This section develops two of the main computational tools that will enable the later arguments. We obtain an analog of the Frattini lemma in Corollary4.8, which shows that ifN is a partial normal subgroup of a locality L, then each element ofL may be written as a product of an elementf∈N and an elementg∈NL(T), whereT=S∩N. The
“splitting lemma” (Lemma??) refines the choice off andg. We end with an important application (Lemma4.9) which provides a criterion for extending an automorphism of a linking system in a finite group to an automorphism of the group itself.
The notation Sf and Sw, defined in Proposition 2.11 and Lemma 2.14, will be employed without further comment.
The following hypothesis (and notation) will be assumed throughout this section.
Hypothesis4.1. There is given a localityL=(L,∆, S) and a partial normal subgroup N ofL. SetT=S∩N.
Lemma4.2. The following hold:
(a) T is strongly closed in FS(L)and T is maximal in the poset of all p-subgroups of N;
(b) If P∈∆ and x∈N with P6Sx, then P T=PxT; (c) If T=1, then NN(P, S)=CN(P)for all P∈∆.
Proof. Let x∈T and let φ∈F:=FS(L) such that x lies in the domain of φ. As φ is a composition of restrictions of conjugation maps between objects, it suffices, in proving (a), to consider only the case wherexφ=xf for some f∈L; and in that case we have xφ∈N. Thus xφ∈S∩N=T, and so T is strongly closed in F. Now let R be a
p-subgroup of N containingT. By Proposition 2.22(b) we may choose R with R6S, and thenR=T. Thus (a) holds.
Next, letg∈P∈∆ and letx∈NN(P, S). Then the wordw=(x−1, g, x, g−1) is inDvia Px, and then Π(w)=x−1xg−1=gxg−1. Thus Π(w)∈N ∩S=T, andgx∈gT. In particular, this proves (c), and it shows thatPx6P T. Upon replacing (P, x) with (Px, x−1), the same argument shows thatP6PxT, and this yields (b).
Definition 4.3. Let L∆ be the set of all pairs (f, P)∈L×∆ such that P6Sf. Define a relation " on L∆ by (f, P)"(g, Q) if there exist elements x∈NN(P, Q) and y∈NN(Pf, Qg) such thatxg=f y.
This relation may be indicated by means of a commutative diagram Q g //Qg
P
x
OO
f //Pf
y
OO
(∗)
of conjugation maps, labeled by the conjugating elements, and in which the horizontal arrows are isomorphisms and the vertical arrows are injective homomorphisms. The relation (f, P)"(g, Q) may also be expressed by
w:= (x, g, y−1, f−1)∈DviaP and Π(w) =1.
It is easy to see that " is a reflexive and transitive relation on L∆. We say that (f, P) ismaximal inL∆ if (f, P)"(g, Q) implies that|P|=|Q|. AsSis finite there exist maximal elements inL∆. Since (f, P)"(f, Sf) for (f, P)∈L∆, we haveP=Sffor every maximal element (f, P). For this reason, we introduce the following terminology.
Definition 4.4. Letf∈L. Thenf is"-maximal in Lif (f, Sf) is maximal in L∆.
The following is the first main result of this section.
Proposition 4.5. Let f∈Land suppose that f is "-maximal. Then T6Sf. The proof requires two preliminary lemmas.
Lemma 4.6. Let (g, Q),(h, R)∈L∆ with (g, Q)"(h, R) and suppose that T6R.
Then there exists a unique y∈N with g=yh. Moreover, (a) y∈NN(Q, R)and Q6S(y,h);
(b) if NT(Qg)∈Sylp(NN(Qg)),then NT(Qy)∈Sylp(NN(Qy)).
Proof. By the definition of", there exist elementsu∈NN(Q, R) andv∈NN(Qg, Rh) such that (u, h, v−1, g−1)∈DviaQ, and such that Π(w)=1, as indicated in the diagram
R h //Rh
Q
u
OO
g //Qg
v
OO
In particular,uh=gv. SinceT6R, points (a) and (b) of Lemma4.2yield T=Th, QuT=QT6R, and QgT=QgvT6Rh. Then
w:= (u, h, v−1, h−1)∈D via (Q, Qu, Quh, Quhv−1=Qg, Qgh−1).
Sety=Π(w). Theny=u(v−1)h−1∈NN(Q, R). Since (u, h, v−1, h−1, h) and (g, v, v−1) are inD(asLis a partial group), we getyh=uhv−1=g. This yields (a), and the uniqueness ofy is given by right cancellation.
