On the supersolubility of a group with semisubnormal factors

Example 1.1.

Let Zn be a cyclic group of order n. A group

is the product of subgroups A≃Z2×Z3 and B≃Z7⋊Z2 that are seminormal in G. But A and B are not mutually permutable since A does not permute with some order 2 subgroups of B.

Definition.

A subgroup A of a group G is called semisubnormal in G if A is subnormal in G or seminormal in G.

Lemma 2.1 ([6, Theorems 1.2, 1.4, 1.6–1.8, Corollary 3.2]).

The following statements hold.

1. The class 𝔘 is a hereditary saturated formation.

2. Every minimal normal subgroup of a supersoluble group has prime order.

3. Let N be a normal subgroup of G, and assume that G/N is supersoluble. If N is either cyclic, or N≤Z⁢(G), or N≤Φ⁢(G), then G is supersoluble.

4. Each supersoluble group has an ordered Sylow tower of supersoluble type.

5. The derived subgroup of a supersoluble group is nilpotent.

6. A group G is supersoluble if and only if every maximal subgroup of G has prime index.

Lemma 2.2 ([20, Lemma 6]).

Let G be a soluble group. Assume that G∉U, but G/K∈U for every non-trivial normal subgroup K of G. Then

1. G contains a unique minimal normal subgroup N,

   N=F⁢(G)=Op⁢(G)=CG⁢(N) for some⁢p∈π⁢(G),

2. Z⁢(G)=Op′⁢(G)=Φ⁢(G)=1,

3. G is primitive, G=N⋊M, where M is maximal in G with trivial core,

4. N is an elementary abelian subgroup of order pn, n>1,

5. if V is a subgroup G and G=V⁢N, then V=Mx for some x∈G.

Lemma 2.3 ([5, Propositions 2.2.8, 2.2.11]).

Let F and H be formations, and let K be normal in G. Then

1. (G/K)𝔉=G𝔉⁢K/K,

2. G𝔉⁢ℌ=(Gℌ)𝔉,

3. if ℌ⊆𝔉, then G𝔉≤Gℌ.

Lemma 2.4.

Let G be siding. Then the following statements hold.

1. If N is normal in G, then G/N is siding.

2. If H is a subgroup of G, then H is siding.

3. G is supersoluble.

Proof.

(1) By [18, Lemma 4.6], (G/N)′=G′⁢N/N. Let A/N be an arbitrary subgroup of (G/N)′. Then

Since A∩G′≤G′, we have A∩G′ is normal in G. Hence (A∩G′)⁢N/N is normal in G/N.

(2) Since H≤G, it follows that H′≤G′. Let A be an arbitrary subgroup of H′. Then A≤G′, and A is normal in G. Therefore, A is normal in H.

(3) We proceed by induction on the order of G. Let N≤G′ and |N|=p, where p is prime. By the hypothesis, N is normal in G. By induction, G/N is supersoluble and G is supersoluble by Lemma 2.1 (3). ∎

Remark 2.5.

By Lemma 2.4, the class of all siding groups is a hereditary homomorph. The supersoluble group G=S3×S3 ([24], IdGroup(G) = [36,10])) is not siding. Really, the derived subgroup G′=〈a〉×〈b〉 is an elementary abelian group of order 9, but the subgroup 〈a⁢b〉 of G′ is not normal in G. Moreover, all primitive quotients of G are isomorphic to either a cyclic group of order 2, or S3, hence are siding. Thus the class of all siding groups is not a Schunck class and formation.

Lemma 3.1.

The following statements hold.

1. If H is a semisubnormal subgroup of G and H≤X≤G, then H is semisubnormal in X.

2. If H is a semisubnormal subgroup of G and N is normal in G, then H⁢N/N is semisubnormal in G/N.

3. If H is a semisubnormal subgroup of G and Y is a non-empty set of elements from G, then HY=〈Hy∣y∈Y〉 is semisubnormal in G. In particular, Hg is semisubnormal in G for any g∈G.

Proof.

If H is subnormal in G, then statements (1)–(3) are true; see [14, Chapter 2]. If H is seminormal, then these statements were proved in [15, Lemmas 2 and 5]. Thus statements (1)–(3) are true. ∎

Lemma 3.2.

The following statements hold.

