Groups with maximal subgroups of Sylow subgroups σ-permutably embedded

A subgroup H of G is said to be σ-quasinormally embedded or σ-permutably embedded in G if H is σ-full and, for every σi∈σ⁢(H), every Hall σi-subgroup of H is also a Hall σi-subgroup of some σ-permutable subgroup of G.

(i) G is said to be σ-nilpotent [7] if G=H1×⋯×Ht, where {1,H1,…,Ht} is a complete Hall σ-set of G. It is clear that every subgroup of a σ-nilpotent group is σ-permutable.

(ii) Let p>q>r be primes, where q divides p-1 and r divides q-1. Let H=Q⋊R be a non-abelian group of order qr, P a simple 𝔽p⁢H-module which is faithful for H, and G=P⋊H.

Let σ={σ1,σ2}, where σ1={p,r} and σ2={p,r}′. Then G is not σ-nilpotent and |P|>p. Since q divides p-1, PQ is supersoluble. Hence for some normal subgroup L of PQ we have 1<L<P. Then for every Hall σ1-subgroup V of G we have L≤P≤V, so L⁢V=V=V⁢L. On the other hand, for every Hall σ2-subgroup Qx of G we have Qx≤P⁢Q, so L⁢Qx=Qx⁢L. Hence L is σ-permutable in G. It is also clear that L is not normal in G, and so L⁢R≠R⁢L, which implies that L is not S-permutable in G.

(iii) Let p>q>r be primes, Cq a group of order q and H=R⋊Cq, where R is a simple 𝔽r⁢Cq-module which is faithful for Cq. Let G=P⋊H, where P is a simple 𝔽p⁢H-module which is faithful for H. Let A be a subgroup of R of order r and σ={σ1,σ2}, where σ1={p} and σ2={p}′. Then ℋ={1,P,H} is a complete Hall σ-set of G, so PA is σ-permutable subgroup of G. This means that A is σ-permutably embedded in G. Since CG⁢(P)=P, HG=1 and so A is not σ-permutable in G by [16, Theorem B and Lemma 2.6 (7)]. It is clear that P⁢Cq is a maximal subgroup of G. Therefore A is not S-permutably embedded in G, otherwise, it is not difficult to show that P⁢Cq<P⁢A⁢Cq<G, which contradicts the maximality of P⁢Cq.

Every maximal subgroup of every Sylow subgroup of G is σ-permutably embedded in G if and only if G=D⋊M, where D and M are σ-Hall subgroups of G, D=GNσ is nilpotent of odd order and every element of M induces a power automorphism on D/Φ⁢(D).

Corollary 3 (Ballester-Bolinches, Pedraza-Aguilera [3])

If every maximal subgroup of every Sylow subgroup of G is S-permutably embedded in G, then G is supersoluble.

Corollary 4 (Srinivasan [18])

If every maximal subgroup of every Sylow subgroup of G is S-permutable in G, then G is supersoluble.

Every subgroup of G is σ-permutably embedded in G if and only if G=D⋊M, where D and M are σ-Hall subgroups of G, D=GNσ is abelian of odd order and every element of M induces a power automorphism on D.

The group G is a soluble PST-group if and only if every subgroup of G is S-permutably embedded in G.

Every maximal subgroup of every Sylow subgroup of G is σ-permutable in G if and only if G=A⋊(B⋊C), where

1. A and BC are σ -Hall subgroups and C is a Hall subgroup of G,

2. A is a normal nilpotent subgroup of G of odd order, B≤Zσ⁢(G) is a normal subgroup of G and C is a σ -nilpotent subgroup of G all of whose Sylow subgroups are cyclic,

3. the generators of Sylow subgroups of C induce power automorphisms on A/Φ⁢(A) and automorphisms of order dividing a prime on A,

4. [V,a]=1 for each σi′-element a∈G, where V is the maximal subgroup of a Sylow p-subgroup P≤C and p∈σi.

Let G=(C7⋊Aut⁡(C7))×A5, where C7 is a group of order 7 and A5 is the alternating group of degree 5. Let σ={σ1,σ2}, where σ1={2,3,5} and σ2={2,3,5}′. Then G is the group in Theorem B, where A=C7, B=A5 and C=Aut⁡(C7).