Suppose now thatNT(Qg)∈Sylp(NN(Qg)). AsNT(Qy)h=NT(Qg), it follows from Lemma2.7(b) thatNT(Qy)∈Sylp(NN(Qy)).
Lemma4.7. Suppose that f is "-maximal and let y∈NN(Sf, S). Then
|T∩Sf|=|T∩(Sf)y| and (f, Sf)"(y−1f,(Sf)y).
In particular, y−1f is "-maximal.
Proof. SetP=Sf. ThenPyT=P T, by Lemma 4.2(b). Thus
|Py:Py∩T|=|PyT:T|=|P T:T|=|P:P∩T|, and so|T∩P|=|T∩Py|. The diagram
Py y
−1f //Pf
P
y
OO
f //Pf
1
OO
shows that (f, P)"(y−1f, Py).
Proof of Proposition4.5. Let f be "-maximal. Set P=Sf and Q=Pf, and sup-pose first thatNT(P)∈Sylp(NN(P)). Then NT(P)f∈Sylp(NN(Q)), by Lemma2.7(b), and there exists x∈NN(Q) such that NT(Q)6(NT(P)f)x. Here (f, x)∈D via P, so (NT(P)f)x=NT(P)f x, and then (f, P)"(f x, NT(P)P). Asf is"-maximal, we conclude thatNT(P)6P, and henceT6P. ThusT6Sf ifNT(P)∈Sylp(NN(P)). Assuming that f provides a counterexample to Proposition4.5, we conclude that
(1) NT(P)∈Syl/ p(NN(P)).
Among all counterexamples to Proposition4.5, choosefso that first|P∩T|and then
|P|are as large as possible. Chooseg∈NL(Q, S) so thatQgis fully normalized inFS(L), and seth=f gandR=Ph. AsR=Qg is fully normalized we haveNS(R)∈Sylp(NL(R)), and thenNT(R)∈Sylp(NN(R)). Let (h−1, R)"(h0, Sh0), whereh0 is "-maximal, and set R0=Sh0 and P0=(R0)h0. Thus, there exist y, z∈N such that yh0=h−1z, Ry6R0, and Pz6P0, as indicated in the diagram
R0 h
0 //P0
R
y
OO
h−1 //P.
z
OO
Then (T∩R)y6T∩R0.
Suppose that TR0. The conditions on the choice of f then yield (T∩R)y=T∩R0 andRy=R0. Since NT(R)∈Sylp(NN(R)), we get NT(R)y∈Sylp(NN(R0)), and so there existsx∈NN(R0) such that (NT(R)y)x=NT(R0). Replacingyandh0 withyxandx−1h0, we then obtain NT(R)y=NT(R0). But then T6R0 by (1), in any case, and then also T6P0.
Evidently (h, P)"((h0)−1, P0), so by Lemma4.6there exists ˜y∈NN(P, S) such that h= ˜y(h0)−1, P6S(˜y,(h0)−1), and NT(Py˜)∈Sylp(NN(Py˜)). Then Lemma 4.7 applies to (f, P) and ˜y, and yields the result that ˜y−1f is "-maximal and Sy˜−1f=Py˜. Thus (1) implies thatT6Py˜, and then alsoT6P.
Recall from Definition2.18 (c) that the partial groupNL(T) is a locality via the set
∆T of objects Q∈∆ withQ6T.
Corollary 4.8. (Frattini lemma) Let L=(L,∆, S)be a locality,let N be a partial normal subgroup of L, and set T=S∩N. Then L=NNL(T) as a product of partial subgroups of L.
Proof. Let f∈L, set P=Sf, and choose (g, Q)∈L∆ so that (f, P)"(g, Q) and so thatg is "-maximal. By transitivity of", we may takeQ=Sg. ThenT6Qby Proposi-tion4.5, and then by Lemma4.6there existsy∈NN(P, Q) with f=yg. Here g∈NL(T) by Lemma4.2(a).
Lemma 4.9. Let M be a finite group, let S be a Sylow p-subgroup of M, and let K be a normal subgroup of M. Set F=FS(M) and let Γ be a non-empty, overgroup closed, F-invariant collection of subgroups of S. Let L:=LΓ(M) be the locality given by Example/Lemma 2.10, and let β be a rigid automorphism of L. Assume that the following three conditions hold:
(1) Q∩K∈Γ for all Q∈Γ;
(2) CM(Op(M))6Op(M)6K;
(3) Γis a set of F-centric subgroups of S.