1. Let p be the greatest prime in π⁢(G), and let P be a Sylow p-subgroup of G. If P is semisubnormal in G, then P is normal in G.

2. If any Sylow subgroup of G is semisubnormal in G, then G is supersoluble.

3. Let H be a maximal subgroup of G. If H is semisubnormal in G, then the index of H in G is prime.

4. If every maximal subgroup of G is semisubnormal in G, then G is supersoluble.

5. If the index of H in G is prime, then H is semisubnormal in G.

Proof.

(1) It is clear that if P is subnormal in G, then P is normal in G. If P is seminormal in G and p is maximal in π⁢(G), then by [17, Lemma 4], P is normal in G.

(2) Suppose that G has at least one subnormal Sylow subgroup P. Then P is normal in G and therefore seminormal in G. Hence any Sylow subgroup of G is seminormal in G. By [17, Corollary 6], G is supersoluble.

(3) If H is subnormal in G, then H is normal in G and |G:H| is prime. Let H be seminormal in G, and let K be a subgroup of G such that H⁢K=G and H⁢K1 is a subgroup of G for every subgroup K1 of K. Let r be a prime which divides the index |G:H|, and let R be a Sylow r-subgroup of K. Then H⁢R=G and G=H⁢〈x〉 for x∈R∖H. We choose an element x of minimal order. Then H⁢〈xr〉=〈xr〉⁢H=H and |G:H|=r.

(4) Let M be a maximal subgroup of G. By (3), the index of M in G is prime. By Lemma 2.1 (6), G is supersoluble.

(5) Let |G:H|=r, and let R be a Sylow r-subgroup of G. Then R is not contained in H and there exists an element x∈∖⁡RH. Let

It is obvious that a>a1; hence

Now xr belongs to H and H is seminormal in G, and therefore is semisubnormal in G. ∎

Example 3.3.

A group with seminormal 2-maximal subgroups is supersoluble; see [23]. A group with subnormal 2-maximal subgroups can be non-supersolvable. Any non-supersoluble Schmidt group (a non-nilpotent group whose proper subgroups are all nilpotent) confirms this fact. Such a group G=P⋊Q, where |P|=pm, |Q|=q and m>1, is the order of p modulo q, for example, A4. Since P is a minimal normal subgroup of G, it follows that P and Qy, y∈P, are all maximal subgroups of G. It is clear that all 2-maximal subgroups are subnormal in G. Hence a group with semisubnormal 2-maximal subgroups can be non-supersolvable.

Lemma 3.4.

The following statements hold.

1. If A is a semisubnormal 2 -nilpotent subgroup of G, then AG is soluble.

2. Let p be the smallest prime divisor of order of G. If A is semisubnormal in G and p does not divide the order of A, then p does not divide the order of AG.

Proof.

(1) If A is subnormal in G, then by [18, Theorem 5.31], AG is soluble. If A is seminormal in G, then AG is soluble by [15, Lemma 10].

(2) If A is a subnormal p′-subgroup of G, then by [18, Theorem 5.31], AG is a p′-subgroup. If A is a seminormal p′-subgroup of G, then AG is a p′-subgroup by [15, Lemma 11]. ∎

Lemma 3.5.

Let G be soluble. If G has a subgroup H of prime index, then G/HG is supersoluble.

Proof.

Suppose that H is not normal in G. Then HG≠H and G/HG is primitive with primitivator H/HG. By [18, Theorem 4.42],

Let |G:H|=p, where p is prime. Then

The subgroup H/HG is cyclic, as the automorphism group of P/HG of prime order. Hence G/HG is supersoluble. If H is normal in G, then H=HG and G/HG is supersoluble. ∎

Lemma 3.6.

Let A be a seminormal subgroup of a soluble group G, and let r be the greatest prime in π⁢(G). If A is r-closed, then Ar is subnormal in G.

Proof.

Let Y be a supersupplement to A in G, and let X be a maximal subgroup of Y of prime index. By hypothesis, A permutes with X. Then by induction, Ar is subnormal in AX. Since AX is a subgroup of G of prime index, it follows that, by Lemma 3.5, G/(A⁢X)G is supersoluble. Let |G:AX|=t. If t=r, then A⁢X/(A⁢X)G is a r′-group. Hence Ar≤(A⁢X)G. Since Ar is subnormal in AX, we have Ar is subnormal in (A⁢X)G and therefore is subnormal in G. If t≠r, then t<r and G/(A⁢X)G is a r′-group. Thus Ar≤(A⁢X)G and Ar is subnormal in G. ∎

Lemma 3.7.

