I am having trouble understand theorem $9.4$ of Chapter $6$ of Lang's Algebra (pg. 300-301).   The setup is a we have a field $k$ of characteristic not dividing $n$. We know that the splitting field of $f=X^n-a$ is $k(\zeta_n,\alpha)$ where $\alpha$ is a root of $f$ and $\zeta$ a primitive $n^{th}$ root of unity. Any automorphism $\sigma$ of the Galois group of $f$ over $k$ maps $\alpha \mapsto \alpha\zeta^b$ where $b$ is unique modulo $n$, and $\sigma$ induces an automorphism of the cyclic group $\mathbf{\mu_n}=\left\langle \zeta \right \rangle $ via $\zeta \mapsto \zeta^d$ where $(d,n)=1$ and $d$ is uniquely determined by $\sigma$.   We then verify that the map  $\sigma \mapsto \begin{pmatrix} \ \ 1 &amp; 0\\ b_\sigma &amp; d_\sigma \end{pmatrix}$ where $b$, $d$ are the integers determined by $\sigma$ in the previous paragraph is an injective homomorphism into the group $G(n)$ of all matrices $\begin{pmatrix} \ 1 &amp; 0\\ a &amp; c \end{pmatrix}$ such that $a \in \mathbf{Z}/n\mathbf{Z}$, $c \in (\mathbf{Z}/n\mathbf{Z})^{*}.$    The question that the theorem addresses is when the above the map is an isomorphism of the Galois group of $f$ and $G(n)$. With $\phi$ being the Euler function, the theorem states:   Suppose $[k(\zeta_n):k]=\phi(n)$ and let $a \in k$. Suppose for each prime $p|n$ that $a$ is not a $p^{th}$ power. Let $K$ be the splitting field of $X^n-a$ over $k$ and $G$ the Galois group. Then the above map is an isomorphism $G \cong G(n)$ with commutator subgroup Gal$(K/k(\zeta_n))$, so $k(\zeta_n)$ is the maximal abelian subextension of $K$.   The proof begins with the case $n=p$ where $p$ is a prime, which I follow. However, following that case Lang writes (bold is what I don't understand):  A direct computation of commutator of elements in $G(n)$ for arbitrary n shows that the commutator subgroup $C$ is contained in the group of matrices $ \begin{pmatrix} \ \ 1 &amp; 0\\  b &amp; 1 \end{pmatrix}$, $b \in \mathbf{Z}/n\mathbf{Z}$ and so must be that subgroup because its factor group is isomorphic to $(\mathbf{Z}/n\mathbf{Z})^{*}$ under the projection on the diagonal.  When $n=p$ is prime I already happened to know that $G \cong \mathbf{Z}/p\mathbf{Z} \rtimes_\varphi (\mathbf{Z}/p\mathbf{Z})^{*}$ from which it is clear (I think?) that the quotient by the image of $\mathbf{Z}/p\mathbf{Z}$ is the maximal abelian quotient, and the fact that the commutator subgroup is nontrivial inside a subgroup of order $p$ means it must be the whole group. However, when $n$ is arbitrary it's not obvious to me why the quotient by the commutator $C$ in $G(n)$ is ismorphic to $(\mathbf{Z}/n\mathbf{Z})^{*}$ nor why $C$ has to be isomorphic to $\mathbf{Z}/n\mathbf{Z}$. If someone could explain what is missing here that would be much appreciated.  I also have a couple of questions on the rest of the argument which I will just link here     1.) On the $3^{rd}-4^{th}$ lines: $\beta$ is a root of $X^m-a$ and by induction we can apply the theorem to $g=X^m-a$.   OK, fine, but what is Lang using the induction for? The splitting field for $g$ is $k(\beta,\zeta_m)$ and has Galois group isomorphic to $G(m)$ and it's maximal abelian extension is $k(\zeta_m)$. I've been staring at this and I don't see what the conclusion is.  2.) lines $3-5$ after the diagram: apply the $1^{st}$ part of proof (case of $n=p$ is prime) to $X^p-\beta$ over $k(\beta)$...shows that $k(\beta,\zeta_n)\cap k(\alpha)=k(\beta)$.  Again I don't know what exactly is being said. The splitting field of $X^p-\beta$ over $k(\beta)$ is $k(\alpha,\zeta_p)$, and it's maximal abelian subextension is $k(\beta,\zeta_p)$. How does Lang's conclusion follow from that?   Thanks to everyone who took the time to read my question, any help is much appreciated.
