On a class of finite soluble groups

Definition 1.

(1) A subgroup H of a group G is said to be c-embedded in G if H is self-normalising in SG⁡(H).

(2) A group G is said to be a C-group if every subgroup is c-embedded in G.

Theorem A.

The class of all C-groups is subgroup-closed formation of soluble groups of p-length at most 1 for all primes p.

Example 2.

(1) Let A be a cyclic group of order 4 and let B be a cyclic group of order 3 on which A acts invertingly. The extension BA has a faithful irreducible module N of dimension 2 over the field of 7 elements. Then the semidirect product G=N⁢B⁢A is a C-group whose nilpotent residual is not nilpotent.

(2) Let G=D26×A4, where D26 is a dihedral group of order 26 and A4 is an alternating group of order 4. Then G is a C-group whose nilpotent residual is not a Hall subgroup of G.

Theorem B.

A group G is a C-group if and only if G is soluble and every Carter subgroup of every subgroup H of G is also a Carter subgroup of SG⁡(H).

Corollary 3.

A group G is a C-group if and only if every nilpotent subgroup of G is self-normalising in its subnormal closure.

Corollary 4.

The following statements hold.

1. A soluble group G is a T-group if and only if every Carter subgroup of every subgroup H of G is also a Carter subgroup of HG, the normal closure of H in G.

2. A soluble group G is a PT-group if and only if every Carter subgroup of every subgroup H of G is also a Carter subgroup of AG⁡(H).

3. A soluble group G is a PST-group if and only if every Carter subgroup of every subgroup H of G is also a Carter subgroup of BG⁡(H).

Lemma 5.

The class of all C-groups is a subgroup-closed class of groups which is closed under taking epimorphic images.

Proof.

By the above observation, the class of all C-groups is subgroup-closed. We prove now that if G is a C-group and N is a normal subgroup of G, then G/N is also a C-group. Let H/N be a subgroup of G/N. Then H is c-embedded in G. Therefore, if X=SG⁡(H), then NX⁢(H)=H. By [3, Lemma A.14.1], X/N is the subnormal closure of H/N in G/N. Moreover, NX/N⁢(H/N)=H/N. Therefore H/N is c-embedded in G/N and hence G/N is a C-group. ∎

Lemma 6.

Let p be a prime and let G be a Cp-group.

1. If A is a subgroup of G, then A is a Cp-group.

2. If L is a normal p-subgroup of G, then G/L is a Cp-group.

3. If L is a normal p′-subgroup of G, then G/L is a Cp-group.

Proof.

Statement (1) follows by the observation at the beginning of the section and statement (2) can be proved by using the same arguments to those used in Lemma 5.

(3) Arguing by induction on |G|, we may assume that L is a minimal normal subgroup of G. Let X/L be a p-subgroup of G/L. Then X=Y⁢L for some Sylow p-subgroup Y of X. Since G is a Cp-group, it follows that NZ⁡(Y)=Y, where Z=SG⁡(Y). Let T/L=SG/L⁡(X/L). Then T=Z⁢L. Assume that T is a proper subgroup of G. By statement (1), T is a Cp-group. Hence T/L is a Cp-group by induction. Since T/L=ST/L⁡(X/L), it follows that NT/L⁡(X/L)=X/L.

Assume now that G=T=Z⁢L. Then [Z,L] is normal in G. It follows from [3, Lemma A.14.3] that L normalises Z. Since L is minimal normal in G, we have either G=Z or L⊆CG⁡(Z).

Let g⁢L∈NG/L⁡(X/L). We may assume that g∈Z. Then X=Yg⁢L and there exists l∈L such that Yg⁢l=Y. Thus g⁢l∈NG⁡(Y). If G=Z, then NG⁡(Y)=Y and so g∈X. If L⊆CG⁡(Z), it follows that g∈NZ⁡(Y)=Y. Hence, in both cases, we have that g⁢L∈X/L.

