Proof. This is certainly true when the corresponding $a_i$ is a primitive root of unity. Otherwise, note that $a_i$ is a root of $x^{n_i}-a_i^{n_i}\in F_{i-1}[x]$. Let $\omega=\omega_{n_i}$ and consider the set $\{a_i,\omega a_i,\omega^2a_i,\ldots,\omega^{n_i-1}a_i\}$. Each of these elements is in $F_i$, each is a root of $x^{n_i}-a_i^{n_i}$, and they are distinct. Thus, $x^{n_i}-a_i^{n_i}$ splits in $F_i=F_{i-1}(a_i)$, so $F_i$ is the splitting field of $x^{n_i}-a_i^{n_i}$ over $F_{i-1}$ and by theorem 1.3.4, $F_i$ is a normal extension of $F_{i-1}$. $\qed$

Proof. There are two cases, depending on whether $a_i$ is a primitive root of unity or not. We do just the latter case; the former is similar. Remember that the roots of unity are all added first, so they are all in $F_{i-1}$.

Suppose that $\sigma\in G(F_i/F_{i-1})$. Since $a_i^{n_i}\in F_{i-1}$, $\sigma(a_i^{n_i})=a_i^{n_i}$, so $(\sigma(a_i))^{n_i}-a_i^{n_i}= \sigma(a_i^{n_i}-a_i^{n_i})=0$, or in other words, $\sigma(a_i)$ is a root of $x^{n_i}-a_i^{n_i}$. By the proof of the previous lemma, $\sigma(a_i)=\omega_{n_i}^{j(\sigma)}a_i$ for some integer $j(\sigma)$. Now if $\tau\in G(F_i/F_{i-1})$, $$\eqalign{(\sigma\circ\tau)(a_i)=&\sigma(\omega_{n_i}^{j(\tau)}a_i)= \omega_{n_i}^{j(\tau)}\sigma(a_i)= \omega_{n_i}^{j(\tau)}\omega_{n_i}^{j(\sigma)}a_i=\cr &\omega_{n_i}^{j(\sigma)}\omega_{n_i}^{j(\tau)}a_i= \omega_{n_i}^{j(\sigma)}\tau(a_i)= \tau(\omega_{n_i}^{j(\sigma)}a_i)=(\tau\circ\sigma)(a_i).}$$ Thus for all $\sigma$ and $\tau$ in $G(F_i/F_{i-1})$, $\sigma\tau$ and $\tau\sigma$ agree on $a_i$.

By theorem 19.3 in Gallian, every $b\in F_i$ can be written as $\sum c_ja_i^j$, where $c_j\in F_{i-1}$. Then $$(\sigma\tau)(b)=\sum c_j((\sigma\tau)(a_i))^j= \sum c_j((\tau\sigma)(a_i))^j=(\tau\sigma)(b).$$ Thus, $\sigma\tau=\tau\sigma$ and so $G(F_i/F_{i-1})$ is abelian as desired. $\qed$

Proof. Let $g_i$ be the minimal polynomial for $a_i$ over $F$, and let $b_i$, $1\le i\le s$ be a list of all roots of all of the $g_i$, that is, $s$ is the sum of the degrees of the $g_i$. For convenience, let $b_1,\ldots, b_r$ be $a_1,\ldots, a_r$. Let $N$ be the splitting field of the product $g=\prod g_i$, that is, $N=F(b_1,\ldots,b_s)\supseteq F_r$. We will extend the tower of fields: $$F=F_0\subseteq F_1\subseteq F_2\subseteq\cdots\subseteq F_r \subseteq F_{r+1}\subseteq\cdots\subseteq F_m=N.$$

For each root $b_i$ of $g_k$, there is an isomorphism $\sigma_i\colon F(a_k)\to F(b_i)$ that fixes $F$, by the lemma on page 372 in Gallian. By theorem 19.4, $\sigma_i$ may be extended to an automorphism $\overline\sigma_i$ on the splitting field $N$ that fixes $F$(note that $\sigma_i(g)=g$). Now $\overline\sigma_i(F_r)$ is a subfield of $N$ that contains $b_i$ and is isomorphic to $F_r$; in particular, $\overline\sigma_i(F_r)$ may be viewed as arising from a tower of fields formed using the elements $\overline\sigma_i(a_j)$, $1\le j\le r$, and $\overline\sigma_i(F_j)=\overline\sigma_i(F(a_1,\ldots,a_j))= F(\overline\sigma_i(a_1),\ldots,\overline\sigma_i(a_j))$.