In Section 9 of Chapter 6, Lang constructrs for the first time a non-abelian extension as a splitting field $K$ of $X^n - a$ when the ground field $k$ does not contain primitive $n$-th roots of unity $\zeta_n$. He constructs an injective homomorphism of the Galois group $G(K/k)$ into the matrices of type   $$ M=\begin{pmatrix} 1 &amp; 0 \\ b &amp; d \end{pmatrix} \quad \hbox{with}\ b\in \mathbb{Z}/n\mathbb{Z} \quad \hbox{and} \quad d\in (\mathbb{Z}/n\mathbb{Z})^*,$$  where $(\mathbb{Z}/n\mathbb{Z})^*$ the multiplicative group of integers modulo n.  Afterwards, he states in Theorem 9.4 that if $n$ is an odd positive integer prime to the characterstic and $[k(\zeta_n):k]=\varphi(n)$ (i.e, the primitive roots of unity are not in the ground field), then the above homomorphism is, in fact, an isomorphism. Besides, the commutator group is $G(K/k(\zeta_n))$, so $k(\zeta_n)$ is the maximal abelian subextension of $K$.  Lang proves the theorem inductively over the number of prime factors in the decomposition of $n$ (i.e. if $n=9= 3\cdot 3$, then the proof "would take two steps" to complete). The induction start with $n$ prime is written clearly and I agree with it.   Now come the questions:   I do not understand Lang's argument to prove that the commutator subgroup is the subgroup $\begin{pmatrix}1 &amp; 0 \\ b &amp; 1\end{pmatrix}$, $b\in \mathbb{Z}/n\mathbb{Z}$. What is this projection on the diagonal and how does one see that the factor group of the commutator group is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$?   By the way, direct computation of $x\, y\, x^{-1}\, y^{-1}$ for two arbitrary matrices $x=\begin{pmatrix} 1 &amp; 0 \\ b &amp; d \end{pmatrix}$, $y=\begin{pmatrix} 1 &amp; 0 \\ c &amp; e \end{pmatrix}$ yields a matrix $\begin{pmatrix} 1 &amp; 0 \\ b(1-e)+c(d-1) &amp; 1 \end{pmatrix}$. As $n$ is odd, $2 \in (\mathbb{Z}/n\mathbb{Z})^*$, so taking $e=1$ and $d=2$, we indeed reconstruct the above-mentioned group.   The second (inductive) part of the proof is a complete mystery to me with virtually every sentence being a Theorem which needs an extra proof. Can someone give me some hints?   2.a. Lang writes $n=p\,m$ and takes $\alpha$ to be the root of $X^n-a$. It is clear that $\beta=\alpha^p$ is a root of $X^m -a$. Now comes the argument "By induction we can apply the theorem to $X^m-a$". I agree that we can apply it here but the result is that $K=k(\beta,\zeta_m)$ is the splitting field of $X^m-a$. In the diagram below, there appears $k(\beta,\zeta_n)$ out of nowhere.   2.b. Now Lang proceeds to apply the theorem on $X^p-\beta$ over $k(\beta)$. Bergman's "Companion" reminds to check that $[k(\beta,\zeta_p):k(\beta)]=\varphi(p)$ before applying the theorem and gives a hint "$[k(\zeta_p):k]$ and $[k(\beta):k]$ are relatively prime". I disagree with this hint. Take $n=21= m p$ where $m=3$ and $p=7$. Then $[k(\zeta_p):k]=\varphi(p)=p-1=6$ but $[k(\beta):k]=m=3$ and both are not co-prime. Can we apply the theorem here at all? Or if so what is the argument? If Bergman's argument is valid, where is the fault in my logic?  2.c. Even assuming that we can apply the theorem in 2.b., then we can say something about $k(\alpha, \zeta_p)$ and not $k(\alpha,\zeta_n)$. How do we proceed to checking everything Lang states in the last section of the proof (basically, everything after the diagram)?  EDIT: I have found the essentially same question
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