Therefore G/L is a Cp-group. ∎

Theorem 7.

Let p be a prime. If G is a Cp-group, then the p-length of G is at most 1.

Proof.

Assume that G is a Cp-group. We prove that the p-length of G is at most 1 by induction on the order of G. Suppose that the largest p′-normal subgroup L=Op′⁡(G) is non-trivial. By Lemma 6, G/L is a Cp-group. Therefore G/L has p-length at most 1 by induction. Thus G has p-length at most 1. Consequently, we may assume that Op′⁡(G)=1.

Let A and B be maximal normal subgroups of G. By Lemma 6, A and B are both Cp-groups. The induction hypothesis implies that A and B are of p-length at most 1. Suppose that A≠B. Then G=A⁢B. Since the class of all groups of p-length at most 1 is closed under taking normal products, it follows that G has p-length at most 1. Consequently, we may assume that G has a unique maximal normal subgroup, M say, and the p-length of M is at most 1. Since Op′⁡(M)⊆Op′⁡(G)=1, it follows that M has a normal Sylow p-subgroup. Let P be a Sylow p-subgroup of G. If P were contained in M, then P would be normal in G and so G would have p-length at most 1. Therefore we may assume that P is not contained in M. Since M is the unique maximal normal subgroup of G, it follows that SG⁡(P)=G and so NG⁡(P)=P.

Suppose that PM is a proper subgroup of G. By Lemma 6, PM is a Cp-group. By induction, PM has p-length at most 1. Then Op′⁡(P⁢M)⊆Op′⁡(M)=1 and so P is a normal subgroup of PM. But P is self-normalising in PM. Hence M is contained in P. Assume that a minimal normal subgroup of G is not contained in M. Then G is a simple group. If G were non-abelian, then the subnormal closure of every non-trivial subgroup of G would be equal to G and so either G is a p′-group (and so G is of p-length at most 1), or every non-trivial p-subgroup of G would be self-normalising. In the latter case, every Sylow p-subgroup of G would be cyclic of prime number. This cannot be happen. Hence G is abelian and so it is cyclic of order p.

Therefore we may assume that Soc⁡(G) is contained in M. Therefore every minimal normal subgroup of G is an elementary abelian p-subgroup of G. Let N be a minimal normal subgroup of G. By Lemma 6, we have that G/N satisfies the hypothesis of the theorem. Hence G/N has p-length at most 1. Since the class of all groups of p-length at most 1 is a saturated formation by [3, Example A. 3.4 (b)], we may assume that G is has unique minimal normal subgroup, N say, and there exists a core-free maximal subgroup X of G such that G=N⁢X, N∩X=1 and N=F⁡(G) (see [3, Theorem A.15.2]). Consequently, M=N and X is a simple group of p-length at most 1. This implies that either X is a p-group or a p′-group. In particular, G has p-length at most 1.

Assume that G=P⁢M. Then G/M is a cyclic group of order p. Since P∩M is normal in G, it follows that G has p-length at most 1.

The proof of the theorem is now complete. ∎

Theorem 8.

Let p be a prime and let G be a group. Then G is a Cp-group if and only if the following two conditions hold:

1. G is p-soluble of p-length at most 1,

2. for any p-subgroup H of G, we have that COp⁡(SG⁡(H))⁡(H)=1.

Proof.

Let H be a p-subgroup of G. Set X=SG⁡(H). Since H is contained in Op′⁡(X) and Op′⁡(X) is subnormal in G, it follows that X=Op′⁡(X). If X has p-length at most 1, it follows that Op⁡(X) is a normal Hall p′-subgroup of X. Note that H is subnormal in a Sylow p-subgroup of X. Therefore Op⁡(X)⁢H is subnormal in X. Thus Op⁡(X)⁢H is subnormal in G and so X=Op⁡(X)⁢H.

