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

Statement
Suppose $$G$$ is a (not necessarily finite) group and $$H$$ is a fact about::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 fact about::permutable subgroup of $$G$$, and $$H$$ is a fact about::subnormal subgroup of $$G$$.

Then, $$H$$ is a fact about::solvable group. Further, if the fact about::subnormal depth of $$H$$ in $$G$$ is $$k$$, the fact about::solvable length of $$H$$ is at most $$k - 1$$.

Facts used

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

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