I'm reading Galois Theory for Beginners by John Stillwell. It's a good introduction, giving the essence of the idea with minimum algebra complexity.
However, I'm a bit lost at his Theorem 2 (the details are at the end).
The proof intend to find a homomorphism of $\texttt{Gal}(E/B)$, with kernel $\texttt{Gal}(E/B(\alpha))$, into an abelian group. John says:
The obvious map with kernel $\texttt{Gal}(E/B(\alpha))$ is restriction to $B(\alpha)$, $\lvert_{B(\alpha)}$, since by definition $$\sigma \in \texttt{Gal}(E/B(\alpha)) \Leftrightarrow \sigma \lvert_{B(\alpha)} \texttt{ is the identity map}.$$
Why restriction to $B(\alpha)$ is a homomorphism from $\texttt{Gal}(E/B)$ to a group with kernel $\texttt{Gal}(E/B(\alpha))$? What does this "restriction" mean?
Trying to understand it, I'm taking $B = \mathbb Q$, $\alpha = \zeta$, $B(\alpha) =\mathbb Q(\zeta)$, $E = \mathbb Q (\zeta, \sqrt{2})$, where $\zeta^5 = 1$.
So $$\texttt{Gal}(E/B(\alpha)) = \texttt{Gal}(\mathbb Q(\zeta, \sqrt{2})/\mathbb Q(\zeta)), $$ in this example it is
$\begin{array}{c|cc} \mathbb Q(\zeta, \sqrt{2}) & \sigma_1 & \sigma_2 \\ \hline \sqrt{2} & \sqrt{2} & -\sqrt{2} \\ -\sqrt{2} & -\sqrt{2} & \sqrt{2} \\ \zeta & \zeta & \zeta \\ \zeta^2 & \zeta^2 & \zeta^2 \\ \zeta^3 & \zeta^3 & \zeta^3 \\ \zeta^4 & \zeta^4 & \zeta^4, \end{array}$
so it's isomorphism to $$S_2 = \{(1), (1,2)\}.$$
Then $$\texttt{Gal}(E/B) = \texttt{Gal} (\mathbb Q(\zeta, \sqrt{2})/\mathbb Q),$$ in this example it is
$\begin{array}{c|cccccccc} \mathbb Q(\zeta, \sqrt{2}) & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4 & \sigma_5 & \sigma_6 & \sigma_7 & \sigma_8 \\ \hline \sqrt{2} & \sqrt{2} & \sqrt{2} & \sqrt{2} & \sqrt{2} & -\sqrt{2} & -\sqrt{2} & -\sqrt{2} & -\sqrt{2}\\ -\sqrt{2} & -\sqrt{2} & -\sqrt{2} & -\sqrt{2} & -\sqrt{2} & \sqrt{2} & \sqrt{2} & \sqrt{2} & \sqrt{2}\\ \zeta & \zeta & \zeta^2 & \zeta^3 & \zeta^4 & \zeta & \zeta^2 & \zeta^3 & \zeta^4 \\ \zeta^2 & \zeta^2 & \zeta^4 & \zeta & \zeta^3 & \zeta^2 & \zeta^4 & \zeta & \zeta^3 \\ \zeta^3 & \zeta^3 & \zeta & \zeta^4 & \zeta^2 & \zeta^3 & \zeta & \zeta^4 & \zeta^2 \\ \zeta^4 & \zeta^4 & \zeta^3 & \zeta^2 & \zeta & \zeta^4 & \zeta^3 & \zeta^2 & \zeta, \end{array}$
so it's isomorphism to $$\{(1), (3,4,5,6), (3,5,6,4), (3,6)(4,5), (1,2), (1,2)(3,4,5,6), (1,2)(3,5,6,4), (1,2)(3,6)(4,5) \}.$$
My guess is, so that the "restriction to $B(\alpha)$" actually means a subset of $\texttt{Gal} (E/B)$ that only change elements in $B(\alpha)$, so itcan be defined as $$\Sigma := \{ \tau \in \texttt{Gal}(E/B) \lvert \tau(\beta) = \beta, \forall \beta \in E \setminus B(\alpha) \},$$ in this example
$\begin{array}{c|c} \mathbb Q(\zeta, \sqrt{2}) & A = f(\texttt{Gal}(\mathbb Q(\zeta, \sqrt{2})/\mathbb Q)) \\ \hline (1) & (1) \\ (1,2) & (1) \\ (3,4,6,5) & (3,4,6,5) \\ (1,2)(3,4,6,5) & (3,4,6,5) \\ (3,5,6,4) & (3,5,6,4) \\ (1,2)(3,5,6,4) & (3,5,6,4) \\ (3,6)(4,5) & (3,6)(4,5) \\ (1,2)(3,6)(4,5) & (3,6)(4,5) \end{array}$
Is my understanding correct?
