subgroups of soluble type

H-Salomon, in his Doctoral Dissertation [Sal87], has introduced and studiednotions of H-projectors and H-covering subgroups different from the usualones which lead to a theory of these su

Subgroups of soluble type

Consider a subgroup H of a soluble group G Since every minimal normal subgroup of G is abelian, the following implication holds:

If G = M H, M is a minimal normal subgroup of G,

and H is a proper subgroup of G, then H ∩ M = 1 (5.1)

It is precisely this property which makes the theory ofprojectors and covering subgroups,H a Schunck class, so much easier in the soluble universethan in the general finite one

H-Salomon, in his Doctoral Dissertation [Sal87], has introduced and studiednotions of H-projectors and H-covering subgroups (different from the usualones) which lead to a theory of these subgroups in arbitrary groups resemblingthe theory ofH-projectors and H-covering subgroups in finite soluble groups.His first basic idea is to give a definition ofH-projectors along the followinglines:

Recall that ifH is a class of groups, a subgroup H of a group G is called

anH-projector of G if H ∈ S(G) ∩ H and if N
G and HN/N ≤ K/N ∈

S(G/N ) ∩ H, then HN/N = K/N HereS(X) is the set of all subgroups of a group X.

Salomon tries to replace the setS(G) of all subgroups of a group G by able subsets d(G) ofS(G) which are such that any element H of d(G) enjoys

suit-the above property (5.1); he also tries to develop a suit-theory of suit-the “projectors”

so obtained by following the classical approach

It is clear that in order to carry out this, one cannot take just any d(G) ⊆

S(G) satisfying (5.1) First of all, since the definition ofH-projector involves

not only the group G itself but also its quotients, such sets d(G) have to be chosen not only for G but at least for all quotients of G; in fact, the classical

theory of projectors suggests that it would be reasonable to demand thatsuch a choice be made for all finite groups simultaneously Put differently, tobegin with, one chooses a “subgroup functor” d which associates with any

group G a set of subgroups d(G) (we refer to elements of d(G) as subgroups

205

of soluble type) subject, of course, to the condition that (5.1) holds for any G and each H in d(G) Moreover, it is also plausible that a subgroup functor d ought to satisfy certain formal properties such as the following: If H ∈ d(G) and N
G, then HN/N ∈ d(G/N) Not surprisingly, it turns out that the

properties relevant here are closely related to properties of ordinary projectorsand covering subgroups

The theory developed in this Chapter is largely the work of Salomon [Sal87]and F¨orster [F¨orb], [F¨ora]

5.1 Subgroup functors and subgroups of soluble type: elementary properties

The purpose of this section is to establish the necessary formal properties

of the various functors for subgroups of soluble type The functors t and t

introduced by Salomon in [Sal87] are studied here Two similar functors r and

r defined by F¨orster [F¨orb] are also studied.

A subgroup functor is a function f which assigns to each group G a sibly empty set f(G) of subgroups of G satisfying θ

pos-f(G)

= f

θ(G)for any

sub-3 Let p be a prime Let Sp be the function assigning to each group G the

set Sp (G) of all subgroups U of G containing a Sylow p-subgroup of G; Sp is

Definition 5.1.2 Let f be a subgroup functor We say that f is inherited if

f enjoys the following properties:

1 If U ∈ f(T ) and T ∈ f(G), U ≤ T ≤ G, then U ∈ f(G).

2 If U ∈ f(G) and N
G, then UN/N ∈ f(G/N).

3 If U/N ∈ f(G/N), N
G, then U ∈ f(G).

If, moreover, f satisfies

4 U ≤ G and T ∈ f(G) implies U ∩ T ∈ f(U),

we say that f is w-inherited.

Obviously the functors Sand Sn are w-inherited Sp is inherited but notw-inherited

5.1 Subgroup functors and subgroups of soluble type 207

Lemma 5.1.3 Let f be an inherited functor Then

1 If U ∈ f(G) and N
G, then UN ∈ f(G).

2 If N is a subnormal subgroup of G and N = N1
N2
· · ·
N k = G

is a chain from N to G such that 1 ∈ f(N i ) for all i ∈ {2, , k}, then

N ∈ f(G).

In particular, Sn (G) ⊆ f(G) for all groups G if 1 ∈ f(X) for all groups X Proof. 1 follows from Definition 5.1.2 (2 and 3)

2 By 1, N i ∈ f(N i+1 ) for all i ∈ {1, , k − 1} Applying

f(G) and N is a subnormal subgroup of G such that U ≤ N G (N ), then U N ∈ f(G).