Assume that G is a Cp-group. By Theorem 7, G is p-soluble of p-length at most 1. Then Op⁡(X) is a normal Hall p′-subgroup of X. Since NX⁡(H)=H, it follows that COp⁡(X)⁡(H)=1.

Assume now that G satisfies statements (1) and (2), and let H be a p-subgroup of G. Then X=Op⁡(X)⁢H. Therefore we have that NX⁡(H)=H⁢NOp⁡(X)⁡(H)=H⁢COp⁡(X)⁡(H)=H. Hence G is a Cp-group. ∎

Proof of Theorem A.

By Lemma 5, the class of all C-groups is closed under taking epimorphic images. We shall prove next that it is closed under taking direct products. Assume this is not true and derive a contradiction. Then there exist C-groups G and H such that E=G×H is not a C-group. Let us take E of minimal order. Then E has a subgroup S such that NX⁡(S)≠S, where X=SE⁡(S).

If either S⊆G or S⊆H, then X=SG⁡(S) or X=SH⁡(S). In both cases, S is self-normalising in X, against the choice of E. Then neither G nor H contains S.

Let U and V be the projections of S onto G and H, respectively. Then we have U≠1≠V and SE⁡(S) is contained in SG⁡(U)×SH⁡(V) by [3, Theorem A.14.4].

If (x,y)∈NSE⁡(S)⁡(S)∖S with x∈SG⁡(U),y∈SH⁡(V), then we have that (s,t)(x,y)=(sx,ty)∈S and so sx∈U for all s∈U and ty∈V for all t∈V. Since G is a C-group, we have x∈U. Analogously, we obtain y∈V and so NSE⁡(S)⁡(S)⊆U×V. Moreover, there exists x′∈U such that (x′,y)∈S. Therefore (x′⁢x-1,1)∈NSE⁡(S)⁡(S)∖S. Write z=x′⁢x-1. Then z∈U.

Set A=SG⁡(U) and B=SH⁡(V). Since S(A×B)⁡(S) is subnormal in E, it follows that S(A×B)⁡(S)=SE⁡(S). The minimality of E gives E=A×B=SE⁡(S) and A=G and B=H. Let C and D be the nilpotent residuals of A and B, respectively. Then C×D is the nilpotent residual of E by [3, Lemma II.2.12]. Since (C×D)⁢S is a subnormal subgroup of E containing S, it follows that E=SE⁡(S)=(C×D)⁢S.

Clearly E is not nilpotent. Thus we can assume without loss of generality that C≠1. Let M be a minimal normal subgroup of G contained in C. Since G is soluble by Theorem 8, it follows that M is an elementary abelian p-group for some prime p. By Lemma 5, G/M is a C-group. If M*=M×1⊆E, then E/M*≅(G/M)×H is a C-group. Since (z,1)⁢M*/M*∈NE/M*⁡(S⁢M*/M*), it follows that (z,1)⁢M*∈S⁢M*/M*. Hence (z,1)∈S⁢M*∖S so that (z,1)=(m,1)⁢s, for some elements m∈M and s∈S.

Suppose that F⁡(G) is not a p-group, and let N be a minimal normal q-subgroup of G for some prime q≠p. Arguing as above, we conclude that (z,1)=(n,1)⁢t, for some elements n∈N and t∈S. Hence we have (m⁢n-1,1)=t⁢s-1∈S, giving 1≠(mq,1)=((m⁢n-1)q,1)∈S and then (z,1)=(m,1)⁢s∈S, contrary to the choice of z. Therefore F⁡(G) is a p-group.

