Prove that $a$ commutes with each of its conjugates in $G$ if and only if a belongs to an abelian normal subgroup of $G$. Let $a$ be an element of a group $G$. Prove that $a$ commutes with each of its conjugates in $G$ if and only if a belongs to an abelian normal subgroup of $G$.
My attempt: Firstly, suppose that $N$ is an abelian normal subgroup of $G$.
Since, $a$ belongs to $N$ and $N$ is a normal subgroup. This implies, $gag^{-1}\in N$, for all $g \in G$.
Now, $a \in N$, and  $gag^{-1}\in N$, this implies $(a)(gag^{-1})=(gag^{-1})(a)$ (because, $N$ is a normal subgroup of $G$).
This shows that $a$ commute with all of each conjugates of $a$.
Converse:Let $a$ commute with all of its conjugate. i.e $(a)(gag^{-1})=(gag^{-1})(a)$ for all $g \in G$.
Let $N=<gag^{-1}| \forall g \in G>$.
Clearly, $N$ is a normal subgroup of $G$.
Abelian: Let $gag^{-1},hah^{-1} \in N$, then
$(gag^{-1})(hah^{-1})=ga(g^{-1}hah^{-1}g)g^{-1}=g(g^{-1}hah^{-1}g)ag^{-1}=(hah^{-1})(gag^{-1})$.
This implies $N$ is a normal abelian subgroup of $G$.
Is my proof correct?
 A: Your proof is essentially correct, as other commenters have pointed out. Mathematically, everything checks out, although there are a few places where you could improve your exposition (or where you need to fix your justification). Read on for my nit-picking!
First, as halrankard says, you reference the object $a$ before you define it. In your starting sentence "Firstly, suppose that $N$ is an Abelian normal subgroup of $G$," you should add something letting the reader know what $a$ is. For example, you might write "Suppose that $N$ is an Abelian normal subgroup of $G$ and let $a\in N.$" It's rude to start talking about your characters before you introduce them!
Second, as Koro pointed out, your reasoning "Now, $a\in N$, and $gag^{-1}\in N,$ this implies $(a)(gag^{-1}) = (gag^{-1})(a)$ (because, $N$ is a normal subgroup of $G$)." is flawed. This equality holds because the subgroup $N$ is Abelian, not because it is normal. (Normality is what implies that $gag^{-1}\in N$ in the first place.)
For the converse, your argument is again correct. However, you might consider elaborating on why $N$ is normal. This is not hard to show, but it's not good practice to claim things are "clear" or "obvious." (Have you read any texts where the author has claimed this about something you thought wasn't? It can be very frustrating and demoralizing!) If it is so obvious, you should be able to justify it quickly.
You might also include a word or two about why to show that $N$ is Abelian it suffices to show that two elements of the form $gag^{-1}$ commute. Again, this is simple to show, but depending on how clear and rigorous you want to be, it might be a helpful detail.
Whether or not you add something like the above depends largely on what the purpose of writing this proof is and who will be reading it. If your reader is supposed to be someone who is new to normal subgroups, this is going to be more important to spell out. If you're writing this as homework for a class, it might be good to include it so that the grader sees all the details and can't complain about anything. If you're writing this for someone who's more familiar with these things or for yourself, it isn't as big of an issue. The larger point is that you should be mindful of your audience and purpose when writing!
Additionally, I would include an extra step in your commutation computation for clarity:
\begin{align*}
(gag^{-1})(hah^{-1})&=ga(g^{-1}hah^{-1}g)g^{-1}\\
&= ga\left((g^{-1}h)a(g^{-1}h)^{-1}\right)g^{-1}\\
&= g\left((g^{-1}h)a(g^{-1}h)^{-1}\right)ag^{-1}\\
&=g(g^{-1}hah^{-1}g)ag^{-1}\\
&=(hah^{-1})(gag^{-1}).
\end{align*}
Finally, you might try to write your proof as a few short paragraphs rather than disjointed sentences/sentence fragments each on their own line. This will help the flow and presentation of your solution. However, mathematically, you're good to go!