In particular, if 1 ∈ f(G), then Sn (G) ⊆ f(G).

Proof We argue by induction on |G| Let N1be the normal closure of N in G and, for i > 1, denote N i =N N i −1 , the normal closure of N in N i −1 Since

N is subnormal in G, there exists n ≥ 1 such that N = N n Suppose that

G = U N1, then N2= N1 because U normalises N Repeating the argument with every N k , it follows that N is normal in G By Lemma 5.1.3 (1), we have U N ∈ f(G) Therefore we may assume that UN1 is a proper subgroup

of G By induction, U N ∈ f(UN1) Now U N1 ∈ f(G) yields UN ∈ f(G) by

f(G) Then the intersection of all subgroups of G belonging to f(G) and taining 

con-i ∈I X i is the smallest subgroup of G in f(G) containing 

i ∈I X i This subgroup is denoted by X i : i ∈ If and called the f-join of {X i : i ∈ I}.

Theorem 5.1.7 Let f be a w-inherited subgroup functor For each group G,

f(G) is a lattice under the operations “ ∩” and “·, ·f.”

If H is a subgroup of G such that H ∈ e(G) for some e ∈ {t, r, t  , r  , t  , r  },

we shall say that H is a subgroup of soluble type.

The functors t and t have been introduced by Salomon in his Dissertation

[Sal87] There are certain problems Salomon encounters with these two choices

of e: the first of these functors does not really give sets t(G) “large enough” whenever G is “highly non-soluble,” while for t  one of the crucial properties,

namely, if H ∈ e(G) and N
G, then H ∈ e(HN), is missing Later, F¨orster

[F¨orb] overcame these problems by introducing the remaining functors As will

be seen in the next section “the r-functors” enjoy the advantage of producingrelevant subgroups of primitive groups

Remarks 5.1.9 1 If G is soluble, then t(G) = r(G) = t  (G) = r  (G) =

t (G) = r  (G) =S(G).

2 Condition “(H ∩ S)T
S” in the definitions of t and r implies that

H either covers or avoids the simple section S/T , that is, (H ∩ S)T ∈ {S, T }.

3 In defining t, r, t, r (t, r, respectively) we might have taken the

strictly semisimple (simple) groups S/T as direct products of non-abelian

simple groups (as non-abelian simple, respectively) Moreover, since everysubnormal subgroup of a direct product of non-abelian simple groups is in

fact normal by [DH92, A, 4.13] the condition “(H ∩ S)T
S” can be replaced

by “(H ∩ S)T sn S.”

4 If H is a subgroup of G such that H ∈ e(G) for some e ∈ {t, r, t  , r  , t  ,

r }, and N is a normal subgroup of G, then HN/N ∈ e(G/N).

5.1 Subgroup functors and subgroups of soluble type 209

Proof Only a proof for Statement 6 is needed Let S/T denote a strictly semisimple section of G, with simple component E, say, and let H ≤ N G (S) Then T ∗, the D0(1, E)-residual of S, is a characteristic subgroup of S con-

tained in T with strictly semisimple quotient S/T ∗ contained in D0(1, E); in

particular, H ≤ N G (T ∗) Now Statement 6 follows directly from this

Proposition 5.1.10 Let H and N be subgroups of a group G such that N

is quasinilpotent and H normalises N Suppose that the following condition holds:

B
A sn N, A/B strictly semisimple, and H ≤ N G (A) implies that

Then H ∩ N is subnormal in N.

Proof Proceeding by induction on |G| + |N|, we may clearly assume that

G = HN , and hence that N is normal in G In the case N = 1 or N is a minimal normal subgroup of G, the claim is immediate from condition (5.2);

so without loss of generality, there is a normal subgroup M of G such that

1 = M < N Taking into account that the class of all quasinilpotent groups

is an Sn -closed homomorph and that condition (5.2) is inherited by M and N/M , we may apply the inductive hypothesis twice to get that H ∩ M is

a subnormal subgroup of M and HM ∩ N is a subnormal subgroup of N;

in particular, HM ∩ N is quasinilpotent Therefore, if HM ∩ N is a proper subgroup of N , then another application of the inductive hypothesis yields that H ∩ N = H ∩(HM ∩ N) is subnormal in HM ∩ N, and hence that H ∩ N

is subnormal in N Thus, without loss of generality, N = HM ∩ N ≤ HM and G = HN = HM Therefore we have:

