Suppose a unique a generates a cyclic subgroup of order. Show ax = xa. - Fraleigh p. 67 6.50 
(1.) I don't understand above. 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?
 A: You're given $\;a\in G\;$ is the unique element of order $\;2\;$ in $\;G\;$ . Now, we have that
$$\forall\,x\in G\;,\;\;(xax^{-1})^2=xax^{-1}xax^{-1}=xa^2x^{-1}=xx^{-1}=1$$
so by the above uniqueness it must be that
$$xax^{-1}=a\iff xa=ax\;,\;\;\text{since it can not be}\;\;xax^{-1}=1\;\;\text{(why?)}$$
The above means that $\;a\in Z(G):=\{g\in G\;;\;gx=xg\;\;\forall\,x\in G\}=\;$ the group's center 
A: I fill in DonAntonio's answer here. 
You're given $\;a\in G\;$ is the unique element of order $\;2\;$ in $\;G\;$ .
How do you envisage and envision to consider $(xax^{−1})^2$ here?
This magically cropped up in DonAntonio's answer.
$$\forall\,x\in G\;,\;\;(xax^{-1})^2=xax^{-1}xax^{-1}=xa^2x^{-1}=xx^{-1}=1$$
so by the above uniqueness of $a\in G$ it must be that
$$xax^{-1}=a\iff xa=ax\;,\;\;\text{since it can not be}\;\;xax^{-1}=1\;\;\color{darkorange}{\text{(why?)}}$$
The above means that $\;a\in Z(G):=\{g\in G\;;\;gx=xg\;\;\forall\,x\in G\}=\;$ the group's center 
Update on (why?) Question postulates: $a \in G$ is the unique element of order 2 in G. Hence only $a^2 = e$. $xax^{-1} \neq a$ hence $xax^{-1} \neq e$. Is this right?
