# 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 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 $k - 1$.

## 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