Set Φ={Q∩K|Q∈Γ},and set K=LΦ(K). Then (a) β restricts to a rigid automorphism of K, and
(b) β extends to an automorphism of M if and only if extends to an automor-phism of K.
Proof. SetY=Op(M) and letP∈Γ. SinceCM(Y)6Y by (2), the ThompsonA×B lemma [11, Theorem 5.3.4] implies that CM(P) is a p-group. Then CM(P)=CL(P)=
Z(P), as P is F-centric by (3). Point (a) then follows from Lemma 3.8. Further, Proposition2.17yields
(4) every f∈L is a product Π(f1, ..., fn), where fi is in a normalizer NL(Ri) for someRi∈Γ.
Let (M, β) be a counterexample to (b) with|M| as small as possible. Let K0 be the subgroup of K generated by the subset K of K. Let g∈K∩L. Then Sg∈Γ and Sg∩K∈Φ, so g∈K. Thus K∩L⊆K. The reverse inclusion is given by the definition of K, so K=K∩L. Then K is a partial normal subgroup of L by Lemma 3.10. Set T=S∩K and observe that for anyh∈NM(T) we have (h−1, g, h)∈D via (Sg∩T)h, and hencegh∈K. Thus,KisNM(T)-invariant, so alsoK0 isNM(T)-invariant.
SetM0=NM(T)K0 and setL0=LΓ(M0). We next show (5) NM(P)6M0for allP∈Γ.
Among allP for which (5) fails to hold, chooseP so that|P|is as large as possible.
Suppose thatP is not fully normalized inF, and letP0be a fully normalizedF-conjugate ofP. Then Alperin’s theorem yields a sequencew=(g1, ..., gn) of elements ofM and a sequence (R1, ..., Rn) of fully normalizedF-centric subgroups ofS, such thatP0:=P6R1, Pi:=Pg1...gi6Ri for alli, andP0=PΠ(w). One may assume thatn is minimal for these conditions, and hence Pi6=Ri for anyi. The maximality of|P| in the choice of P then yieldsNM(Ri)6M0 for alli, and hence Π(w)∈M0. Without loss, then, we may assume thatP=P0.
WithPfully normalized inFwe obtainNT(P)∈Sylp(NK(P)). AsNK(P)ENM(P), the Frattini lemma yields
NM(P) =NK(P)(NM(NT(P))∩NM(P)). (∗) If T6P then (∗) yields NM(P)6K0NM(T)=M0, contrary to the choice of P. Thus TP, and henceNT(P)P. SetQ=NT(P)P. ThenNM(Q)6M0by the maximality of
|P|, and then (∗) again implies thatNM(P)6M0. This completes the proof of (5). Now (4) yieldsL⊆M0. Thus,
(6) L0=L.
Suppose next that K0 is a proper subgroup of K. Then K∩M0=NK(T)K0=K0, and soM0is a proper subgroup ofM. SinceL0=L, the minimality of|M|then yields an extension ofβ to an automorphismγ ofM0. The condition (2), together with Proposi-tion1.10(c) then implies thatγ=czis conjugation by somez∈Z(S). Sincecz is also an automorphism ofM, we have an extension of β to an automorphism ofM in this case, so we conclude thatK0=K.
Let h,¯h∈K, let x,x∈N¯ M(T), and suppose that (h, x) and (¯h,x) are in¯ D(L) with hx=¯h¯x. SetP=Sh∩KandP=S¯h∩K, and setQ=PhandQ= P¯h. Then (h, x,x¯−1)∈D viaP, and Π(h, x,x¯−1)=¯h. It follows thatP=Pand that (h−1,¯h)∈DviaQ. Then
Π(h−1β,¯hβ) = (h−1¯h)β= (x¯x−1). (∗∗) By hypothesis, there exists an extension η of to an automorphism of K. It follows from (∗∗) that there is a well-defined mapping γ:M!M given by γ:hx7!(hη)(xβ) for h∈K andx∈NM(T).
In order to show that γ is a homomorphism, it suffices to show that (hη)xβ=(hx)η for allh∈Kandx∈NM(T). AsK0=Kwe may writehas a product ΠK(h1, ..., hn) with hi∈K. Then
(hx)η= (hx1... hxn)η= (hx1)η ...(hxn)η= (h1η)xβ...(hnη)xβ= (h1η ... hnη)xβ= (hη)xβ, as required.
We check that Ker(γ)=1. Namely, if (hη)(xβ)=1 with h and x as above, then x∈NK(T) and xβ=xη, and then hx=1 as η is injective. Thus γ is injective, and is therefore an automorphism ofM.