Component commutes with or is contained in subnormal subgroup

Source

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

Statement

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

Proof

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 $X$ $Y$ as a subgroup of $G$. Then, either $L$ is a component of $X$ $Y$ or $[L, X$ $Y]$ is trivial.

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

• Suppose we take the first case, viz $L$ is a component of $X$ $Y$. 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 $X$ $Y$. Then using the three subgroup lemma, we obtain $[Y, L] = 1$.