Component commutes with or is contained in subnormal subgroup

Source
The proof is from Aschbacher's book titled Finite group theory.

Symbolic statement
Let $$G$$ be a finite group, $$H$$ a subnormal subgroup of $$G$$ and $$L$$ a component of $$G$$. Then, one of the following is true:


 * $$L$$ is contained in $$H$$ (equivalently, $$L$$ is a component of $$H$$)
 * $$[L,H]$$ is the trivial group

Setup of the proof
We prove this statement by induction on the order of the groups involved. Induction is on the order of $$G$$.

Let $$G$$ be a counterexample of minimal order. Let $$L$$ and $$H$$ be subgroups of $$G$$ which do not satisfy either of the conditions. In other words:


 * $$L$$ is not a component of $$H$$
 * Every element of $$L$$ does not commute with every element of $$H$$

We now try to derive a contradiction.

Note that both the conditions describe a local relation between $$L$$ and $$H$$.

Steps of the proof

 * Let $$X$$ denote the normal closure of $$L$$ and $$Y$$ denote the normal closure of $$H$$. Then, $$Y$$ is a proper subgroup of $$G$$

Proof: It suffices to show that $$H$$ is a proper subgroup of $$G$$, because the normal closure of any proper subnormal subgroup must be proper. The properness of $$H$$ follows from the fact that $$L$$ is not contained in $$H$$.


 * Consider the group $$X$$ &cap; $$Y$$ as a subgroup of $$G$$. Then, either $$L$$ is a component of $$X$$ &cap; $$Y$$ or $$[L, X$$ &cap; $$Y]$$ is trivial.

Proof: Since $$Y$$ is a proper subgroup of $$G$$, so is $$X$$ &cap; $$Y$$. Hence, by the minimality of $$G$$ as a counterexample, we conclude that within $$X$$ &cap; $$Y$$, the result must hold, and that's precisely what we have written.


 * Suppose we take the first case, viz $$L$$ is a component of $$X$$ &cap; $$Y$$. Then $$L$$ is also a component of $$Y$$ and this leads to a contradiction


 * Suppose we take the second case, viz $$L$$ commutes element-wise with $$X$$ &cap; $$Y$$. Then using the three subgroup lemma, we obtain $$[Y, L] = 1$$.