H < G = HM whenever M
G and 1 = M < N. (5.3)

Assume that the Fitting subgroup of N , F(N ), is non-trivial Then N contains

an abelian minimal normal subgroup M of G; since N = (H ∩ N)M, we can deduce that H ∩ N is normal in N because N centralises M This is due to the fact that M is a direct product of minimal normal subgroups of M which are central in N because N is quasinilpotent It remains to deal with the case when F(N ) = 1 Then N is a direct product of non-abelian simple groups by Proposition 2.2.22 (3) By Condition (5.2), H ∩ M is normal in M for every minimal normal subgroup of G contained in N Therefore we obtain:

N is a direct product of non-abelian simple groups, and H ∩ M = 1, whenever M is a minimal normal subgroup of G contained in N (5.4)

Now, N cannot have two non-isomorphic simple direct factors E and F : otherwise for each X in {E, F } we could find minimal normal subgroups M X

of G contained in N with X as composition factor, so by Condition (5.3) and Condition (5.4) both M and M should be normal complements of H in G,

leading to the contradiction that M E ∼ = (M E ×M F)∩ H ∼ = M F Consequently,

N is strictly semisimple, and our claim holds by assumption 

w-inherited.

Proof We give a proof for t.

1 Let G be a group and let X and U be subgroups of G such that X ∈ t(U) and U ∈ t(G) Consider a strictly semisimple section S/T of G We prove that (S ∩ X)T is normal in G Clearly we may assume that S/T is a direct product

of non-abelian simple groups Since U ∈ t(G), it follows that (S ∩ U)T is normal in S Hence (S ∩U)/(T ∩ U) is either 1 or a strictly semisimple section

of U If S ∩ U = T ∩ U, then S ∩ X = T ∩ X and (S ∩ X)T = T Thus we may assume that S ∩ U = T ∩ U Since X ∈ t(U), we have that (S ∩ X)(T ∩ U)

is normal in S ∩ U This implies that (S ∩ X)T is normal in (S ∩ U)T
S Since S/T is a direct product of non-abelian simple groups, we have (S ∩X)T

3 Consider a normal subgroup N of G Let U/N ∈ t(G/N) Suppose that S/T is a non-abelian and strictly semisimple section of G If T N =

SN , then S = (S ∩ N)T = (S ∩ U)T and (S ∩ U)T is normal in S Hence

we may assume that T N = SN In particular, SN/T N ∼ = S/

(S ∩ N)T ) and (SN/N )

(T N/N ) is a strictly semisimple section of G/N Since U/N ∈ t(G/N ), it follows that 

(SN/N ) ∩ (U/N)(T N/N ) is normal in SN/N and

so (SN ∩ U)T is normal in SN This implies that (SN ∩ U)T ∩ S = (S ∩ U)T

is a normal subgroup of S Therefore U ∈ t(G).

The same statements can be obtained for the functors t and t just by

adding the assumption S subnormal in G for t  , and S subnormal in G and

S/T simple for t in the above proof.

Finally, it is clear that t is w-inherited because every strictly semisimplesection of every subgroup of a group is actually a strictly semisimple section

Example 5.1.12 Let E be a non-abelian simple group and let K = E E be the regular wreath product Then K can be written as semidirect product

K = F E, where F is the base group and E is its canonical complement in

K Consider now G = E E K, the wreath product of E with K with respect

to the transitive permutation representation of K on the right cosets of E Let B be the base group of G Then B = E1× · · · × E n , n = |K : E|, where

E i ∼ = E (i = 1, , n), E1 is E-invariant and E1E = E1× E It is rather easy

to see that F (B, respectively) is the unique minimal normal subgroup of K (G, respectively) Consequently, G has a unique maximal normal subgroup, namely BF

5.1 Subgroup functors and subgroups of soluble type 211

Let H be the diagonal subgroup of E1E We claim that H ∈ t  (G) Let T

S sn G with S/T non-abelian and strictly semisimple If S = G, S is contained

in the unique maximal normal subgroup BF of G Then (by construction of

H and BF ) H ∩BF = 1, whence H ∩S = 1 and (H ∩S)T = T
S If S = G, then T ∈ {G, BF } and

(S ∩ H)T = HT = HBF = G = S.

Assume, by way of contradiction, that H ∈ r  (HB) Then HN/N ∈ r  (HB/N )

whenever N
HB But B = E1× C, where C = E2× · · · × E n and HB/C = (E ×E1)C/C ∼ = E×E1via the canonical isomorphism This maps the diagonal

subgroup H of E ×E1onto the non-normal subgroup HC/C of EB/C, which contradicts HC/C ∈ r  (EB/C) (see Proposition 5.1.10).