Let A and B be semisubnormal in G and G=A⁢B. If A and B have an ordered Sylow tower of supersoluble type, then G has an ordered Sylow tower of supersoluble type.

Proof.

We proceed by induction on |G|. Since A is 2-nilpotent, it follows, by Lemma 3.4 (1), that AG is soluble and G=AG⁢B is soluble. Let r∈π⁢(G) with r maximal. It is clear that a Sylow r-subgroup Ar is normal in A. Let A be seminormal in G. Then by Lemma 3.6, Ar is subnormal in G. If A is subnormal in G, then Ar is subnormal in G. Similarly, a Sylow r-subgroup Br of B is subnormal in G. Since R=Ar⁢Br is a Sylow subgroup of G, we have that G is r-closed. The subgroups A⁢R/R≃A/A∩R and B⁢R/R≃B/B∩R are semisubnormal in G/R=(A⁢R/R)⁢(B⁢R/R) and have an ordered Sylow tower of supersoluble type. By induction, G/R has an ordered Sylow tower of supersoluble type; hence G has an ordered Sylow tower of supersoluble type. ∎

Theorem A.

Let A and B be semisubnormal subgroups of G and G=A⁢B. If A is nilpotent and B is supersoluble, then G is supersoluble.

Proof.

If A is subnormal in G, then AG is the nilpotent normal subgroup of G. Hence, in the factorization G=A⁢B, we can replace the subgroup A by the nilpotent normal subgroup AG. Further, we assume that A is seminormal in G. Let Y be a supersupplement to A in G.

Assume that the claim is false, and let G be a minimal counterexample. If N is a non-trivial normal subgroup of G, then the subgroups A⁢N/N and B⁢N/N are semisubnormal in G/N by Lemma 3.1 (2), A⁢N/N≃A/A∩N is nilpotent and B⁢N/N≃B/B∩N is supersoluble. Then by induction,

is supersoluble. By Lemmas 2.1 (4) and 3.7, G has an ordered Sylow tower of supersoluble type, and therefore we apply Lemma 2.2. We keep for G the notation of this lemma; in particular, N=Gp is the Sylow p-subgroup for the greatest p∈π⁢(G). Since G=A⁢B, it follows that N=Ap⁢Bp, where Ap and Bp are Sylow p-subgroups of A and B, respectively; see [13, Theorem VI.4.6].

Suppose that Ap=1. Then N=Bp≤B. We choose a minimal normal subgroup N1 of B such that N1≤N. Since B is supersoluble, we have |N1|=p by Lemma 2.1 (2). Since N1≤N≤Y, it follows that there exists a subgroup A⁢N1=[N1]⁢A and N1 is normal in G; this contradicts Lemma 2.2 (4). Thus the assumption Ap=1 is false and Ap≠1.

Assume that Bp=1. Hence N=Ap≤A and N=A by Lemma 2.2 (1). Then B∩N=1 and B is maximal in G. Since B is semisubnormal in G, we have, by Lemma 3.2 (3), the index of B in G is prime; this contradicts Lemma 2.2 (4).

Let Y1 be a Hall p′-subgroup of Y. Then A⁢Y1 is a subgroup of G and Ap is normal in A⁢Y1. Since N is abelian, a Sylow p-subgroup Yp of Y centralizes Ap and Ap is normal in G; hence Ap=N. Because A is nilpotent and by Lemma 2.2 (1), it follows that A=N. Since B is supersoluble, we have Bp is normal in B. In this case, Bp is normal in N=A and therefore is normal in G. Thus Bp=N and G=A⁢B=N⁢B=B is supersoluble, a contradiction. The theorem is proved. ∎

Corollary 4.1.

Let A and B be subgroups of G and G=A⁢B. Suppose that A is nilpotent and B is supersoluble. Then G is supersoluble in each of the following cases.

