Wielandt's first maximizer lemma

Statement

Let $G$ be a finite group and let $H$ be a subgroup of $G$ such that $H$ is not subnormal in $G$ but $H$ is subnormal in every proper subgroup of $G$ containing it. Then:

1. $H$ is contained in a unique maximal subgroup $M$ of $G$, termed its Wielandt maximizer.
2. The conjugate of $H$ by $g\in G$ is contained in $M$ iff $g\in M$.

References

Textbook references

• Subnormal subgroups of groups by John C. Lennox and Stewart E. Stonehewer, Oxford Mathematical Monographs, ISBN 019853552X, Page 222, Lemma 7.3.1, ^{More info}