Application of automorphisms in constructing groups This is an excerpt from Herstein's book. There are some moments which are unclear to me and I would like to clarify them.

They take two elements of set $\langle G,x\rangle$ (this set is not yet a group!) which consists of $21$ elements, namely, $xa$ and $xa^2$ and multiply them: $(xa)(xa^2)=x(ax)a^2$.
Q1: How did the author manage to write the product as $x(ax)a^2$? Why associativity is correct in this set $\langle G,x\rangle$?
Q2: We have that $x^{-1}ax=a^2$ in set $\langle G,x\rangle$. How did they get $ax=xa^2$ in $G$ which is not yet a group? If $\langle G,x\rangle$ is a group, then OK. But $\langle G,x\rangle$ is not a group.
Q3: Why the exponent laws hold in set $\langle G,x\rangle$? For example, Herstein uses that $xx=x^2$ and $a^2a^2=a^4$
Please explain all these questions. The same question occured in my head couple weeks ago but I did not get persuasive. I hope that this time the questions will be explained properly.
 A: I think that you are right, and that Herstein goes a bit too fast here for a beginner. (I've had many students who found this puzzling.)
But you are also going too fast by writing $\langle G, x\rangle$ too soon, that only really makes sense when we are working in some overgroup -- which as you recognise, we are not.
We have, Herstein says, a set, let's call it $\hat{G}$, of 21 elements whose names are $x^i a^j$ where $i\in \{0,1,2\}$ and $j\in \{0,1,2,3,4,5,6\}$. We want to define a group operation on this set. 
[Now this is not part of the definition, it's just the motivation: we want the operation of conjugation by $x$ in $\hat{G}$, $a\mapsto x^{-1}ax$ to be the automorphism $\phi:a\mapsto a^2$. What would that force the multiplication in $\hat{G}$ to be?
Well if we are to be successful, and get a group, we will have associativity and so the exponents laws, and we will need to have
$$(x^ i a^j)(x^m a^n)
=x^{i+m}x^{-m}a^{j}x^{m}a^{n}
=x^{i+m}(x^{-m}ax^{m})^{j}a^{n}
=x^{i+m}(a^{2^m})^{j}a^{n}
=x^{i+m}a^{2^{m}j+n}
$$
where of course we have to calculate the value of the exponents modulo $3$ and $7$ respectively.]
With the motivation in our minds we define a binary operation $\cdot$ on $\hat{G}$ by 
$$(x^ i a^j)(x^m a^n)
:=x^{i+m}a^{2^{m}j+n} \text{ where the exponents are reduced modulo 3 and 7.}
$$
We are now faced with the tedious (but essentially trivial) task of proving that this operation is associative; that $x^0 a^0$ is a two-sided identity; and that each $x^i a^j$ has a two-sided inverse (I'll leave you to find out what it is!)
Once we've done this we have that $\hat{G}$ is indeed a group, and all is well. 
If you do a few examples you'll see how this procedure will "always work" and then be happy to use Herstein's very abbreviated account for yourself. 
A: Q1 : We are told that $G$ is a cyclic group of order 7 generated by $a$. I think the unstated assumption here is that the larger object generated by $a$ and $x$ is also a group, and inherits its group operation properties from $G$.
Q2 : We are told that $G=<a>$ is a group, and we are assuming that $<a,x>$ is also a group. So $xa^2 = xx^{-1}ax = eax = ax$.
Q3 : Integer exponents are simply a short hand, so $x^2$ stands for $xx$; $x^3$ stands for $xxx$ etc. More formally, we could define $x^1=x$ and $x^{n+1} = x(x^n)$ for $n \ge 1$. The exponent law $x^ax^b = x^{a+b}$ then follows from associativity of the group operation.