Corollary 8 (Walls [20])

Every maximal subgroup of every Sylow subgroup of G is normal in G if and only if G=H⋊〈x〉, where

1. H is a normal nilpotent Hall subgroup of G,

2. the generators of Sylow subgroups of 〈x〉 induce power automorphisms on H/Φ⁢(H) and automorphisms of order dividing a prime on H.

Lemma 1 (see [16, Lemmas 2.8 and 3.2])

Let H and R≤K be subgroups of G, where H is σ-permutable in G and R is normal in G.

1. The subgroup H⁢R/R is σ -permutable in G/R.

2. If G is a σ -full group of Sylow type, then H∩K is σ -permutable in K.

3. If G is a σ -full group of Sylow type and K/R is σ -permutable in G/R, then K is σ -permutable in G.

Let H≤E and R be subgroups of G, where H is σ-permutably embedded in G and R is normal in G.

1. The subgroup H⁢R/R is σ -permutably embedded in G/R.

2. If G is a σ -full group of Sylow type, then H is σ -permutably embedded in E.

Proof.

Let {1,H1,…,Ht} be a complete Hall σ-set of H. Let W be a σ-permutable subgroup of G such that Hi is a Hall σi-subgroup of W.

(1) Since Hi is a Hall σi-subgroup of H, we conclude that |HR:RHi|=|H:Hi||R∩Hi|:|R∩H| is a σi′-number. Hence R⁢Hi/R is a Hall σi-subgroup of R⁢H/R. Then {R,H1⁢R/R,…,Ht⁢R/R} is a complete Hall σ-set of H⁢R/R. Similarly, R⁢Hi/R is a Hall σi-subgroup of R⁢W/R, where W⁢R/R is a σ-permutable subgroup of G/R by Lemma 1 (1). Hence H⁢R/R is σ-permutably embedded in G/R.

(2) By Lemma 1 (2), W∩E is σ-permutable in E and Hi≤W∩E, so Hi is a Hall σi-subgroup of W∩E. The lemma is proved. ∎

Let N be a normal σi-subgroup of G. Then N≤Zσ⁢(G) if and only if Oσi⁢(G)≤CG⁢(N).

Proof.

If Oσi⁢(G)≤CG⁢(N), then, for every chief factor H/K of G below N, both H/K and G/CG⁢(H/K) are σi-groups since G/Oσi⁢(G) is a σi-group, so N≤Zσ⁢(G).

Now assume that N≤Zσ⁢(G). Let 1=Z0<Z1<…<Zt=N be a chief series of G below N and Ci=CG⁢(Zi/Zi-1). Let C=C1∩⋯∩Ct. Then G/C is a σi-group. On the other hand, C/CG⁢(N)≃A≤A⁢u⁢t⁢(N) stabilizes the series 1=Z0<Z1<…<Zt=N, so C/CG⁢(N) is a π⁢(N)-group by [5, Theorem 0.1]. Hence G/CG⁢(N) is a σi-group, and so Oσi⁢(G)≤CG⁢(N). The lemma is proved. ∎

Lemma 4 (see [16, Lemma 2.6])

Let A and K be subgroups of a σ-full group G. Suppose that A is σ-subnormal in G. Then:

1. A∩K is σ -subnormal in K.

2. If K≤A and A is σ -nilpotent, then K is σ -subnormal in G.

3. If H≠1 is a Hall Π -subgroup of G and A is not a Π′-group, then A∩H≠1 is a Hall Π -subgroup of A.

Lemma 5 (see [16, Lemma 3.1])

Let H be a σi-subgroup of a σ-full group G. Then H is σ-permutable in G if and only if Oσi⁢(G)≤NG⁢(H).

The following statements hold.

1. Fσ⁢(G) is σ -nilpotent.

2. If A is a σ-subnormal subgroup of a σ -full group G and A is σ -nilpotent, then A is contained in Fσ⁢(G). Hence for any two σ -nilpotent σ -subnormal subgroups A and B of any σ -full group G, the subgroup 〈A,B〉 is σ -nilpotent and it is also σ -subnormal in G.