1. A and B are mutually permutable; see [2, Theorem 3.2].

2. A and B are subnormal in G; see [20, Lemma 10].

3. A and B are seminormal in G.

4. One of the subgroups A or B is seminormal in G; the other is subnormal in G.

5. The indices of A and B in G are prime; see [21, Theorem A].

6. One of the subgroups A or B is semisubnormal in G; the index of the other in G is prime.

Corollary 4.2.

If G=A⁢B, A is a supersoluble semisubnormal subgroup of G and B is a normal siding subgroup of G, then G is supersoluble.

Proof.

We use induction on the order of G. If N is a non-trivial normal subgroup of G, then A⁢N/N is semisubnormal in G/N by Lemma 3.1 (2) and is supersoluble, B⁢N/N is a normal siding subgroup of G/N by Lemma 2.4 (1). By induction, G/N=(A⁢N/N)⁢(B⁢N/N) is supersoluble.

Let A be seminormal in G, and let U be a supersupplement to A in G. Since G is soluble, then U has a subgroup U1 of prime index, and hence M=A⁢U1 is a subgroup of prime index in G. If A is subnormal in G, then the segment of a subnormal series between the subgroups A and G can be compacted to a composition series. Because G is soluble, it follows that there exists a maximal subgroup M of G of prime index such that A≤M. By Dedekind’s identity, M=A⁢(M∩B). Since A is semisubnormal in M by Lemma 3.1 (1) and M∩B is a normal siding subgroup of M, we have that M is supersoluble by induction. If B is nilpotent, then by Theorem A, G is supersoluble. Hence B is non-nilpotent and B′≠1.

Let N be a minimal normal subgroup of G such that N≤B′. If N is not contained in M, then G=N⋊M and |N| is prime. By Lemma 2.1 (3), G is supersoluble. Suppose that N is contained in M and N1 is a subgroup of N of prime order such that N1 is normal in M. Then N1 is normal in B and therefore is normal in G. By Lemma 2.1 (3), G is supersoluble. ∎

Example 4.3.

The non-supersoluble group

([24], IdGroup = [216,157]) is the product of a normal supersoluble subgroup A≃S3×S3 and a subnormal siding subgroup

A subgroup that is isomorphic to Z4 is a supplement to B. Therefore, in Corollary 4.2, the condition of normality of the siding factor cannot be weakened to subnormality and seminormality.

Corollary 4.4.

Let A and B be subgroups of G, and let G=A⁢B. Suppose that A is supersoluble and that B is normal and siding. Then G is supersoluble in each of the following cases.

1. A is subnormal in G; see [22, Corollary 3.3].

2. A is normal in G.

3. A is normal in G, and B is a soluble t-group; see [19, Theorem 3].

4. A is seminormal in G.

5. The indices of A and B in G are prime; see [21, Theorem B].

Theorem B.

Let A and B be semisubnormal supersoluble subgroups of G, and let G=A⁢B. If the derived subgroup G′ is nilpotent, then G is supersoluble.

Proof.

Assume that the claim is false, and let G be a minimal counterexample. If N is a non-trivial normal subgroup of G, then the subgroups A⁢N/N and B⁢N/N are semisubnormal in G/N by Lemma 3.1 (2) and are supersoluble. Since

it follows that the derived subgroup (G/N)′ is nilpotent. Consequently, G/N satisfies the hypothesis of the theorem, and by induction, G/N is supersoluble. By Lemmas 2.1 (4) and 3.7, G has an ordered Sylow tower of supersoluble type, and therefore we apply Lemma 2.2. We keep for G the notation of this lemma; in particular, N=Gp is the Sylow p-subgroup for the greatest p∈π⁢(G). By hypothesis, G′ is nilpotent; hence N=G′ and G/N is abelian.

Suppose that A⁢N=G. Then A∩N=1, and A is a maximal subgroup of G. Since A is semisubnormal in G, then by Lemma 3.2 (3), the index of A in G is prime; this contradicts Lemma 2.2 (4). Thus the assumption is false, and A⁢N<G. By Lemma 3.1 (1), A is semisubnormal in AN, and AN is supersoluble by induction. Similarly, we have B⁢N<G and BN is supersoluble. Now G=(A⁢N)⁢(B⁢N) is the product of normal supersoluble subgroups AN and BN, and G′ is nilpotent. By Baer’s theorem [3], G is supersoluble. The theorem is proved. ∎

Corollary 4.5.