Therefore H / ∈ r  (HB) and, in particular, H / ∈ t  (HB) This shows that t 

is not w-inherited

We do not know, however, whether r is w-inherited

Remark 5.1.13 Since t is w-inherited, then, by Theorem 5.1.7, t(G) is closed

under the operations “∩” and “·, ·t.” In general,U, V t = U, V : let n ≥ 5 and let G = Σ n be the symmetric group of degree n Let X1 =(1, 2) and

X2 =(2, 3) Then X i ∈ t(G), i = 1, 2, X1, X2 ∼ = Σ3, and X1, X2t= G,

as we can see by looking at the sections Alt(n)/1 of Sym(n).

Proposition 5.1.14 If H is a proper subgroup of a group G = HM , with

H ∈ r  (G), and M is a minimal normal subgroup of G, then H ∩ M = 1 Proof By Proposition 5.1.10, H ∩ M is normal in M Since H is a proper subgroup of G and M is a minimal normal subgroup of G, we have that

Remark 5.1.15 The statement of Proposition 5.1.14 does not hold for

ele-ments of t (G) Consider the regular wreath product G = E T of a non-abelian simple group E and a group T of order 2 Denote by D the T -invariant diag- onal subgroup of the base group B, and put H = T D Then G = HB, B is a minimal normal subgroup of G, H ∈ t  (G), but H ∩ B = 1.

Since the functors tand r do not have the crucial property described in

Proposition 5.1.14, they are not so interesting for us as the functors t, r, t,

and r Nevertheless, we shall see at the end of the section (Theorem 5.1.25)

that the composition factors of the subgroups in r (G) are composition factors

of the whole group

Definition 5.1.16 A subgroup functor f is called inductive (respectively,

weakly inductive) if it satisfies Conditions 1 and 2 (respectively, 1 and 2  ):

1 G ∈ f(G)

2 If H ≤ K ≤ G, N
G, H ∈ f(K), N ≤ K, and K/N ∈ f(G/N), then

H ∈ f(G).

2  H ≤ G, N
G, H ∈ f(HN), and HN/N ∈ f(G/N) implies that H ∈ f(G).

It is clear that w-inherited functors f such that f(X) is non-empty for all groups X are inductive and inductive functors are also weakly inductive.

Proof We see that t is inductive The same proof applies to t.

First of all, the defining Condition 1 for inductivity of tholds trivially To

verify Condition 2, let H ≤ K ≤ G and N
G such that N ≤ K, H ∈ t  (K)

and K/N ∈ t  (G/N ) We show that (S ∩ H)T is normal in G whenever S/T

is a non-abelian simple section of G such that S is subnormal in G Suppose that SN = T N Then S = (S ∩ N)T = (S ∩ K)T and (S ∩ K)/(T ∩ K) is a simple section of K such that S ∩ K is subnormal in K Since H ∈ t  (K), it

follows that (S ∩ H)(T ∩ K) is normal in S ∩ K Hence (S ∩ H)T is normal in (S ∩ K)T = S Suppose that SN = T N Then (SN/N)(T N/N ) is a simple section of G/N and SN/N is subnormal in G/N Since K/N ∈ t  (G/N ), we

have that (K ∩SN)T is normal in SN Hence (K ∩S)T =S ∩(K ∩SN)T =

S ∩ (K ∩ SN)T is normal in S ∩ SN = S Further, if K ∩ T = K ∩ S, then (S ∩ K)/(T ∩ K) is a simple section of K and S ∩ K is subnormal in K As

H ∈ t  (K), this gives that

H ∩ (K ∩ S)(K ∩ T ) is normal in K ∩ S This is also true if K ∩ T = K ∩ S Consequently we obtain:

(H ∩ S)T = (K ∩ H ∩ S)(K ∩ T )T
(K ∩ S)T
S.

Note that, in the special case when K = HN , the same argument still

works for the functors r, r and r , for H ≤ N G (S) and N
G imply that

HN ≤ N G (SN ); hence we get:

r(G) and L is a subgroup of G containing H, then H ∈ r(L).

1 If H is a subgroup of G and N is a soluble subgroup of G normalised by

H, then H ∈ f(HN).

2 Let H ≤ K ≤ G = HN, where N is a direct product of non-abelian simple groups If H ∈ f(G) and K ∩ N is normal in N, then H ∈ f(K).

