The solvability group criteria in Cox's Galois Theory It's a long time I'm studying group theory in many level of depth, heading to Co.0 and Monster groups construction. Time by time, I come in contact with Galois Theory, where the solvability strictly connects to simple groups, of course.
Actually, reading Cox's text Galois Theory, I came across a different definition than the one I ever seen, which sounds like this:
given a chain/sequence of normal subgroups from the full group G up to {e}, where Gn relates to {e}, a group is solvable if:


*

*descending in the sequence, each subgroup is normal in the following one;

*the index of this normal subgroup in the following one is prime.


Actually, I'm puzzled by point 2): I've always seen this point as "the quotient of neighbour subgroups is abelian".
I mean, I know every finitely generated abelian group is a direct product of cyclic groups, but I cannot see how this relates to the index to be a prime.
More than this: I've read many books on this and splitting fields and cascade and correspondence of extensions and subgroups make sense to me, but I always loose the main point, when coming to the meat:
how this abelian criteria is guaranteeing to have roots with radicals?
Every book depicts the usual example of A5, but I'm lost every time to get an intuitive view of this abelian --> radical correspondence.
Thanks as always for your precious help and support.
 A: Essentially, both definitions are equivalent (at least, for finite nontrivial groups)- clearly, if $N$ is a normal subgroup of $M$ such that $|M/N|$ is prime, then $M/N$ will be cyclic so that $M/N$ is abelian.
For the other direction:

First, we shall indulge an examination of maximal normal subgroups. A maximal normal subgroup (MNS) is defined here as some subgroup, $N$ of $G$, satisfying $N\trianglelefteq G$, $N\neq G$, and for any $M$ such that $N\trianglelefteq M\trianglelefteq G$, either $M=N$ or $M=G$.
A conclusion that directly follows from this definition is that, in a finite group $G$ any normal subgroup $\neq G$, $E$, is contained in some MNS. If this were not the case, we could find at least one $F_1$ such that $E\trianglelefteq F_1\trianglelefteq G$ with $E\neq F_1$ and $F_1\neq G$, as otherwise $E$ itself would be an MNS containing $E$. Now it must not be the case that $F_1$ is contained in an MNS as otherwise $E$ would be too, so we must have $F_1\trianglelefteq F_2\trianglelefteq G$ with $F_1\neq F_2$ and $F_2\neq G$. We can carry on concluding the existence of infinitely many normal subgroups $F_k$, each strictly between $F_{k-1}$ and $G$. But $G$ is finite and since each $F_k\supsetneq F_{k-1}$ we have for each $k$, $|F_k|>|F_{k-1}|$ which would mean that for some $r\leq |G|-|E|$ we will have $F_r=G$. This contradiction leads us to conclude that each normal subgroup of some finite $G$ is contained in an MNS.
The next step is to note that if $f:G\rightarrow H$ is a homomorphism and $J\trianglelefteq H$, then $f^{-1}(J)\trianglelefteq G$ (this is fairly straightforward). If $N$ is an MNS of $G$ and $G$ is nontrivial, consider the canonical homomorphism $f:G\rightarrow G/N$- if $M$ is any proper normal subgroup of $G/N$, $f^{-1}(M)$ will be a proper normal subgroup of $G$ such that $f^{-1}(M)\supsetneq N$ which is impossible. So if $N$ is an MNS of $G$, $G/N$ is simple. In particular, if $G/N$ is finite and abelian $G/N$ has no proper subgroups and must be of prime order.
Finally, let us turn our attention towards solvable groups. Let $G$ be a finite nontrivial solvable group- it has a solvable series with at least two members (namely, $\{e\}$ and $G$ itself). Let this series be $\{e\}=H_0\subset H_1\subset H_2\subset...\subset H_{m-1}\subset H_m=G$ where for each $i$, $H_i\neq H_{i+1}$. Consider for some arbitrary $0\leq i\leq m-1$, $H_i$ and $H_{i+1}$.
Since $H_i\trianglelefteq H_{i+1}$ there is an MNS, $K_{i,1}$ of $H_i$ in $H_{i+1}$. Since $(H_{i+1}/H_i)/(K_{i,1}/H_i)\cong H_{i+1}/K_{i,1}$ we have that  $H_{i+1}/K_{i,1}$ is abelian- since $K_{i,1}$ is an MNS of $H_{i+1}$, $H_{i+1}/K_{i,1}$ is a group of prime order. And it is clear that $K_{i,1}/H_i$ is abelian as it is a subgroup of $H_{i+1}/H_i$. We then repeat this process generating $K_{i,j+1}$ as an MNS in $K_{i,j}$  containing $H_i$. This process terminates eventually as $|H_i|\leq|K_{i,j+1}|<|K_{i,j}|$ for each $j$. We repeat this procedure till termination for each $i$ and so we have generated a "prime series" of $G$.

Edit to answer comment (because it was too long to be a comment): Well, suppose $K$ is a field extension of $F$ (a field of characteristic $0$) such that $Gal(K:F)$ is solvable. We know, then (by the above answer) that $Gal(K:F)$ has a sequence of subgroups, each a normal subgroup of the following term such that the order of each quotient is a prime. Suppose $H_i$ and $H_{i+1}$ are two adjacent subgroups in the "prime series". By the fundamental theorem of Galois theory, there exist fields between $F$ and $K$, $F_i$ and $F_{i+1}$ that correspond to $H_i$ and $H_{i+1}$ such that one is a field extension of prime degree over the other. The condition that $H_i$ be a normal subgroup of $H_{i+1}$ implies that $F_i$ is a splitting field over $F_{i+1}$. Those are all the facts one can extract rather immediately from the setup of the problem.
It turns out that under the right conditions on $F$, (i.e., $F$ having the 'right' roots of unity) any pair of extensions of $F$, $B$ and $C$, such that $[B:C]$ is prime and $B$ is a splitting field over $C$ immediately gives us that $B$ is a radical extension over $C$- this is the core of the proof. Now (again, by the fundamental theorem of Galois theory) each normal subgroup of $Gal(K:F)$ in its "prime series" corresponds to a splitting field between $F$ and $K$- in particular, $Gal(K:F)$ corresponds to $F$, the "smallest field" and the trivial group corresponds to $K$, the "largest field". Since each field is a radical/ root extension over a "smaller" field in the series of intermediate fields and $F$ corresponds to the smallest field and $K$ to the largest, $K$ is a radical extension over $F$ (as if $A$ is a radical extension over $B$ and $B$ is a radical extension over $C$, $A$ is a radical extension over $C$ as well).
So the core idea is to fill in the space between $K$ and $F$ with intermediate fields, $L_0,L_1,...,L_n$, such that $L_0=F$, $L_n=K$, and to use the conditions on $Gal(K:F)$ to show that each $L_{i+1}$ is a radical extension of $L_i$- this is enough to prove $K$ is a radical extension over $F$
(this is just the basic idea- I have obviously skimmed over many of the details and intricacies of the proof)
