![enter image description here](https://i.stack.imgur.com/sHkpP.png)

(1.) I don't understand [above](http://math.berkeley.edu/~ogus/old/Math_113/sol3.pdf). How do you magically envisage and envision to let $b = xax^{-1}$? 

What I did was to start from the answer and see if I can get a chain of equivalences. $ax = xa \iff ax\color{darkcyan}{x^{-1}} = xa\color{darkcyan}{x^{-1}} \iff a = xax^{-1}$
$\begin{align} \Leftarrow or \iff a^2 & = (xax^{-1})^2 
\\ e & = \end{align} $. 

Hence $ax = xa \Leftarrow e = (xax^{-1})^2.$ Hence if I sally forth from the $RHS$ to $LHS$ then I'm done?  

 But I never used $a$ was unique? 

(2.) What's the intuition? $ax = xa$ looks like commutativity. Hence $G$ is Abelian? 

**Update Jan. 7 2014:** The comments and answers involve 'center, 'conjugate' which aren't covered up to this time. I didn't know they are needed here. Hence I'll return. Are there simpler answers?