If a group $G$ contains an element a having exactly two conjugates, then $G$ has a proper normal subgroup $N \ne \{e\}$ 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.
 A: 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.
A: Here is another argument.
Let $\{a,b\}$ be a conjugacy class. Notice that if $aba^{-1}=a$, then $a=b$. Contradiction. So it has to be that $aba^{-1}=b$, which means that $a$ and $b$ commute.
Since $\{a,b\}$ is a conjugacy class, for any $g\in G$, $ga=ag$ or $ga=bg$. Also, $gb=ag$ or $gb=bg$.
Now, consider the subgroup $N$ generated by $a$ and $b$. $N$ is of the form $\{a^{h_1}b^{k_1}...a^{h_n}b^{k_n}|h_i, k_i\in \mathbb{Z}\}$. 
Since we can replace "$g$ times $a$ or $b$" by "$a$ or $b$ times $g$," $ga^{h_1}b^{k_1}...a^{h_n}b^{k_n}g^{-1}=a^{h_1'}b^{k_1'}...a^{h_m'}b^{k_m'}gg^{-1}=a^{h_1'}b^{k_1'}...a^{h_m'}b^{k_m'}$ is again in $N$. So $N$ is normal.
If $N=G$, then everything in $G$ is a product of powers of $a$ and $b$. This gives us that everything commutes with one another. So any conjugacy class can only have one element. Contradiction.
So $N$ is a proper normal subgroup. 
A: Let $g$ be an element with two conjugates then $\mid N(g)\mid\,=\frac{\mid\mathbb{G}\mid}{2}$ where $N(g)$ is normaliser subgroup with respect to $g$ and any subgroup with index 2 is normal hence $N(g)$ is a non-trivial normal group.
A: 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 $C_G(g) \ne \{e\}$. 