Proof 1 For a proof of H ∈ f(H N ), we may use induction on |G| and the

properties of f verified earlier in this section (Propositions 5.1.17 and 5.1.18)

to see that without loss of generality 1 = G = HN, N is a minimal normal subgroup of G, H ∩ N = 1, and H is a core-free maximal subgroup of G Hence G is a primitive group of type 1 and N is the unique minimal normal subgroup of G Let T
S be a non-abelian strictly semisimple section of G

5.1 Subgroup functors and subgroups of soluble type 213

such that S is subnormal in G Assume that S is a proper subgroup of G Then S is contained in some maximal normal subgroup M of G Since M = 1,

it follows that N is contained in M and M = N (H ∩ M) Since M is a proper subgroup of G, H ∩M ∈ f(M) by induction Hence (S ∩H)T = (S ∩H ∩M)T

is normal in S Therefore we may assume that S = G Then T = 1 because otherwise G would be non-abelian and simple Consequently N is contained

in T and so S = (S ∩ H)T This means that H ∈ f(G).

2 By hypothesis K ∩ N is normal in N, so K ∩ N is a normal subgroup

of KN = G Since N is a direct product of non-abelian simple groups, it follows that N = (K ∩ N) × M for a normal subgroup M of G Then G/M = KM/M ∼ = K It is clear therefore that H ∈ f(K) because HM/M ∈ f(G/M).



product of a subgroup H ∈ f(G) and F ∗ (G), the generalised Fitting subgroup of

G If M is a normal subgroup of G and M is quasinilpotent, then H ∈ f(HM) Proof We argue by induction on |G| If M is soluble, the result follows from Lemma 5.1.19 (1) We may suppose that M is not contained in F = GS,

the soluble radical of G If F = 1, then F ∗ (G) and M are both direct

products of non-abelian simple groups by Proposition 2.2.22 (3) Moreover

HM ∩ F ∗ (G) is normal in F ∗ (G) because H ∩ F ∗ (G) is normal in F ∗ (G)

by Proposition 5.1.10 and [DH92, A, 4.13] Applying Lemma 5.1.19 (2), we

conclude that H ∈ f(HM) Therefore we can suppose that F = 1 By tion HF/F ∈ f(HF/F )(M F/F )

induc- This yields H(HM ∩ F )/(HM ∩ F ) ∈

f

HM/(HM ∩ F ) Since H ∈ fH(HM ∩ F )by Lemma 5.1.19 (1), we

non-abelian simple group E Let H be a subgroup of G such that H covers or avoids every simple section of G Then H is a normal subgroup of G Proof We argue by induction on |G| It is clear we can assume Core G (H) = 1 Since either E j ∩ H = 1 or E j ≤ H for all j ∈ {1, , n}, it follows that

E j ∩ H = 1 because H does not contain any normal subgroup of G Denote

N j =Xi =j E i, 1≤ j ≤ n.

Suppose that G = HN j for each j ∈ {1, , n} In particular, H is a subdirect subgroup of G Hence H = R1× · · · × R t , where R k ∼ = R l ∼ = E,

1 ≤ k, l ≤ t Let g = 1 be an element of R1 Without loss of generality, we

may assume that π1(g) = 1, where π1: G −→ E1 is the projection of G over

its first component Then 1 = π1(R1) is a normal subgroup of π1(H) = E1,

and so π1(R1) = E1 Let 1 = h ∈ E1 such that g h = g Since g ∈ R1∩ R h

1, it

follows that R h

1 is contained in H.

Assume [h, R1] = 1 Then 1 = π1([h, R1]) = [h, E1] and h ∈ Z(E1) This

contradiction shows that [h, R1] = 1 Let t be an element of R1 such that

t h = t Then 1 = t h t −1 ∈ H ∩ E1 = 1 (note that if t = (e1, e2, , e n),

t h = (e h , e2, , e n))

This contradiction proves that HN j is a proper subgroup of G for some

j ∈ {1, , n} Then G/N j is a simple section of G which is not covered by

H Hence H is contained in N j By induction H is normal in N j and so H

is subnormal in G This implies that H is a normal subgroup of G [DH92, A,

if, every simple section of G is covered or avoided by H.

Proof The cover-avoidance property dealt with here is obviously a special case of the defining property of H ∈ t(G).