Proof.

(i) It is enough to prove that if G=A⁢B, where A and B are normal σ-nilpotent subgroups of G, then G is σ-nilpotent. Moreover, in this case, in view of [16, Proposition 2.3], it is enough to show that every chief factor H/K of G is σ-central in G. Since the hypothesis holds for G/K, it is enough to consider the case when H is a minimal normal subgroup of G. Let D=A∩B, H a σi-group and C=CG⁢(H). If H≤D, then A/C∩A≃A⁢C/C and B/C∩B≃B⁢C/C are σi-groups, so G/C=(A⁢C/C)⁢(B⁢C/C) is a σi-group. Finally, if H≤A and H≰B, then B≤C and as above we again obtain that G/C is a σi-group. This shows that H/1 is σ-central in G. Therefore in view of the Jordan–Hölder theorem, every chief factor of G is σ-central.

(ii) It is enough to consider the case when A is a σi-group for some i∈I. By hypothesis, G has a Hall σi-subgroup, say H. Then, by Lemma 4 (3), for every x∈G we have A≤Hx. Hence AG≤HG≤Fσ⁢(G). The second assertion of (ii) is a corollary of (i). ∎

Lemma 7 (see [16, Theorem B])

Let H be a subgroup of a σ-full group G. If H is σ-permutable in G, then H is σ-subnormal in G and H/HG is σ-nilpotent.

Let H be a normal subgroup of G. If H/H∩Φ⁢(G) is a Π-group, then H has a Hall Π-subgroup, say E, and E is normal in G. Hence, if the subgroup H/H∩Φ⁢(G) is σ-nilpotent, then H is σ-nilpotent.

Proof.

Let D=OΠ′⁢(H). Since H/(H∩Φ⁢(G)) is a Π-group, D≤H∩Φ⁢(G). Now since H∩Φ⁢(G) is nilpotent, D is a Hall Π′-subgroup of H. By the Schur–Zassenhaus theorem, H has a Hall Π-subgroup, say E. It is clear that H is Π′-soluble, so any two Hall Π-subgroups of H are conjugate. Now by the Frattini argument, we have G=H⁢NG⁢(E)=(E⁢(H∩Φ⁢(G)))⁢NG⁢(E)=NG⁢(E). Thus E is normal in G. The lemma is proved. ∎

Let σ0 be a partition of P such that σ0≤σ. Suppose that G has a complete Hall σ0-set H0={1,H1,…,Ht} such that every maximal subgroup of every member of H0 is σ-permutably embedded in G. If G is a σ0-full group of Sylow type, then G is σ-soluble.

Proof.

Suppose that this theorem is false and let G be a counterexample of minimal order. Then t>1 since σ0≤σ. Let σ0={σi0:i∈J⊆ℕ}. We can assume without loss of generality that Hi is a σi0-group for all i=1,…,t. Assume also that Hi is a σji-group for all i=1,…,t.

Let R be a minimal normal subgroup of G. Then the hypothesis holds for G/R by Lemma 2 (1), so G/R is σ-soluble by the choice of G. It is easy to see that the class of all σ-soluble groups is closed under taking direct products, homomorphic images and subgroups, and that the extension of a σ-soluble group by a σ-soluble group is a σ-soluble group. Hence R is the unique minimal normal subgroup of G and R is not σ-primary.

(a) For all i, either Hi≤R or R∩Hi≤Φ⁢(Hi).

Indeed, suppose that Hi≰R. Assume that R∩Hi≰Φ⁢(Hi). Let V be a maximal subgroup of Hi such that R∩Hi≰V. Then Hi=(R∩Hi)⁢V. Let W be a σ-permutable subgroup of G such that V is a Hall σji-subgroup of W. If WG=1, then W≃W/WG is σ-nilpotent and so 1<W≤Fσ⁢(G) by Lemmas 6 and 7. But then R≤Fσ⁢(G), so R is σ-primary. This contradiction shows that WG≠1. Hence R≤W and so Hi=(R∩Hi)⁢V≤W, which implies that |Hi|=|V|, a contradiction. Hence we have (a).

