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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.

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    $\begingroup$ The basic idea is correct: there is a subgroup of index $2$, and subgroups of index $2$ are always normal. However, I find your explanation a little confusing in places. The subgroup of index $2$ is the centralizer of an element $x$ such that $x$ has precisely two conjugates in $G$. $\endgroup$ – Geoff Robinson Mar 11 '15 at 1:23
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    $\begingroup$ Instead of considering $C_G(S)$, streamline it by considering $C_G(x)$ where $x$ is the element of $G$ possessing exactly two conjugates. $\endgroup$ – David Wheeler Mar 11 '15 at 1:25
  • $\begingroup$ Shouldn't the order of $G$ be $2<|G|<\infty$? $\endgroup$ – big-lion May 27 at 2:57
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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.

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Simplifying $a,b$ argument a little:

Let $g$ be the element with exactly two conjugates. Suppose $C_G(g) = \{e\}$. Since $g \in C_G(g)$, this means $g = e$.

By Lagrange's Theorem, $[G : C_G(g)] = 2$ imples $\lvert G \rvert = 2$, so $G = \{e, h\}$. The two conjugates of $g$ are itself and $hgh^{-1}$, but both are $e$, contradicting $g$ having two conjugates.

Conclude $C_G(g) \ne \{e\}$.

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