Conversely, let H be a subgroup of G enjoying this cover-avoidance erty, and let T
S ≤ G and S/T strictly semisimple and non-abelian Let X = (H ∩ S)T/T Then X is a subgroup which covers or avoids every simple section of S/T By Lemma 5.1.21, X is normal in S/T Consequently

subnormal subgroup of G such that M = Xn

i=1 E i , where the E i (i = 1, , n) are pairwise isomorphic non-abelian simple groups; further assume that H normalises M Let π i : M −→ E i (i = 1, , n) denote the projection of M over E i Put I = i ∈ {1, , n} : (H ∩ M)π i = 1 Then H ∩ M is subdirect

inXi ∈I E i

Proof Note that M/ Ker(π i ) is a simple section of G Since H ∈ r  (G), it

follows that H covers or avoids M/ Ker(π i ) If M ∩ H = M ∩ Ker(π i), we

have π i (M ∩ H) = 1 Therefore I = i ∈ {1, , n} : π i (M ∩ H) = 1 =

i ∈ {1, , n} : M = (M ∩ H) Ker(π i) It follows that H ∩ M is a subdirect

Theorem 5.1.25 Let G be a group Every composition factor of a subgroup

H ∈ r  (G) is isomorphic to a composition factor of G.

Proof Proceeding by induction on |G|, we consider a minimal normal group of G Since HM/M ∈ r  (G/M ), we may assume that our claim holds

sub-for H/(H ∩ M) ( ∼ = HM/M ) If M is abelian, the result follows Hence we may assume that M is non-abelian By Lemma 5.1.24, H ∩ M is a subdirect subgroup of a suitable normal subgroup of M Then H ∩M ∈D0(1, E), where

the direct product of two copies of a non-abelian simple group E and a

diag-∈/ r(G) by Proposition

5.2 Existence criteria 215

5.2 Existence criteria

The main goal in this section is to prove some existence results for various groups of soluble type We begin with a result on t-subgroups which may beviewed as a non-existence result: as a consequence of this theorem, in a mono-lithic primitive group with non-abelian socle the minimal normal subgroupcannot be complemented by a t-subgroup unless the corresponding quotient

sub-of the group is soluble In fact, the results on this type sub-of subgroups show thatthe non-soluble t-subgroups share some properties with non-soluble subnor-mal subgroups This is already apparent from the first theorem of this section,which may be thought as a partial generalisation of the following theorem ofWielandt ([Wie39], see Theorem 2.2.19):

If H and K are subnormal subgroups of a group G and H is a perfect comonolithic group, then either H is contained in K or K normalises H.

tence of t-complements of the minimal normal subgroup in primitive groups

of type 1; it turns out that in order that such a complement exists, the

cor-responding quotient of the group has to be p-soluble, p the prime dividing the

order of the minimal normal subgroup

The remaining part of this section deals with results on the functors r, t,

and r.

semisimple normal subgroup M of G Then

[H, M ] ≤ H ∩ M
HM.

Proof Let (H, M, G) represent a counterexample such that |G| + |H| is

min-imal Then, since t is w-inherited, it follows that:

1 G = HM and H is a proper subgroup of G.

Now, since H ∈ t(G) and M is strictly semisimple, H ∩ M
HM = G Since G is a counterexample, [H, M ] is not contained in H ∩ M.

2 H ∩ M = 1, M is a minimal normal subgroup of G and Core G (H) =

j ∈ {1, , s} Therefore [H, M] = 1 This contradiction proves that M is a minimal normal subgroup of G If Core G (H) = 1, then the minimality of G yields [H, M ] ≤ H ∩M = 1, and we have a contradiction Thus Core G (H) = 1

and C (M ) = 1 because C (M ) is a normal subgroup of HM = G Now put

We supplement this result by a theorem which leads to a criterion for the

exis-C = exis-C G (M ) and D = H ∩CM Then CM = CM ∩ HM = DM Suppose that

D = 1 Then C is contained in M and so C = 1 because M is non-abelian Assume now that D = 1 By Lemma 5.1.3, HC ∈ t(G) Hence HC ∩ M

is normal in (HC)M = G Since M is a minimal normal subgroup of G, either HC ∩ M = 1 or HC ∩ M = M If HC ∩ M = 1, then HC = H This implies C = 1 Suppose that HC ∩ M = M Then G = HC = HM,

DC = (H ∩ CM)C = HC ∩ CM = CM, and

C ∼ = CM/M = DM/M ∼ = D ∼ = DC/C = CM/C ∼ = M,

whence CM ∼ = M ×M is a strictly semisimple subgroup of G Since H ∈ t(G),

D = H ∩ CM is a normal subgroup of CM Hence D is normal in (CM)H = G.

