I have trouble understanding Serge Lang's proof that a solvable extension is solvable by radicals. More precisely, I cannot figure out why did he pick such a number.
Proof. Assume that $E/k$ is solvable, and let $K$ be a finite solvable Galois extension of $k$ containing $E$. Let $m$ be the product of all primes unequal to the characteristic dividing the degree $[K:k]$, and let $F=k(\zeta)$ where $\zeta$ is a primitive $m$-th root of unity. Then $F/k$ is abelian. We lift $K$ over $F$. Then $KF$ is solvable over $F$. There is a tower of subfields between $F$ and $KF$ such that each step is cyclic of prime order, because every solvable group admits a tower of subgroups of the same type, and we can use theorem 1.10. By theorem 6.2 and theorem 6.4, we conclude that $KF$ is solvable by radicals over $F$, and hence is solvable by radicals over $k$. This proves that $E/k$ is solvable by radicals.
I understand that $F/k$ is solvable by radical by definition. However, why did he even define $F$, or more precisely, $m$ in such a manner? Avoiding the characteristic of $k$ makes sense, and taking $m$ to be a factor of $[K:k]$ also makes sense, but why does it have to be the radical of $[K:k]$ (divided by $p$ when $p|[K:k]$)? For example if $[K:k]=24$ and the charactersitic is $0$, why would we pick $m=2 \times 3$ instead of $4$ or $12$ or $8$? Does it have anything to do with $KF/F$?
Here are theorems mentioned in Lang's proof.
Theorem 1.10. Let $K$ be a Galois extension of $k$ with group $G$. Let $F$ be a subfield, $k \subset F \subset K$, and let $H=G(K/F)$. Then $F$ is normal over $k$ if and only if $H$ is normal in $G$. If $F$ is normal over $k$, then the restriction map $\sigma \mapsto \sigma|_F$ is a homomorphism of $G$ onto the Galois group of $F$ over $k$, whose kernel is $H$. We thus have $G(F/k) \cong G/H$.
Theorem 6.2. Let $k$ be a field, $n$ an integer $>0$ prime to the characteristic of $k$, and assume that there is a primitive $n$-th root of unity in $k$.
(i) Let $K$ be a cyclic extension of degree $n$. Then there exists $\alpha \in K$ such that $K=k(\alpha)$ and $\alpha$ satisfies an equation $X^n-a=0$ for some $a \in k$.
(ii) Conversely, let $a \in k$. Let $\alpha$ be a root of $X^n-a$. Then $k(\alpha)$ is cyclic over $k$, of degree $d$, $d|n$, and $\alpha^d$ is an element of $k$.
Theorem 6.4. Let $k$ be a field of characteristic $p$.
(i) Let $K$ be a cyclic extension of $k$ of degree $p$. Then there exists $\alpha \in K$ such that $K=k(\alpha)$ and $\alpha$ satisfies an equation $X^p-X-a=0$ with some $a \in k$.
(ii) Conersely, given $a \in k$, the polynomial $f(X)=X^p-X-a$ either has one root in $k$, in which case all its roots are in $k$< or it is irreducible. In this latter case, if $\alpha$ is a root then $k(\alpha)$ is cyclic of degree $p$ over $k$.