Let A and B be supersoluble subgroups of G, and let G=A⁢B. If G′ is nilpotent, then G is supersoluble in each of the following cases.

1. A and B are mutually permutable; see [2, Theorem 3.8].

2. A and B are subnormal in G; see [20, Theorem 3].

3. A and B are seminormal in G.

4. One of the subgroups A or B is seminormal in G; the other is subnormal in G.

5. The indices of A and B in G are prime; see [21, Corollary 3.6].

6. One of the subgroups A or B is semisubnormal in G; the index of the other in G is prime.

Corollary 4.6.

Let A and B be semisubnormal supersoluble subgroups of G with relatively prime indices in G. If G is metanilpotent, then G is supersoluble.

Proof.

Since (|G:A|,|G:B|)=1, we have G=A⁢B. We use induction on the order of G. From the proof of Theorem B, we obtain that G=N⋊M is a primitive group, where M is a maximal subgroup of G and N=F⁢(G)=Gp is a unique minimal normal subgroup of G, where p is the greatest in π⁢(G). Besides, AN and BN are proper subgroups of G. The subgroups A and N are semisubnormal in AN and supersoluble; moreover, (|AN:A|,|AN:N|)=1. Since AN is metanilpotent, by induction, AN is supersoluble. Similarly, BN is supersoluble. Since G/N is nilpotent, AN and BN are subnormal in G. Now G=(A⁢N)⁢(B⁢N) is the product of subnormal supersoluble subgroups AN and BN such that its indices in G are relatively prime. By [20, Corollary 3.1], G is supersoluble. ∎

Corollary 4.7.

Let A and B be semisubnormal supersoluble subgroups of G with relatively prime indices in G. If |π⁢(G)|≤2, then G is supersoluble.

Proof.

By Lemma 3.7, G has an ordered Sylow tower of supersoluble type. By hypothesis, |π⁢(G)|≤2; hence G is metanilpotent. By Corollary 4.6, G is supersoluble. ∎

Example 4.8.

Huppert [12] and Baer [3] gave the first examples of non-supersoluble groups that are products of two normal supersoluble subgroups. The groups in these examples are metanilpotent and have orders 23⋅52 and 23⋅p2. Hence we cannot omit the condition (|G:A|,|G:B|)=1 in Corollaries 4.6 and 4.7.

Theorem C.

Let A and B be semisubnormal supersoluble subgroups of G and G=A⁢B. Then

1. G𝔘=(G′)𝔑,

2. if (|G:A|,|G:B|)=1, then G𝔘=G𝔑2∩𝔅⁢(G)=(G′)𝔑.

Proof.

(1) If G is supersoluble, then G𝔘=1 and G′ is nilpotent by Lemma 2.1 (5). Consequently, (G′)𝔑=1=G𝔘, and the statement is true. Further, we assume that G is non-supersoluble. By Lemmas 2.1 (4) and 3.7, G has an ordered Sylow tower of supersoluble type. Since 𝔘⊆𝔑⁢𝔄, we have

by Lemma 2.3 (2) and (3). Next we check the reverse inclusion. For this, we prove that G/(G′)𝔑 is supersoluble. The derived subgroup

is nilpotent. The quotients

hence the subgroups A⁢(G′)𝔑/(G′)𝔑 and B⁢(G′)𝔑/(G′)𝔑 are supersoluble, and by Lemma 3.1 (2), these subgroups are semisubnormal in G/(G′)𝔑. By Theorem B, G/(G′)𝔑 is supersoluble. Thus G𝔘≤(G′)𝔑, and (1) is proved.

(2) First we prove that G𝔘=G𝔑2∩𝔅⁢(G). It is obvious that all quotients of G satisfy the hypothesis of the theorem. Since G/G𝔑2∈𝔑2, it follows that G/G𝔑2 is supersoluble by Corollary 4.6. Hence G𝔘≤G𝔑2. Since G/G𝔖{p,q}∈𝔖{p,q}, we have G/G𝔖{p,q} is supersoluble by Corollary 4.7. By Remak’s lemma [18, Lemma 2.33], G/𝔅⁢(G) is isomorphic to a subgroup of direct product

Thus G/𝔅⁢(G) is supersoluble, and G𝔘≤G𝔑2∩𝔅⁢(G).

