30.01.2013 Views

SATURATED FUSION SYSTEMS OF ESSENTIAL RANK 1

SATURATED FUSION SYSTEMS OF ESSENTIAL RANK 1

SATURATED FUSION SYSTEMS OF ESSENTIAL RANK 1

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

Lemma 2.2.3. For subgroups X, Y of G, [X, Y ] is a normal subgroup of 〈X, Y 〉.<br />

Proof. See for example [15, 1.5.5].<br />

Lemma 2.2.4 (Three-Subgroup Lemma). Let X, Y, Z ≤ G and let N be a normal<br />

subgroups of G. Assume [X, Y, Z] ≤ N and [Y, Z, X] ≤ N. Then [Z, X, Y ] ≤ N.<br />

Proof. This follows from [15, 1.5.6].<br />

If a group A acts on the group G, then we can consider at A and G as subgroups of<br />

the semidirect product A ⋉ G. Thus, the definitions of commutators and commutator<br />

subgroups can be inherited and the results hold accordingly.<br />

In particular, the Three-Subgroup Lemma holds, i.e. if each of X, Y, Z is a sub-<br />

group of either G or A then<br />

2.3 Coprime action<br />

[X, Y, Z] = [Y, Z, X] = 1 =⇒ [Z, X, Y ] = 1.<br />

In this section A will always be a group which acts on G. We call the action of A on<br />

G coprime if<br />

(1) |A| and |G| are coprime, and<br />

(2) A or G is soluble.<br />

We remind the reader here that a finite group G is called soluble if [U, U] < U for<br />

all subgroups U of G. Thus abelian groups and nilpotent groups provide examples of<br />

soluble groups. In particular, p-groups are soluble.<br />

By a Theorem of Feit-Thompson, every finite group of odd order is soluble. Hence<br />

if |K| and |G| are coprime then either K or G is soluble. Therefore assumption (2)<br />

is in fact redundant. However we prefer to put it this way since the Theorem of Feit-<br />

Thompson is a deep result of finite group theory and in all the situations we are going<br />

19

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!