-- Theorem 2 details --
Any radical extension $F(\alpha_1, \dots, \alpha_k)$ is the union of an ascending tower of fields $F=F_0 \subseteq F_1 \subseteq \cdots \subseteq F_k = F(\alpha_1, \dots, \alpha_k)$ where each $F_i = F_{i-1}(\alpha_i)$, $\alpha_i$ is the $p_i$-th rot of an element in $F_i{i-1}$, $p_i$ is prime, and $F_i$ contains no $p_i$-th roots of unity not in $F_i{i-1}$ unless $\alpha_i$ is itself a $p_i$-th root of unity.
Theorem 2. If $E\supseteq B(\alpha) \supseteq B$ are fields with $\alpha^p \in B$ for some prime $p$, and if $B(\alpha)$ contains no $p$th roots of unity not in $B$ unless a itself is a $p$th root of unity, then $\texttt{Gal}(E/B(\alpha))$ is a normal subgroup of $\texttt{Gal}(E/B)$ and $\texttt{Gal}(E/B)/\texttt{Gal}(E/B(\alpha))$ is abelian.
Proof: By the homomorphism theorem for groups, it suffices to find a homomorphism of $\texttt{Gal}(E/B)$, with kernel $\texttt{Gal}(E/B(\alpha))$, into an abelian group (i.e., onto a subgroup of an abelian group, which of course is also abelian). The obvious map with kernel $\texttt{Gal}(E/B(\alpha))$ is restriction to $B(\alpha)$, $\lvert_{B(\alpha)}$, since by definition $$\sigma \in \texttt{Gal}(E/B(\alpha)) \Leftrightarrow \sigma \lvert_{B(\alpha)} \texttt{ is the identity map}.$$
The homomorphism property, $$\sigma' \sigma \lvert_{B(\alpha)}=\sigma'\lvert_{B(\alpha)} \sigma\lvert_{B(\alpha)}, \forall \sigma', \sigma \in \texttt{Gal}(E/B),$$ is automatic provided $\sigma \lvert_{B(\alpha)}(b) \in B(\alpha)$ for each $b \in B(\alpha)$, i.e. provided $B(\alpha)$ is closed under each $\sigma \in \texttt{Gal}(E/B)$.
Since $\sigma$ fixes $B$, $\sigma \lvert_{B(\alpha)}$ is completely determined by the value $\sigma(\alpha)$. If $\alpha$ is a $p$th root of unity then $$(\sigma(\alpha))^p = \sigma(\alpha^p) = \sigma(\zeta^p) = \sigma(1) = 1,$$ hence $\sigma(\alpha) = \zeta^i= \alpha^i \in B(\alpha)$, since each $p$th root of unity is some $\zeta^i$. If $\alpha$ is not a root of unity then $$ (\sigma(\alpha))^p = \sigma(\alpha^p) = \alpha^p \texttt{ since } \alpha^p \in B,$$ hence $\sigma(\alpha) = \zeta^ia$ for some $p$th root of unity $\zeta$; and $\zeta \in B$ by hypothesis, so a\texttt{Gal}in $\sigma(\alpha) \in B(\alpha)$. Thus $B(\alpha)$ is closed as required.
This also implies that $I_{B(\alpha)}$ maps $\texttt{Gal}(E/B)$ into $\texttt{Gal}(B(\alpha)/B)$, so it now remains to check that $\texttt{Gal}(B(\alpha)/B)$ is abelian. If $\alpha$ is a root of unity then, as we have just seen, each $\sigma \lvert_{B(\alpha)} \in \texttt{Gal}(B(\alpha)/B)$ is of the form $\sigma_i$, where $\sigma_i(\alpha) = \alpha^i$, hence $$\sigma_i\sigma_j(\alpha) = \sigma_i(\alpha^j) = \alpha^{ij} = \sigma_j\sigma_i(\alpha).$$ Likewise, if $\alpha$ is not a root of unity then each $\sigma \lvert_{B(\alpha)} \in \texttt{Gal}(B(\alpha)/B)$ is of the form $\sigma_i$ where $\sigma_i(\alpha) = \zeta^i\alpha$, hence $$\sigma_i\sigma_j(\alpha) = \sigma_i(\zeta^j\alpha) = \zeta^{i+j}\alpha = \sigma_j\sigma_i(\alpha)$$ since $\zeta \in B$ and therefore $\zeta$ is fixed. Hence in either case $\texttt{Gal}(B(\alpha)/B)$ is abelian.
