Component commutes with or is contained in subnormal subgroup

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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

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

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

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