, then
For every $a \ne b \in G \ \{e\}, bxb^{-1} =axa^{-1}$ but $bxb^{-1}=axa^{-1} \to (a^{-1}b)xb^{-1}=xa^{-1} \to (a^{-1}b)x(a^{-1}b)^{-1}=x \to a^{-1}b \in C_G(S)$
Which means that $C_G(S)$ is actually nontrivial or that $a^{-1}b=e$ if and only if $a=b$, which would be a contradiction. Thus $C_G(S)$ is a nontrivial proper subgroup. Since there are exactly 2 conjugacy classes of S and they are in one to one correspondence with cosets of S, its centralizers' index $[G:C_G(S)]=s$. Subgroups of index $2$ are normal, so $C_G(S)$ is a proper nontrivial normal subgroup.
This approach seemed very different from other examples I have seen so I guess I am wondering if this approach makes sense.
Best Answer
Let $a$ be an element with exactly two conjugates, itself and some other element $b$.
Let $G$ act on the $X = \{a,b\}$ by conjugation. Consider the class equation for this action: $$ |X| = |X_0| + \sum |G/G_x| = 2. $$
We notice that $|X_0| = 1$ (since the identity actions lies in $X_0$, and $a$ isn't fixed by the action of $b$). Hence there is a unique element $x\in X$ such that $[G:G_x] = 2$, and then $G_x$ must be normal since every subgroup of index 2 is normal.