Editing Solvable group (section)

==Definition==
A group ''G'' is called '''solvable''' if it has a [[subnormal series]] whose [[factor group]]s (quotient groups) are all [[Abelian group|abelian]], that is, if there are [[subgroup]]s
:<math>1 = G_0\triangleleft G_1 \triangleleft \cdots \triangleleft G_k=G</math>
meaning that ''G''<sub>''j''−1</sub> is [[normal subgroup|normal]] in ''G<sub>j</sub>'', such that ''G<sub>j</sub>&thinsp;''/''G''<sub>''j''−1</sub> is an abelian group, for ''j'' = 1, 2, ..., ''k''.

Or equivalently, if its [[derived series]], the descending normal series
:<math>G\triangleright G^{(1)}\triangleright G^{(2)} \triangleright \cdots,</math>
where every subgroup is the [[commutator subgroup]] of the previous one, eventually reaches the trivial subgroup of ''G''. These two definitions are equivalent, since for every group ''H'' and every [[normal subgroup]] ''N'' of ''H'', the quotient ''H''/''N'' is abelian [[if and only if]] ''N'' includes the commutator subgroup of ''H''. The least ''n'' such that ''G''<sup>(''n'')</sup> = 1 is called the '''derived length''' of the solvable group ''G''.

For finite groups, an equivalent definition is that a solvable group is a group with a [[composition series]] all of whose factors are [[cyclic group]]s of [[prime number|prime]] [[order (group theory)|order]]. This is equivalent because a finite group has finite composition length, and every [[simple group|simple]] abelian group is cyclic of prime order. The [[Jordan–Hölder theorem]] guarantees that if one composition series has this property, then all composition series will have this property as well. For the Galois group of a polynomial, these cyclic groups correspond to ''n''th roots (radicals) over some [[Field (mathematics)|field]]. The equivalence does not necessarily hold for infinite groups: for example, since every nontrivial subgroup of the group '''Z''' of [[integer]]s under addition is [[group isomorphism|isomorphic]] to '''Z''' itself, it has no composition series, but the normal series {0, '''Z'''}, with its only factor group isomorphic to '''Z''', proves that it is in fact solvable.