(b) If Hi≤R, then Hi is of prime order. Hence a Sylow 2-subgroup G2 of G is not contained in R.

Let V be a maximal subgroup of Hi and W a σ-permutable subgroup of G such that V is a Hall σji-subgroup of W. Then R≰W. Hence WG=1 and so W is σ-subnormal in G and W is σ-nilpotent by Lemma 7. But then V is σ-subnormal in G by Lemma 4 (2). Hence by Lemma 4 (3), V∩Hx=V and so V≤Hx for all x∈G. This implies that VG≤Hi. If V≠1, then R≤VG≤Hi, a contradiction. Hence V=1 and Hi is of prime order.

(c) There exists i such that Hi≰R. (Since R is not σ-primary, this follows from Claim (b) and [9, Chapter IV, Section 2.8].)

Without loss of generality we can assume that Hi≰R for all i=1,…,r, and Hj≤R for all j>r. Let Π={σ10,…,σr0} and π=⋃σi∈Πσi.

(d) Any supplement N to R in G possesses a σ-soluble Hall Π-subgroup L such that some conjugate of Hi is contained in L for all i=1,…,r. Hence R⁢L=G and 2 divides |L|.

Let L be a minimal supplement to R in G contained in N. Then L∩R≤Φ⁢(L), so L is σ-soluble since G/R≃L/L∩R is σ-soluble. Let 1≤j≤r. Then we have Hj≰R, so σj0∈σ0⁢(L). Since G is a σ0-full group of Sylow type, L possesses a Hall σj0-subgroup Lj and for some x∈G we have Lj≤Hjx. Suppose that Lj<Hjx. Then |Hjx| is not prime, so Hjx∩R≤Φ⁢(Hj) by Claims (a) and (b). Since

and, clearly, |RL:(Hjx∩L)(Hjx∩R)| is a (σj0)′-number, we have that

since (Hj∩R)x≤Φ⁢(Hjx). This contradiction shows that Lj=Hjx≤L. Since L∩R≤Φ⁢(L), π⁢(L/L∩R)=π⁢(L). Therefore π⁢(L)=π, so L is a Hall Π-subgroup of G and 2 divides |L|. Hence we have (d).

(e) We have r<t, so R is a non-abelian simple group. (This follows from Claim (d) and the choice of G.)

Let P be a Sylow 2-subgroup of R. Then P is not of prime order. Hence there is x∈G such that P≤Lx and so there is a σ-soluble Hall Π-subgroup L of G such that P≤L by Claim (d).

(f) The group R has a Hall {2,p}-subgroup for each p dividing |R|.

First assume that p∈π and let Gp be a Sylow p-subgroup of G. The Frattini argument implies that G=R⁢NG⁢(P), so NG⁢(P) possesses a σ-soluble Π-Hall subgroup L such that some conjugate of Hi is contained in L for all i=1,…,r by Claim (d). Hence for some x∈G we have Gpx≤L≤NG⁢(P). Then P⁢Gpx∩R=P⁢(Gpx∩R) is a Hall {2,p}-subgroup of R. Now assume that p∈π⁢(Hi) for some i>r. Then Hi is a Sylow subgroup of R by Claims (a) and (b), and there is a Hall Π-subgroup L of G such that L≤NG⁢(Hi) by Claim (d). But for some x we have Px≤L, so Px≤NG⁢(Hi) and hence Px⁢Hi∩R=Px⁢(Hi∩R) is a Hall {2,p}-subgroup of R.

(g) A Sylow 2-subgroup R2 of R is non-abelian.

Assume this is false. Then by Claims (e) and [10, Chapter XI, Theorem 13.7], R is isomorphic to one of the following groups:

1. PSL⁡(2,2f),

2. PSL⁡(2,q), where 8 divides q-3 or q-5,

3. the Janko group J1,

4. a Ree group.

But with respect to each of these groups it is well known that the group has no Hall {2,q}-subgroup for at least one odd prime q dividing its order (see, for example [19, Theorem 1]), which contradicts (f). Hence we have (g).

(h) If k>r and Hk is a p-group, p≠2, then p does not divide |R:NR((P′)|. Hence G=NG((P′)L for each Hall Π-subgroup L of G.

