Subset has 2 Conjugates then Normal Subgroup

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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


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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.


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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:

\(\ds b x b^{-1}\) \(=\) \(\ds a x a^{-1}\)
\(\ds \leadsto \ \ \) \(\ds \paren {a^{-1} b} x b^{-1}\) \(=\) \(\ds x a^{-1}\)
\(\ds \leadsto \ \ \) \(\ds \paren {a^{-1} b} x \paren {a^{-1} b}^{-1}\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds a^{-1} b\) \(\in\) \(\ds \map {C_G} S\)


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 $G$.

$\blacksquare$


Sources

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