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
Take $C_{G}(S)$ the centralizer of $S$. This is a subgroup of $G$.

If $C_{G}(S)=G$, then $S$ has no conjugate but itself, so $C_{G}(S)$ is a proper subgroup.

If $C_{G}(S)=\{e\}$ the identity of $G$, then for there to be exactly two conjugates of $S$:

$\forall a \ne b \in G\setminus \{e\}, bxb^{-1}=axa^{-1}$

but $bxb^{-1}=axa^{-1} \implies (a^{-1}b)xb^{-1}=xa^{-1} \implies (a^{-1}b)x(a^{-1}b)^{-1}=x \implies a^{-1}b \in C_{G}(S)$

This either implies that $C_{G}(S)$ is actually nontrivial, or that $a^{-1}b=e \iff a=b$, 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 centralizer's index $[G\ :\ C_{G}(S)]=2$.

Subgroups of index 2 are normal, so $C_{G}(S)$ is a proper nontrivial normal subgroup.