This yields a contradiction against CoreG (H) = 1 Therefore C G (M ) = 1.

3 Let HS denote the soluble radical of H Then H/HS is a non-abelian

simple group Moreover there is no proper t-subgroup K of G with H = KHS.

Let A/HS be a minimal normal subgroup of H/HS It is clear that A is

not soluble Hence there exists an integer n such that (A (n)) = A (n) = 1 Moreover A (n) ∈ t(A) and so A ∈ t(G) By the inheritness of t, it follows

A (n) ∈ t(G) If A were a proper subgroup of H, then A (n) < H The minimal choice of (G, H) would yield A (n) ≤ C G (M ) = 1, contrary to our supposition Therefore A = H and H/HS is a non-abelian simple group.

Suppose there exists a subgroup K ∈ t(G) such that H = KHS Then K

is not soluble and so there exists an integer b such that 1 = K (b) is a perfect

subgroup of K Moreover K (b) ∈ t(G) Since |G|+|K (b) | < |G|+|H|, it follows

K (b)is contained in CG (M ) = 1 This is a contradiction.

Applying [Hup67, VI, 4.7], there exists a Sylow 2-subgroup H2 of H and

a Sylow 2-subgroup M2of M such that G2= M2H2is a Sylow 2-subgroup of

G By the Odd Order Theorem ([FT63]), H2 = 1 and M2 = 1 Moreover M2

is a normal subgroup of G2and HSis a proper subgroup of H2HS.

4 [M2, H] = 1.

By way of contradiction, assume that T = C M2(H) is a proper subgroup

of M2 Then T is a proper subgroup of N = N M2(T ) It is clear that N/T

is a normal subgroup of H2N/T because H2 normalises N and T Hence Z(H2N/T ) ∩ (N/T ) = 1 Let g ∈ N such that 1 = gT ∈ Z(H2N/T ) ∩ (N/T ) Clearly HST /T < (H2T /T )(HST /T ) ≤ (H g T ∩ HT )/THST /T

Thus

HSis a proper subgroup of HS(H ∩ H g T ) We prove that H ∩ H g T ∈ t(G) Let C = N M (T ) Then T is normal in C and C is H-invariant Since t is w-inherited, it follows that HT /T ∈ t(HN/T ) and H g T/N ∈ t(HN/T) Hence (HT /T ) ∩(H g T /T ) ∈ t(HN/T ) because t has the intersection property Then (HT ∩ H g T )/T ∈ t(HT/T ) and hence HT ∩ H g T ∈ t(HT ) This implies that

H ∩ H g T ∈ t(H) By the inheritness of t, H ∩ H g T ∈ t(G) Now H ∩ H g T does not avoid the simple section H/HS Hence H = HS(H ∩ H g T ) by Proposition 5.1.22 Applying Step 3, H ∩ H g T = H Therefore g ∈ N G (HT ).

Moreover [g, H, H] = [H, g, H] ≤ [HT, g, H] ≤ [HT ∩ C, H] = [T, H] = 1

by Step 1 By the Three Subgroups Lemma ([Hup67, III, 1.10]), we infer that

5.2 Existence criteria 217

[H, g] = [H  , g] = [H, H, g] = 1 Thus g ∈ C N (H) ≤ C M2(H) = T This contradiction against the choice of g establishes Step 4.

5 Final contradiction.

Since CG (M ) = 1, it follows that G is, up to isomorphism, a subgroup

of Aut(M ) Now O2
(M ) = 1 Hence a theorem of G Glauberman [Gla66]

applies It says that CAut(M ) (M2) is 2-nilpotent In particular, CAut(M ) (M2)

is soluble by the Odd Order Theorem [FT63] By Step 4, H is a subgroup

of CG (M2)≤ C Aut(M ) (M2) This is a contradiction because H is perfect and

Before proving a result similar to Theorem 5.2.1 for abelian normal

sub-groups M , we have to recall some terminology and results.

Let G be a group If H ∗ is a normal subgroup of a proper subgroup H

of G and H g ∩ H is a subgroup of H ∗ for all g ∈ G \ H, then G is called a Frobenius-Wielandt group with respect to H and H ∗ H Wielandt [Wie58]

(cf [Hup67, V, 7.5]) has shown that the Frobenius-Wielandt kernel, defined by

G ∗ = G \g ∈G (H \H ∗)g , is a normal subgroup of G = HG ∗ and H ∩ G ∗ = H ∗.

