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 $$g \in M$$.