# Component commutes with or is contained in subnormal subgroup

## Contents

## Source

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

## Statement

### Symbolic statement

Let be a finite group, a subnormal subgroup of and a component of . Then, one of the following is true:

- is contained in (equivalently, is a component of )
- 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 .

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

- is
*not*a component of - Every element of does
*not*commute with every element of

We now try to derive a contradiction.

Note that both the conditions describe a *local* relation between and .

### Steps of the proof

- Let denote the normal closure of and denote the normal closure of . Then, is a proper subgroup of

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

- Consider the group ∩ as a subgroup of . Then, either is a component of ∩ or ∩ is trivial.

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

- Suppose we take the first case, viz is a component of ∩ . Then is also a component of and this leads to a contradiction

- Suppose we take the second case, viz commutes element-wise with ∩ . Then using the three subgroup lemma, we obtain .