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 $\map {C_G} S$ of $S$ in $G$.

From Centralizer of Group Subset is Subgroup, $\map {C_G} S$ is a subgroup of $G$.

If $\map {C_G} S = G$, then $S$ has no conjugate but itself.

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

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

If $\map {C_G} S = \set e$, then for there to be exactly two conjugates of $S$:


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

But:

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

Thus $\map {C_G} S$ 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 $\index G {\map {C_G} S}$ of the centralizer is:
 * $\index G {\map {C_G} S} = 2$

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