If H ∗ = 1, then G is called simply a Frobenius group.

Let Ω = {xH : x ∈ G} and let H act on Ω by left multiplication We have

that |Ω| = |G : H| and if x ∈ G \ H, then the orbit of xH has size divisible

by|H : H ∗ | Consequently gcd(|H/H ∗ |, |G : H|) = 1.

subgroup of G Further, assume that H is comonolithic, with maximal normal subgroup L = Cosoc(H), and that there is no a proper subgroup K of H such that K ∈ t(H) and H = KL Then one of the following holds:

1 [H, M ] is contained in H ∩ M.

2 H ∩ M is a subgroup of C H (M/N ), with N defined by N/(H ∩ M) =

CM/(H ∩M) (H) Moreover C H (M/N ) is a proper subgroup of H and HM/N is a Frobenius-Wielandt group with respect to HN/N and LN/N , with Frobenius-Wielandt kernel LM/N

Proof Assuming that [H, M ] is not contained in H ∩ M, we will verify

State-ment 2 As in the proof of Theorem 5.2.1, an induction arguState-ment yields that

without loss of generality G = HM and N = H ∩ M = 1 Consider g ∈ G such that H g ∩ H is not contained in L; put g = hm, where h ∈ H and

m ∈ M Then H = L(H ∩ H g ) Moreover H ∩ H g ∈ t(H) The hypothesis of the Lemma yields H ∩ H g = H Hence H = H m ∩ H = C H (m) by [DH92,

A, 16.3]; that is, m ∈ C M (H) = N = 1 Thus g ∈ H We have shown that

H g ∩ H is a subgroup of L for all g ∈ G \ H, and G is a Frobenius-Wielandt group with respect to H and L Note that, if G ∗ is the Frobenius-Wielandt

kernel, then M L is contained in G ∗ Therefore G ∗ = M (G ∗ ∩ H) = ML and

Lemma 5.2.3 Let G = HM be a group, where M is an abelian normal

p-subgroup of G for some prime p Assume that H is a subgroup of G such that

H ∩ M = 1 If H is p-soluble and H p
∈ t(H p
M ) for every Hall p  -subgroup

SM is p-soluble and every Hall p  -subgroup B of H ∩ SM belongs to t(BM) Suppose that S = G Then either T = 1 or M is contained in T In both cases (H ∩ S)T ∈ {S, T } Hence we may assume that S is a proper subgroup of G Applying Remark 1.1.11 (1), it follows that S = H m for some m ∈ M On the other hand, S/T is a p  -group because G is p-soluble This implies that T

contains every Sylow p-subgroup of S Let K be a Hall p  -subgroup of H such

that K m contains a Hall p  -subgroup of H ∩S, (H ∩S) p
say Since S = K m T ,

it follows that K m /(K m ∩ T ) is a simple section of K m M = KM Since

K ∈ t(KM) either (K ∩ K m )(K m ∩ T ) = K m or K ∩ K m ≤ K m ∩ T Assume the latter holds Then (S ∩ H)T = (S ∩ H) p
T ≤ (K ∩ K m )T = T Suppose that (K ∩ K m )(K m ∩ T ) = K m Then S = S m T = (K ∩ K m )T ≤ (H ∩ S)T ≤ S and so S = (H ∩ S)T Applying Proposition 5.1.22, H ∈ t(G) and the proof

Theorem 5.2.4 Let G = HM be a factorised group, where M is an abelian

normal p-subgroup of G for some prime p Assume that H ∩M = C H (M ) = 1 Then the following three conditions are pairwise equivalent:

b) H p
∈ t(H p
M ) for every Hall p  -subgroup H p
of H.

Proof 1 implies 2 First of all, if K is a subgroup of H, then K = (H ∩ M)K =

H ∩ MK ∈ t(MK) as t is w-inherited Hence, if L
K ≤ H, K/L non-abelian simple, and m ∈ M, then (K ∩ K m )L ∈ {L, K} because K m ∈ t(MK) Next we see that H is p-soluble Let S/T be a non-abelian simple section

of H Then S ∈ t(SM) by the above argument Choose R as a supplement

of T in S contained in t(SM ) of minimal order As Sn ≤ t, R ∩ T is the unique maximal normal subgroup of R; the corresponding quotient is a non- abelian simple group: R/(R ∩ T ) ∼ = RT /T = S/T In particular, R is perfect Hence the above lemma applies to R, R ∩ T , and SM Since the hypothesis