Core-free permutable subnormal implies solvable of length at most one less than subnormal depth
From Groupprops
Statement
Suppose is a (not necessarily finite) group and is a Core-free permutable subnormal subgroup (?) of . In other words, is core-free in (its normal core in is trivial), is a Permutable subgroup (?) of , and is a Subnormal subgroup (?) of .
Then, is a Solvable group (?). Further, if the Subnormal depth (?) of in is , the Solvable length (?) of is at most .
Facts used
- Modular property of groups
- Cyclicity is subgroup-closed
- Commutator of the whole group and a subgroup implies normal
- Permutability is strongly join-closed
- Cocentral implies abelian-quotient
- Characteristic of normal implies normal
- Permutability satisfies image condition
- Subnormality satisfies image condition
Proof
We induct on .
Given: A group , a core-free permutable -subnormal subgroup .
To prove: is solvable of solvable length at most .
Proof: Consider a counterexample, i.e., a situation where is not trivial. Then, there exists such that is not contained in .
- It suffices to consider the case where : Suppose is a counterexample group with a counterexample subgroup . This means that there exists a conjugateThen, let be the normal core of in the subgroup . Then, is also a counterexample.
- Let be the term in the unique fastest descending subnormal series for (i.e., the normal closure of ). Then, there exists such that : We have by fact (1). Since is a subgroup of a cyclic group , it is cyclic on some element (Fact (2)).
- The commutator of and is contained in : We have . The last step follows from being normal in .
- The commutator of and is normal in : This follows from fact (3).
- is trivial, so is in the center of : This follows from the previous two steps, and the fact that is core-free.
- is a permutable subgroup of : Since is central, it is normal and hence permutable. By fact (4), is also permutable.
- If is the normal core of in , then is abelian: Again using fact (1), . Since is central, is cocentral in , and hence by fact (5), is abelian, so . On the other hand, is characteristic in which is normal in , so is normal in . Again using that is core-free, we obtain that must be trivial, so is abelian.
- The term of the derived series of is in : is a core-free -subnormal subgroup of (also fact (7)). Further, by step (6), is permutable in , so fact (6) yields that is permutable in . Thus, by induction on , we have that has solvable length at most , which is reinterpreted as saying that the term of the derived series of is in .
- The term of the derived series of is in : This is immediate from the previous step and the fact that .
- has solvable length at most : This follows from steps (7) and (9). Step (9) guarantees that the derived series reaches inside in steps, and step (7) yields that is abelian, so the next step lands at the identity.
References
Journal references
- Subnormal, core-free, quasinormal subgroups are solvable by Fletcher Gross, , Volume 7, Page 93 - 95(Year 1975): ^{{{{weblinkdescriptor}}}}^{More info}
===Textbook references---
- Subnormal subgroups of groups by John C. Lennox and Stewart E. Stonehewer, Oxford Mathematical Monographs, ISBN 019853552X, Page 214, Theorem 7.1.5, ^{More info}