Thus, for each root $b_i$ we have a corresponding sequence of elements $\overline\sigma_i(a_j)=a_{i,j}$, $1\le j\le r$, one of which is $b_i$. Then we see that $N=F(a_1,a_2,\ldots,a_r,a_{r+1,1},a_{r+1,2},\ldots,a_{s,r})$, and $$\eqalign{F=&F_1\subseteq F_2\subseteq\cdots\subseteq F_r\subseteq N_{r+1,1}=F_r(a_{r+1,1})\subseteq N_{r+1,2}=N_{r+1,1}(a_{r+1,2}) \subseteq\cdots\cr &\subseteq N_{s,r-1}=N_{s,r-2}(a_{s,r-1})\subseteq N_{s,r}=N_{s,r-1}(a_{s,r})=N.\cr}$$ Note that we add all of the elements $\{a_{i,1}, a_{i,2},\ldots,a_{i,r}\}$ just to get $b_i=a_{i,k}$; we do this so that each field is obtained from the previous one by adding a root. Now we need to verify that each $a_{i,j}$ is a root of some element in the previous field. There are two cases:

$j=1$: We need to show some power of $a_{i,1}$ is in $N_{i-1,r}$: $$a_{i,1}^{n_1}=\left(\osigma_{i}(a_1)\right)^{n_1}=\osigma_{i}(a_1^{n_1}) \in \osigma_{i}(F)=F\subseteq N_{i-1,r}.$$

$j>1$: We need to show some power of $a_{i,j}$ is in $N_{i,j-1}$: $$\eqalign{ a_{i,j}^{n_j}&=\left(\osigma_{i}(a_j)\right)^{n_j}=\osigma_{i}(a_j^{n_j}) \in \osigma_{i}(F_{j-1})=\osigma_i(F(a_1,\ldots,a_{j-1}))\cr &=F(\osigma_i(a_1),\ldots,\osigma_i(a_{j-1}))= F(a_{i,1},\ldots,a_{i,j-1}) \subseteq N_{i,j-1}.\cr}$$ $\qed$

Proof. Let $H_0\nsg H_1\nsg H_2\nsg\cdots\nsg H_k=G$ be a tower of groups that illustrates that $G$ is solvable, and let $N\nsg G$. Consider the groups $$\gmod H_0N/N. \le \gmod H_1N/N. \leq\cdots\leq \gmod H_kN/N. = \gmod G/N..$$ Suppose that $x\in H_i$, $y\in H_{i+1}$, $n,m\in N$, $xnN=xN\in H_iN/N$, $ymN=yN\in H_{i+1}N/N$. Then $yNxN(yN)^{-1}=yxy^{-1}N$. Since $H_i\nsg H_{i+1}$, $yxy^{-1}\in H_i$, so $yxy^{-1}N\in H_iN/N$ and $H_iN/N\nsg H_{i+1}N/N$.

Now consider two elements of $(H_{i+1}N/N)/(H_iN/N)$, $(gnN)(H_iN\!/\!N)=(gN)(H_iN\!/\!N)$ and $(hmN)(H_iN/N)=(hN)(H_iN/N)$. We want to show these elements commute, that is, $(ghN)(H_iN/N)=(hgN)(H_iN/N)$. It suffices to show that $(hgN)^{-1}(ghN)=g^{-1}h^{-1}ghN\in H_iN/N$. Since $H_{i+1}/H_i$ is abelian, $g^{-1}h^{-1}gh\in H_i$, this is true. $\qed$