Claim (d) implies that for some element x∈G we have P≤NG⁢(Hkx). Let W=Hkx⋊P. Then from Claims (a) and (b) we have that |Hkx|=p, and so W/CW⁢(Hkx)≃P/P∩CW⁢(Hkx) is abelian. Hence Hkx≤NG⁢(P′). This implies that p does not divide |G:NG((P′)|.

Final contradiction. In view of Claim (d), there is a σ-soluble Hall Π-subgroup L of G such that P≤L. On the other hand, G=NG⁢(P′)⁢L by Claim (h). Hence

But by Claim (g), P′≠1. Hence R≤L. But then R is σ-primary. The final contradiction completes the proof. ∎

Proof of Theorem A.

Let D=G𝔑σ and ℋ={1,H1,…,Ht} a complete Hall σ-set of G. We can assume without loss of generality that Hi is a σi-group for all i=1,…,t.

Necessity. Suppose that this is false and let G be a counterexample of minimal order. Then D≠1, and so t>1.

(1) The hypothesis holds on every quotient of G. (This directly follows from Lemma 2 (1).)

(2) G is σ-soluble. (This directly follows from Theorem 1.)

(3) D is soluble.

Let R be a minimal normal subgroup of G. Then D/D∩R≃D⁢R/R=(G/R)𝔑σ is nilpotent by Claim (1) and the choice of G. Therefore R≤D and R is the unique minimal normal subgroup of G. Assume that R is non-abelian. Since G is σ-soluble by Claim (2), R is σ-primary. Let R≤Hi and Rp be a Sylow p-subgroup of R, where p∈π⁢(R). Then Rp=R∩P≰Φ⁢(P), where P is a Sylow p-subgroup of G containing Rp by the Tate theorem [9, Chapter IV, Section 4.7]. Let V be a maximal subgroup of P such that P=Rp⁢V. If P is a σi-group, then by hypothesis there is a σ-permutable subgroup W of G such that V is a Hall σi-subgroup of W. Hence P≰W, and so R≰W. It follows that WG=1. Therefore V is σ-subnormal in G by Lemma 7.

We show that V is σ-permutable in G. First note that V≤Hix for all x∈G by Lemma 4 (3), so V⁢Hix=Hix⁢V. Now let j≠i. Then V is a σ-Hall subgroup of W⁢Hjx and V is σ-subnormal in W⁢Hjx by Lemma 4 (1). Hence V is normal in W⁢Hjx by Lemma 4 (3), so V⁢Hjx=Hjx⁢V. This shows that V is σ-permutable in G. Therefore R≤NG⁢(V) by Lemma 5 since R≤D≤Oσi⁢(G), and so V∩Rp=V∩R∩P=V∩R is normal in R, which implies that V∩R=1, Thus |Rp|=p. This shows that every Sylow subgroup of R is cyclic, and so R is abelian by [9, Chapter IV, Section 2.11]. This contradiction completes the proof of (3).

(4) D is a σ-Hall subgroup of G.

Suppose that this is false and let U be a Hall σi-subgroup of D such that 1<U<Hi. Without loss of generality, we can assume that i=1. Then:

(a) Let R be a minimal normal subgroup of G contained in D. Then R=U is a Sylow p-subgroup of D for some prime p∈σ1 and a p-complement of D is a σ-Hall subgroup of G. Hence R is the unique minimal normal subgroup of G contained in D and R=H1∩D=Gp∩D, where Gp is a Sylow p-subgroup of G contained in H1.

Since D is soluble by Claim (3), R is a p-group for some prime p. Moreover, D/R=(G/R)𝔑σ is a σ-Hall subgroup of G/R by Claim (1) and the choice of G. Suppose that U⁢R/R≠1, then U⁢R/R is a Hall σ1-subgroup of G/R. If p∉σ1, then U is a Hall σ1-subgroup of G by order considerations. This contradicts the fact that U<H1. If p∈σ1, then R≤U and so U/R is a Hall σ1-subgroup of G/R. It follows that U is a Hall σ1-subgroup of G, which contradicts that U<H1. Therefore U⁢R/R=1. Consequently, U≤R and U=R. But, clearly, we have H1≰U⁢R≤D. Thus R=U=H1∩D is a Sylow p-subgroup of D. It is also clear that a p-complement of D is a σ-Hall subgroup of G.

