Component commutes with or is contained in subnormal subgroup
The proof is from Aschbacher's book titled Finite group theory.
- is contained in (equivalently, is a component of )
- 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 .
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
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 .