$G$ is a finite group and $a \in G$ s.t. $a$ has exactly $2$ conjugates. Then $G$ contains a non-trivial, proper normal subgroup.
This question proved to be more difficult than I had expected, and as a result I had to do a lot of reading. By reading various proofs, I compiled the full proof into the following and I have some questions about it:
Proof:
Given $2$ conjugates of $a$ means $|G|/|N(a)| = 2$ where $N(a)$ is the normalizer (or centralizer) of $a$. So $N(a) \neq G$ which means $N(a)$ is a proper subgroup of $G$.
Note: Suppose if $N(a) = \{e\}$. For any $g \in G$ $gag^{-1}$ is either $a$ or the other conjugate. If $gag^{-1} = a \implies ga = ag \implies g \in N(a) \implies g = e$.
So let's consider $x,y \in G$ s.t. $x \neq y$.
Then $xax^{-1} = yay^{-1} \implies a(x^{-1}y) = (x^{-1}y) a \implies (x^{-1}y) \in N(a) \implies (x^{-1}y) = e \implies x=y$ which is a contradiction.
So $N(a)\neq \{e\}$. Since $[G:N(a)]=2$ then $N(a)$ is normal in $G$.
$\Box$
Now I have a problem with this proof. I don't like that there are 2 hypothesis floating around. That is, $N(a) = \{e\}$ and $x \neq y$ especially because I didn't really "use" the fact $x \neq y$ in an effective way. My only reason for writing "...$x=y$ which is a contradiction" is that I just didn't want that to be true. Beyond that, I don't see how I can effectively argue it.
Can I please get some help with filling holes in my proof?