(b) R≰Φ⁢(G), so for some maximal subgroup M of G we have G=R⋊M.

Assume that R≤Φ⁢(G). Then D≠R by Lemma 8. On the other hand, D/R is a σ1′-group by Claim (a). Hence Oσ1′⁢(D)≠1 by Lemma 8. But Oσ1′⁢(D) is characteristic in D and so it is normal G, which contradicts (a). Thus R≰Φ⁢(G). The second assertion of (b) follows from Claim (3).

(c) If G has a minimal normal subgroup L≠R, then H1=R×L and Gp=R×(L∩Gp). Hence Oσ1′⁢(G)=1.

Indeed, L≰D by Claim (a). On the other hand, D⁢L/L≃D is a σ-Hall subgroup of G/L by Claim (1) and the choice of G. Hence L≤H1 and 1<R⁢L/L≤(H1⁢L/L)∩(D⁢L/L). Consequently, H1/L≤D⁢L/L and so H1=L⁢(H1∩D)=L×R, which implies that Gp=R×(L∩Gp).

(d) CG⁢(R)=R×V, where V=CG⁢(R)∩M≤H1.

In view of Claims (3) and (b), CG⁢(R)=R×V, where V=CG⁢(R)∩M is a normal subgroup of G. By Claim (a), V∩D=1. Hence V≃D⁢V/D is σ-nilpotent. Let W be a σ1-complement of V. Then W is characteristic in V and so it is normal in G. Therefore we have (d) by Claim (c).

(e) |π⁢(H1)|>1.

Assume that H1=Gp. Claim (b) implies that R≰Φ⁢(H1). Let V be a maximal subgroup of H1 such that H1=R⁢V. Let W be a σ-permutable subgroup of G such that V is a Hall σ1-subgroup of W. Then H1≰W, so V=H1∩W. Hence R∩V=R∩H1∩W=R∩W is σ-permutable in G by [16, Theorem C]. It is clear that H1≤NG⁢(R∩V), so G=H1⁢Oσ1⁢(G)≤NG⁢(R∩V) by Lemma 5. The minimality of R implies that R∩V=1, so H1=R⋊V. First assume that WG≠1 and let L be a minimal normal subgroup of G contained in WG. Then H1=R×L by Claim (c), so |V|=|L| and hence V=L is normal in G. If WG=1, then arguing as in the proof of Claim (3), one can show that V is σ-permutable in G and so V is normal in G by Lemma 5. Hence H1=R×V is an elementary abelian p-group, where |R|=p and V is a minimal normal subgroup of G. Note that since H1≤D, we have that V≰D. The G-isomorphism D⁢V/D≃V implies that V≤Zσ⁢(G). Hence G=H1⁢Oσ1⁢(G)≤CG⁢(V), and so |V|=p. Now let R=〈a〉, V=〈b〉 and L=〈a⁢b〉. Then, arguing as above, one can get that L is normal in G. Clearly, L≰D, so in view of the G-isomorphisms D⁢L/D≃L we get that L≤Zσ⁢(G). Hence Gp=H1=V⁢L≤Zσ⁢(G). But then G/CG⁢(R) is a p-group, so G=H1. This contradiction shows that we have (e).

Final contradiction for (4). By [17, Theorem B], H1 has a complement E in G such that E⁢Gp=Gp⁢E. Let S=(Gp⁢E)𝔑σ. By Claim (e), E⁢Gp≠G. By Lemma 2 (2), the hypothesis holds for Gp⁢E, so the choice of G implies that S is a nilpotent σ-Hall subgroup of Gp⁢E. But since D⁢Gp⁢E/D≃Gp⁢E/Gp⁢E∩D is σ-nilpotent, S≤Gp⁢E∩D=(Gp∩D)⁢(E∩D)=R⁢(E∩D) by Claim (a). Then, since R<Gp, it follows that S is a p′-group. Now since R≤D≤E⁢Gp by Claim (a), S≤CG⁢(R)≤H1. Hence S≤D∩H1=R, and so V=1. Therefore E⁢Gp is σ-nilpotent and thereby E≤CG⁢(R)≤H1. Thus E=1 and so t=1, a contradiction. Hence we have (4).

