Core-free permutable subnormal implies solvable of length at most one less than subnormal depth

From Groupprops
Jump to: navigation, search

Statement

Suppose G is a (not necessarily finite) group and H is a Core-free permutable subnormal subgroup (?) of G. In other words, H is core-free in G (its normal core in G is trivial), H is a Permutable subgroup (?) of G, and H is a Subnormal subgroup (?) of G.

Then, H is a Solvable group (?). Further, if the Subnormal depth (?) of H in G is k, the Solvable length (?) of H is at most k - 1.

Facts used

  1. Modular property of groups
  2. Cyclicity is subgroup-closed
  3. Commutator of the whole group and a subgroup implies normal
  4. Permutability is strongly join-closed
  5. Cocentral implies abelian-quotient
  6. Characteristic of normal implies normal
  7. Permutability satisfies image condition
  8. Subnormality satisfies image condition

Proof

We induct on k.

Given: A group G, a core-free permutable k-subnormal subgroup H.

To prove: H is solvable of solvable length at most k - 1.

Proof: Consider a counterexample, i.e., a situation where H^{(k-1)} is not trivial. Then, there exists x in G such that H^{(k-1)} is not contained in H^x.

  1. It suffices to consider the case where G = H \langle x \rangle: Suppose G is a counterexample group with a counterexample subgroup H. This means that there exists a conjugateThen, let M be the normal core of H in the subgroup H \langle x \rangle. Then, H \langle x \rangle/M is also a counterexample.
  2. Let L be the (k-1)^{th} term in the unique fastest descending subnormal series for H (i.e., the (k-1)^{th} normal closure of H). Then, there exists y \in \langle x \rangle such that L = H \langle y \rangle: We have H(\langle x \rangle \cap L) = H \langle x \rangle \cap L = G \cap L = L by fact (1). Since \langle x \rangle \cap L is a subgroup of a cyclic group \langle x \rangle, it is cyclic on some element y (Fact (2)).
  3. The commutator of G and \langle y \rangle is contained in H: We have [G, \langle y \rangle] = [H \langle x \rangle, \langle y \rangle] = [H, \langle y \rangle] \le [H,L] \le H. The last step follows from H being normal in L.
  4. The commutator of G and \langle y \rangle is normal in G: This follows from fact (3).
  5. [G,\langle y \rangle] is trivial, so y is in the center of G: This follows from the previous two steps, and the fact that H is core-free.
  6. L is a permutable subgroup of G: Since \langle y \rangle is central, it is normal and hence permutable. By fact (4), L = H \langle y \rangle is also permutable.
  7. If K is the normal core of L in G, then K is abelian: Again using fact (1), K = \langle y \rangle(H \cap K). Since y is central, H \cap K is cocentral in K, and hence by fact (5), K/(H \cap K) is abelian, so [K,K] \le H \cap K. On the other hand, [K,K] is characteristic in K which is normal in G, so [K,K] is normal in G. Again using that H is core-free, we obtain that [K,K] must be trivial, so K is abelian.
  8. The (k-2)^{th} term of the derived series of L is in K: L/K is a core-free k-1-subnormal subgroup of G (also fact (7)). Further, by step (6), L is permutable in G, so fact (6) yields that L/K is permutable in G/K. Thus, by induction on k, we have that L/K has solvable length at most k-2, which is reinterpreted as saying that the (k-2)^{th} term of the derived series of L is in K.
  9. The (k-2)^{th} term of the derived series of H is in K: This is immediate from the previous step and the fact that H \le K.
  10. H has solvable length at most k-1: This follows from steps (7) and (9). Step (9) guarantees that the derived series reaches inside K in k-2 steps, and step (7) yields that K is abelian, so the next step lands at the identity.

References

Journal references

===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