Next we prove the reverse inclusion. By Lemma 2.1 (5), every supersoluble group is metanilpotent. Hence 𝔘⊆𝔑2 and G𝔑2≤G𝔘 by Lemma 2.3 (3). Therefore,

So the equality G𝔘=G𝔑2∩𝔅⁢(G) is proved. By (1), we have

The theorem is proved. ∎

Theorem D.

Let A and B be subgroups of G and G=A⁢B. If all Sylow subgroup of A and of B are semisubnormal in G, then G is supersoluble.

Proof.

We use induction on the order of G. Let t be the smallest in π⁢(G). We consider r≠t. By Lemma 3.4 (2), t does not divide the order of ArG. Similarly, t does not divide the order of BrG, and the subgroup

is a normal t′-subgroup of G, i.e. H=Gt′. In particular, G is soluble. By [13, Theorem VI.4.6], Gt′=At′⁢Bt′. By induction, Gt′ is supersoluble; hence G has an ordered Sylow tower of supersoluble type.

Let N be a non-trivial proper normal subgroup of G. Then

Let S/N be a Sylow s-subgroup of A⁢N/N, and let T be a Sylow s-subgroup of S∩A. Then T⁢N/N=S/N, and T is a Sylow s-subgroup of A.

Since T is semisubnormal in G, by Lemma 3.1 (2), T⁢N/N=S/N is semisubnormal in G/N. Similarly, if K/N is a Sylow subgroup of B⁢N/N, then K/N is semisubnormal in G/N. By induction, G/N is supersoluble. By Lemma 2.2, G is primitive, and we use for G the notation of this lemma. In particular, N=Gp, and p is the greatest in π⁢(G).

Let N1≤N and |N1|=p. Since M is a Hall p′-subgroup of G, we have M=Ap′⁢Bp′.

Let π⁢(Ap′)=π1∪π2. If r∈π1, then Ar is seminormal in G, and if r∈π2, then Ar is subnormal in G. It is obvious that π1 and π2 can be chosen so that π1∩π2=∅. Suppose that r∈π1. Then Ar is seminormal in G, and there is a subgroup U such that G=Ar⁢U and Ar permutes with every subgroups of U. Because N≤U, it follows that Ar permutes with N1. Since it is true for any r∈π1, we have Aπ1 permutes with N1. Let r∈π2. Then Ar is subnormal in G, and (Ar)G is a r-group. Hence N≤(Ar)G. It is impossible because p≠r. Thus π2=∅ and π⁢(Ap′)=π1. Therefore, Ap′ permutes with N1.

Similarly, Bp′ permutes with N1. Hence M permutes with N1. Now M⁢N1 is a subgroup of G, and N1 is normal in M⁢N1. Since N is abelian, then N1 is normal in N⁢M=G; this contradicts |N|>p. The theorem is proved. ∎

Example 4.9.

The alternating group G=A4 of degree 4 is a product of subgroups A=Z3 and B=Z2×Z2. It is clear that all maximal subgroups of A and of B are semisubnormal in G. But G is non-supersoluble.

Lemma 5.1 ([13, Theorem VI.9.1]).

The following statements hold.

1. The class p⁢𝔘 is a hereditary saturated formation.

2. Each minimal normal subgroup of a p-supersoluble group is either a p′-subgroup or a group of order p. In particular, the p-rank of a p-supersoluble group is equal to 1.

3. Let N be a normal subgroup of G and G/N∈p⁢𝔘. If N is cyclic or

   N∈{Z⁢(G),Op′⁢(G),Φ⁢(G)},

   then G∈p⁢𝔘.

4. The derived subgroup of a p-supersoluble group is p-nilpotent.

Lemma 5.2 ([19, Lemma 4]).

Let G be a p-supersoluble group, let P be a Sylow p-subgroup of G, and let H be a Hall p′-subgroup of G. If Op′⁢(G)=1, then the following statements hold.

1. P is normal in G and F⁢(G)=P.

2. If Φ⁢(G)=1, then P=P1×P2×…×Pt, where Pi is a normal subgroup of G of prime order for any i. In particular, P is elementary abelian.

3. H is abelian of exponent dividing p-1.

4. G is supersoluble.

Lemma 5.3 ([19, Lemma 5]).

Suppose that a p-soluble group G does not belong to p⁢U, but G/K∈p⁢U for every non-trivial normal subgroup K of G. Then the following hold.