Since G has p-length at most 1 and Op′⁡(G)=1, we have that F⁡(G) is a Sylow p-subgroup of G centralising M. Thus E/CE⁢(M*)≅G/CG⁡(M) is a p′-group. In particular, S/CS⁢(M*) is a p′-group so that M* is a semisimple module by [3, Theorem A.11.4]. Then M*=(M*∩S)×T* for some subgroup T*=T×1 normalised by S. Clearly we may assume that (z,1)∈T*. Therefore we have [(z,1),S]≤T*∩S=1. In particular, z∈CG⁡(U). Assume that U is a p-group. Then U is contained in F⁡(G) and so it is subnormal in G. Since A=G, it follows that G=U=F⁡(G), contrary to the choice of G. Consequently, U has a non-trivial Hall p′-subgroup, X say. Then U=(U∩F⁡(G))⁢X. Let D=SG⁡(X). Then F⁡(G)⁢D is a subnormal subgroup of G containing U. Therefore G=F⁡(G)⁢D. Applying [3, Lemma II.2.12], C is the nilpotent residual of D and, since CX is subnormal in G, it follows that D=C⁢X. Thus z∈ND⁡(X)∖X and X is not c-embedded in G. This final contradiction confirms our claim.

Assume now that G is a group with two normal subgroups E and F such that G/E and G/F are C-groups and E∩F=1. Then G is isomorphic to a subgroup of the direct product G/E×G/F which is a C-group by the above argument. Applying Lemma 5, we have that G is a C-group.

Assume now that G is a C-group. Then G has p-length at most 1 for all primes p. Hence G is p-soluble for all primes p so that G is soluble. ∎

Proof of Theorem B.

Assume that G is a C-group. By Theorem A, G is soluble. Let C be a Carter subgroup of a subgroup H of G. Then C is self-normalising in H and so if L is the nilpotent residual of H, we have H=L⁢C. Let X=SG⁡(H) and let N be the nilpotent residual of X. Then X contains Y=SG⁡(C). Note that H is not contained in any proper normal subgroup of X so that X=H⁢N. Hence X=N⁢C and hence C is not contained in any proper normal subgroup of X. In particular, X=Y. Since NY⁡(C)=C, it follows that C is a Carter subgroup of X.

Assume now that G is a soluble C-group such that every Carter subgroup of every subgroup is also a Carter subgroup of its subnormal closure. Let H be a subgroup of G. Then H contains a Carter subgroup C of X=SG⁡(H). Let g∈NX⁡(H). Then C and Cg are Carter subgroups of H. By [3, Theorem III.4.6], there exists h∈H such that Cg⁢h=C. Hence g⁢h∈NX⁡(C)=C. In particular, g∈H. Therefore NX⁡(H)=H and G is a C-group. ∎

Proof of Corollary 3.

Only the sufficiency of the condition is in doubt. Assume that every nilpotent subgroup of a group G is self-normalising in its subnormal closure. Applying Theorem 8, G is soluble. Let H be a subgroup of G. Arguing as in the proof of Theorem B, we have that X=SG⁡(H)=SG⁡(C) for all Carter subgroups C of H. Since C is nilpotent, it follows that NX⁡(C)=C. Hence C is also a Carter subgroups of X. By Theorem B, G is a C-group. ∎

Proof of Corollary 4.

We give a proof for the third statement. If G is a soluble PST-group, then SG⁡(H) is S-permutable for all subgroups H of G and so we have SG⁡(H)=BG⁡(H). By Theorem B, every Carter subgroup of H is also a Carter subgroup of BG⁡(H).

Conversely, let H be a subgroup of G and let C a Carter subgroup of H. Then C is a Carter subgroup of Z=BG⁡(H). If z∈NZ⁡(H), then C and Cz are Carter subgroups of H. By [3, Theorem III.4.6], there exists h∈H such that Cz⁢h=C. Hence z⁢h∈NX⁡(C)=C. In particular, z∈H. Applying [2, Theorem 13], we conclude that G is a PST-group.

Note that if G is a soluble T-group, then every subnormal subgroup is normal. Hence SG⁡(H)=BG⁡(H). If G is a soluble PT-group, then every subnormal subgroup is permutable and SG⁡(H)=AG⁡(H) by [2, Theorems 4 and 8]. Hence statements (1) and (2) follow using similar arguments. ∎