The definition of solvable groups can be regarded as two constraints, one is that there must be a sequence of normal subgroups, and the other is that the quotient groups between these sequences are Abelian groups. The first one is well understood, but I don't quite understand the second constraint. To be specific, why must the commutative law hold? I know that the root formula of the quintic equation can be expressed in the Bring radicals, but A5 does not satisfy the commutative law (non-Abelian group). So what is the special property that binds the algebraic radicals and commutative law of quotient groups? And what makes algebraic radicals different from Bring radicals?
$\begingroup$
$\endgroup$
3
-
6$\begingroup$ What do you mean by “the first one is well understood”? Did you try to read a proof in a Galois theory book that solvable equations (in characteristic 0) have solvable Galois groups? Or do you know why the splitting field of $x^n-a$ (in characteristic 0) has a solvable Galois group? $\endgroup$– KConradCommented Aug 7, 2022 at 7:39
-
5$\begingroup$ "one is that there must be a sequence of normal subgroups": no , this is not a constraint. Every group has a normal series. And your second "constraint" "the quotient groups between these sequences are Abelian groups" is senseless by itself, since it makes reference to the normal series of the "first constraint". So you should view the whole definition as a single constraint. $\endgroup$– YCorCommented Aug 7, 2022 at 8:03
-
$\begingroup$ Maybe OP would be satisfied with an explanation of "what makes algebraic radicals different from Bring radicals," but I agree that this website is not the appropriate place for such an explanation. $\endgroup$– Gerry MyersonCommented Aug 8, 2022 at 0:58
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
A solvable group is, by definition, a group with a finite series of normal subgroups such that the successive factor groups are abelian. It is the content of the Fundamental Theorem of Galois Theory that a finite Galois extension $L\supset K$ of fields has solvable Galois group (the group of field automorphisms of $L$ that pointwise fix $K$) if and only if the elements of $L$ can be expressed in terms of the elements of $K$ using only the extraction of $n$th roots for each $n$ together with the standard field operations of addition, subtraction, multiplication, and division.
Regarding your comment about $A_5$, note that not only is $A_5$ non-abelian but it is not solvable. By contrast, the alternating group $A_4$, which is also non-abelian, is solvable.