1. Z⁢(G)=Op′⁢(G)=Φ⁢(G)=1.

2. G has a unique minimal normal subgroup N, N=F⁢(G)=Op⁢(G)=CG⁢(N).

3. G is a primitive and G=N⋊M, where M is a maximal subgroup of G with trivial core.

4. N is an elementary abelian group of order pn, n>1.

5. If M is abelian, then M is cyclic of order dividing pn-1, and n is the smallest positive integer such that pn≡1(mod|M|).

Lemma 5.4 ([19, Corollary 1.1]).

Let A and B be normal p-supersoluble subgroups of G, and let G=A⁢B. If G′ is p-nilpotent, then G is p-supersoluble.

Theorem E.

Let A and B be semisubnormal subgroups of a p-soluble group G and G=A⁢B. If A is p-nilpotent and B is p-supersoluble, then G is p-supersoluble.

Proof.

We use induction on the order of G. Let N be a non-trivial normal subgroup of G. The quotients

hence A⁢N/N is p-nilpotent and B⁢N/N is p-supersoluble. By Lemma 3.1 (2), these subgroups are semisubnormal in G/N. Consequently, G/N satisfies the hypothesis of the theorem, and by induction, G/N is p-supersoluble. By Lemma 5.3, G has a unique minimal normal subgroup N, N=F⁢(G)=Op⁢(G)=CG⁢(N) and N is an elementary abelian group of order pn, n>1.

Suppose that A⁢N=G. Then A∩N=1, and A is a maximal subgroup of G. By hypothesis, A is semisubnormal in G, and by Lemma 3.2 (3), the index of M in G is prime; this contradicts |N|>p. Thus the assumption is false, and A⁢N<G. Similarly, B⁢N<G.

The subgroups A and N are semisubnormal in AN, A is p-nilpotent, and N is p-supersoluble. By induction, AN is p-supersoluble. Similarly, BN is p-supersoluble since N and B are semisubnormal in BN, N is p-nilpotent and B is p-supersoluble.

Since Op′⁢(G)=1 and N=CG⁢(N)=Op⁢(G), we have

By Lemma 5.2, AN and BN are supersoluble and p-closed; besides, its Hall p′-subgroups are abelian. Hence A is nilpotent, and B is supersoluble. By Theorem A, G is supersoluble. ∎

Theorem F.

Let A and B be semisubnormal p-supersoluble subgroups of a p-soluble group G and G=A⁢B. If G′ is p-nilpotent, then G is p-supersoluble.

Proof.

By induction and by Lemma 5.3, we obtain that G=N⋊M is a primitive group, where M is a maximal subgroup of G and N=F⁢(G)=Op⁢(G) is a unique minimal normal subgroup of G for some prime p∈π⁢(G). By hypothesis, the derived subgroup of G is p-nilpotent. Since Op′⁢(G)=1, it follows that G′ is a p-group. Thus N=G′ and G/N is abelian.

Besides, AN and BN are proper subgroups of G. The subgroups A and N are semisubnormal in AN and p-supersoluble. Since (A⁢N)′ is p-nilpotent, by induction, AN is p-supersoluble. Similarly, BN is p-supersoluble. Since G/N is abelian, we have AN and BN are normal in G. Now G=(A⁢N)⁢(B⁢N) is the product of normal p-supersoluble subgroups AN and BN. By Lemma 5.4, G is p-supersoluble, a contradiction. ∎

Theorem G.

Let G=A⁢B be a p-soluble group, and let A and B be semisubnormal p-supersoluble subgroups of G. Then Gp⁢U=(G′)Ep′⁢Np.

Proof.

If G is p-supersoluble, then we have Gp⁢𝔘=1 and G′ is p-nilpotent by Lemma 5.1 (4). Consequently, Gp⁢𝔘=1=(G′)𝔈p′⁢𝔑p, and the statement is true. Further, we assume that G is non-p-supersoluble.

Since p⁢𝔘⊆𝔈p′⁢𝔑p⁢𝔄, we have

by Lemma 2.3 (2) and (3). Next we check the reverse inclusion. For this, we prove that G/(G′)𝔈p′⁢𝔑p is p-supersoluble. The derived subgroup

is p-nilpotent. The quotients