# On the equivalent definitions for solvable groups

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## Main Question or Discussion Point

We have the 3 equivalent definition for solvable groups:

There exists a chain of subgroups

1 < G1 ....< Gi + < G i+1 < Gr = G

such that Gi is normal in Gi+1 and Gi+1/Gi is abelian.

Another definition is

there exists

1 < H1 ....< Hi + < H i+1 < Hs = H

such that Hi is normal in Hi+1, and Hi+1/Hi is cyclic

, the last definition is similar but the quotient group is isomorphic to Z/p.

so my question is, does s need to be equal r?

thanks

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andrewkirk
Homework Helper
Gold Member
There is no reason why they should be the same. The definition involving r (which, according to Wikipedia, is only equivalent for finite groups) requires that each ##H_i## be a maximal normal subgroup of ##H_{i+1}##. The steps in the H sequence are smaller than those in the G sequence, since the factor groups (quotients of successive elements) are cyclic, which is 'smaller' than Abelian, so one would expect the H sequences to generally be longer than the G sequences. That is, generally r would be longer than s. It may even be the case that s never exceeds r, but I have no proof of that.

fresh_42
Mentor
That is, generally r would be longer than s. It may even be the case that s never exceeds r, but I have no proof of that.
This cannot be proven. E.g. if ##\mathbb{Z}_6## is a factor, then it's both Abelian and cyclic, but there is still a refinement possible. The maximal length will be reached, if all factors are simple.

There exists a chain of subgroups

1 < G1 ....< Gi + < G i+1 < Gr = G

such that Gi is normal in Gi+1 and Gi+1/Gi is abelian.
This is the correct definition.

There are several theorems which deal with normal sequences and composition sequences. The latter is a normal sequence whose factors are simple, such that no refinement is possible.

Two normal sequences have isomorphic refinements. (O. Schreier)
Is there a composition sequence, then each normal sequence can be refined to a composition sequence and two composition series are isomorphic. (Jordan, Hölder)

Two sequences are isomorphic, if the sets of factors are the same.

mathwonk
Homework Helper
as a consequence of the results stated by fresh_42, one knows that a finite group has associated to it a unique collection of "simple constituent groups", namely the simple factors of any composition series. Then one way to say a group is solvable is to say that its simple constituents are all cyclic of prime order. Since this is true of any abelian group, the two definitions you gave are equivalent.

Just which characterization you wish to use, may depend on what goal you have in mind, as usual. For example if you want to prove that the Galois group of a solvable polynomial is a solvable group, you want to be able just to show as a key step say, that the Galois group of X^n-1 is abelian. But if you want to show conversely that a polynomial (over the rationals say), is solvable if it has a solvable Galois group, then it is useful to begin from a decomposition of its splitting field into a series of field extensions of prime degree, obtained via the fundamental theorem of Galois theory from a composition series for the group.

These two results are explained in detail in my free class notes:

http://alpha.math.uga.edu/~roy/843-2.pdf

http://alpha.math.uga.edu/~roy/844-2.pdf

• lavinia and jim mcnamara