(5) D is nilpotent and every maximal subgroup of every Sylow subgroup of D is normal in G.

Firstly, we show that D is nilpotent. Assume that this is false and let R be a minimal normal subgroup of G. Then R⁢D/R=(G/R)𝔑σ is nilpotent by Claim (1) and the choice of G. Hence R≤D and so R is a p-group for some prime p by Claim (3). Moreover, R is the unique minimal normal subgroup of G and for some Hi we have R≤Hi≤D by Claim (4). Clearly R≰Φ⁢(G), so R=CG⁢(R)=F⁢(G) by [4, Chapter A, Section 15.2]. Let P be a Sylow p-subgroup of G contained in Hi. Since Hi/R a nilpotent Hall subgroup of D/R, P/R is normal in D/R and hence P is normal in G. But then P=F⁢(G)=R. Let V be a maximal subgroup of P. By hypothesis, there exists a σ-permutable subgroup W of G such that V is a Hall σi-subgroup of W, so V=W∩P=W∩R is σ-permutable in G by [16, Theorem C]. Hence G=Hi⁢Oσi⁢(G)=Oσi⁢(G)≤NG⁢(V) by Lemma 5 since Hi≤D≤Oσi⁢(G). It follows that V=1, consequently |R|=p. Therefore G/CG⁢(R)=G/R is an abelian group. This implies that G is supersoluble and so D is nilpotent. Finally, note that we, in fact, have already proved that if P is a normal Sylow subgroup of G contained in D, then every maximal subgroup of P is normal in G.

(6) If p is a prime such that (p-1,|G|)=1, then p does not divide |D|. In particular, |D| is odd.

Assume that this is false. Then, by Claims (4) and (5), D has a maximal subgroup E such that |D:E|=p and E is normal in G. Then CG⁢(D/E)=G. Since D is a Hall subgroup of G, it follows that G/E=(D/E)×(M⁢E/E), where M⁢E/E≃M≃G/D is σ-nilpotent. Therefore G/E is σ-nilpotent. But then D≤E, a contradiction. Hence p does not divide |D|. In particular, |D| is odd.

(7) Every subgroup H of D satisfying Φ⁢(D)≤H is normal in G.

In view of Claim (5), Φ⁢(D)=Φ⁢(P1)×⋯×Φ⁢(Pr), where {P1,…,Pr} is the set of all different Sylow subgroups of D. Assume that for some i, Φ⁢(Pi)≠1 and let R be a minimal normal subgroup of G contained in Φ⁢(D). Then Φ⁢(D)/R=Φ⁢(D/R)≤H/R. The choice of G implies that H/R is normal in G/R. It follows that H is normal in G. Now assume that Φ⁢(Pi)=1 for all i. Then every subgroup of Pi is normal in G by Claim (5). But H=(H∩P1)×⋯×(H∩Pr), so H is normal in G.

From Claims (3)–(7) we get that the necessity holds for G, which contradicts the choice of G.

Sufficiency. Let V be a maximal subgroup of Pi, where Pi is a Sylow pi-subgroup of G. Assume that Pi≤D, without loss of generality, we may assume that i=1. Then V⁢Φ⁢(D)=V⁢Φ⁢(P1)×⋯×Φ⁢(Pr)=V×Φ⁢(P2)×⋯×Φ⁢(Pr) is normal in G by hypothesis. Clearly, V is characteristic in V⁢Φ⁢(D), so V is normal in G. Finally, suppose that Pi≰D, and let Pi is a σj-group. Then D⁢V/D is a subgroup of the σ-nilpotent group G/D, so D⁢V/D is σ-permutable in G/D. Hence DV is σ-permutable in G by Lemma 1 (3), where V is a Hall σj-subgroup of DV since D is a σ-Hall subgroup of G by hypothesis. Hence V is σ-permutably embedded in G. The theorem is proved. ∎