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SATURATED FUSION SYSTEMS OF ESSENTIAL RANK 1

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If G is dihedral, then D = 〈x 2 , x l t〉 for some arbitrary integer l. If l is even then<br />

D = 〈x 2 , t〉, and if l is odd then D = 〈x 2 , xt〉. Since 〈x 2 , t〉 �= 〈x 2 , xt〉, this shows (e).<br />

Let Y be a subgroup of 〈x〉 of order 4. Using (b), an elementary calculation shows<br />

that every element of order 4, which is inverted by an involution from G, is contained in<br />

Y . So every subgroup of G isomorphic to D8 is of the form 〈s〉⋉Y for some involution<br />

s ∈ G\〈x〉. Note that Y is characteristic in G. Hence, for the proof of (f), it is suffi-<br />

cient to show that Aut(G) acts transitively on the set of involutions in G\Z(G). Let<br />

s ∈ G\Z(G) be an involution. Note that C〈x〉(s) = Z(G) and thus CG(s) = Z(G)〈s〉.<br />

So |x G | = |G|/4. If G is semidihedral then it follows from (b) that G acts transitively<br />

on the set of involutions in G\Z(G). This shows (f) in the semidihedral case. If G is<br />

dihedral then G acts transitively on the set of involutions in D\Z(G) for any dihedral<br />

subgroup D of G of index 2. Since G embeds into a dihedral group of order 2|G|, this<br />

implies (f).<br />

Now let D1 ≤ G be isomorphic to D8 and A = Aut(D1). Note that by (d), A is<br />

a 2-group. By (e), D1 has exactly two subgroups Q, R that are fours groups. Then A<br />

acts on {Q, R} and thus |A/NA(Q)| = 2. Moreover, Aut(Q) ∼ = Aut(R) ∼ = S3. Thus,<br />

|NA(Q)/CA(Q)| ≤ 2 and |CA(R)| = |CA(R)/CA(R)∩CA(Q)| ≤ |NA(Q)/CA(Q)| ≤<br />

2, since CA(R) ∩ CA(Q) = CA(D1) = 1. This shows |A| ≤ 8. Now note that<br />

NG(D1)/CG(D1) embeds into A. Moreover, if D1 �= G, NG(D1) is dihedral or semidi-<br />

hedral of order 16 and CG(D1) = Z(NG(D1)) has order 2. This implies (h).<br />

Lemma 2.4.4. Let G be a 2-group and D0 ≤ G be dihedral of order 2 n , n ≥ 2.<br />

Assume D1 := NG(D0) is dihedral of order 2 n+1 and normal in G. Then G is dihedral<br />

or semidihedral and |G/D1| ≤ 2.<br />

Proof. The group G acts on the set of dihedral subgroups of D1. Therefore, |G/D1| ≤<br />

2. We may assume that |G/D1| = 2. Then there exists an element y ∈ G\D1. Note<br />

that y 2 ∈ D1.<br />

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