Subset has 2 Conjugates then Normal Subgroup

Theorem
Let $G$ be a group.

Let $S$ be a subset of $G$.

Let $S$ have exactly two conjugates in $G$.

Then $G$ has a proper non-trivial normal subgroup.

Proof
Consider the centralizer $C_G \left({S}\right)$ of $S$ in $G$.

From Centralizer of Group Subset is Subgroup, $C_G \left({S}\right)$ is a subgroup of $G$.

If $C_G \left({S}\right) = G$, then $S$ has no conjugate but itself.

So, in order for $S$ to have exactly two conjugates in $G$, it is necessary for $C_G \left({S}\right)$ to be a proper subgroup.

Let $e$ be the identity of $G$.

If $C_G \left({S}\right) = \left\{{e}\right\}$, then for there to be exactly two conjugates of $S$:


 * $\forall a \ne b \in G \setminus \left\{{e}\right\}: b x b^{-1} = a x a^{-1}$

But:

This implies either that $C_G \left({S}\right)$ is actually nontrivial, or that $a^{-1}b = e \iff a = b$, a contradiction.

Thus $C_G \left({S}\right)$ is a nontrivial proper subgroup of $G$.

We have that there are exactly $2$ conjugacy classes of $S$.

These are in one-to-one correspondence with cosets of $S$.

Thus the index $\left[{G : C_G \left({S}\right)}\right]$ of the centralizer is:
 * $\left[{G : C_G \left({S}\right)}\right] = 2$

From Subgroup of Index 2 is Normal:
 * $C_G \left({S}\right)$ is a proper nontrivial normal subgroup of $G$.