Proposition 7.2. of Adkins' Algebra I don't understand some of the following text (from Adkins' Algebra):

1- Why if $q$ does not divide $p-1$ then $\phi_h = 1_N$ (i.e. $H$ is normal) ? Exercise 11 doesn't help since it says "... if $nh=hn$ ..." and not conclude that!
2- If $q|p-1$ why there are nontrivial homomorphisms, especially when in both cases $\phi : Z_q \to Aut(N) \cong Z^*_p$? 
3- Why $b^{-1}ab$ have to be in a form of $a^r$?! And why $r^q \equiv 1 \ mod \ p$?
Thank you. 
 A: (1) There can't be a nontrivial homomorphism $G\to H$ if $|G|$ and $|H|$ are coprime. This follows from the first isomorphism theorem (the image is isomorphic to a quotient of $G$) and Lagrange's theorem (which implies quotients of $G$ have size dividing $|G|$). In particular, if $q$ and $p-1$ are coprime (equivalently $q\nmid p-1$), then the homomorphism $\mathbb{Z}_q\to\mathrm{Aut}(\mathbb{Z}_p)$ must be trivial, since the order of $\mathrm{Aut}(\mathbb{Z}_p)$ is just $p-1$.
(2) If $q\mid p-1$ then $\mathrm{Aut}(\mathbb{Z}_p)$ must have an element of order $q$, and then there is a homomorphism which maps $\mathbb{Z}_q$'s generator to that element of order $q$. This will be a nontrivial homomorphism.
(3) If $N=\langle a\rangle$ is normal then $bab^{-1}\in\langle a\rangle$. Since all elements of $\langle a\rangle$ are of the form $a^r$ for some exponent $r$, that means $bab^{-1}=a^r$ for some $r$. (I don't know why the text is writing the conjugation backwards.) Notice that $bab^{-1}$ is none other than $\phi_b(a)$. Applying $\phi_b$ over and over again is tantamount to conjugating by powers of $b$. That is, 
$$\underbrace{\phi_b\circ\cdots\circ\phi_b}_n(a)=b(\cdots bab^{-1}\cdots)b^{-1}=b^nab^{-n}=\phi_{b^{\large n}}(a).$$
(Of course, $\phi_b^n=\phi_{b^{\large n}}$ is just a corollary to the fact that $x\mapsto \phi_x$ is a group homomorphism.)
But notice that, if $\phi_b(a)=a^r$ and $\phi_b$ is itself a group homomorphism, then
$$\begin{array}{llll} \phi_b(\phi_b(a)) & = & \phi_b(a^r) & = & \phi_b(a)^r & = &(a^r)^r & = & a^{r^{\large 2}}, \\ 
\phi_b(\phi_b(\phi_b(a))) & = & \phi_b(a^{r^{\large 2}}) & = & \phi_b(a)^{r^{\large 2}} & = & (a^r)^{r^{\large 2}} & = & a^{r^{\large 3}}, \\
& & & \vdots \\ \underbrace{\phi_b\circ\cdots\circ\phi_b}_n(a) & = & \phi_b(a^{r^{\large n-1}}) & = & \phi_b(a)^{r^{\large n-1}} & = & (a^r)^{r^{\large n-1}} & = & a^{r^{\large n}} \end{array}$$
In particular, this means $\phi_{b^q}(a)=a^{r^{\large q}}$. But $\phi_{b^q}(a)=\phi_e(a)=a$. Therefore
$$a^{r^{\large q}}=a \quad\Rightarrow\quad a^{r^{\large q}-1}=e \quad \Rightarrow \quad p\mid (r^q-1) \quad\textrm{i.e.}\quad r^q\equiv 1\bmod{p}. $$
