On a theorem of Ito and Szép

Definition 1.1 ([8, Definition 1.1]).

Let H,T be subgroups of G.

1. H and T are said to be conditionally permutable (or in brevity, c-permutable) in G if, for some x∈G, we have H⁢Tx=Tx⁢H.

2. H is said to be c-permutable in G if H is c-permutable with all subgroups of G.

Definition 1.2 ([8, Definition 1.2]).

Let H,T be subgroups of G.

1. H and T are said to be completely c-permutable in G if H and T are c-permutable in 〈H,T〉.

2. H is said to be completely c-permutable in G if, for every subgroup T of G, the subgroups H and T are c-permutable in 〈H,T〉.

Problem ([7, Chapter 2, Problem 6.5]).

Let H be a completely c-permutable subnormal subgroup of a group G. Is H/HG nilpotent then?

Theorem 1.3.

Let H be a completely c-permutable subnormal subgroup of G. Then H/HG is nilpotent.

Corollary 1.4 ([12, Theorem 1]).

If H is a quasinormal subgroup of G, then H/HG is nilpotent.

Theorem 1.5.

If G is p-soluble and every 2-maximal subgroup having a composite index in G which is divisible by p is completely c-permutable in G, then G is p-supersoluble.

Corollary 1.6 ([8, Theorem 3.10]).

If G is soluble and every 2-maximal subgroup having a composite index in G is completely c-permutable in G, then G is supersoluble.

Lemma 2.1.

Let K⁢⊴⁢G and H≤G. Then the following statements hold:

1. if H is completely c-permutable in G, then H⁢K/K is completely c-permutable in G/K;

2. if T≤M≤G, H≤M and H is completely c-permutable with T in G, then H is completely c-permutable with T in M;

3. if H≤M and H is completely c-permutable in G, then H is completely c-permutable in M.

Proof.

(1) is from [8, Corollary 2.2 (i)], and (2) is [8, Lemma 2.1 (iii)]. Clearly, (3) directly follows from (2). ∎

Lemma 2.2.

Let A≤K≤G, B≤G.

1. Suppose that A is normal in G. Then K/A is subnormal in G/A if and only if K is subnormal in G (see [5, Chapter A, Lemma 14.1]).

2. If A is subnormal in G, then A∩B is subnormal in B (see [5, Chapter A, Lemma 14.1]).

3. If A is subnormal in G and B is a Hall π -subgroup of G, then A∩B is a Hall π -subgroup of A (see [22]).

4. If A is subnormal in G and the index |G:A| is a p′-number, then A contains all Sylow p-subgroups of G (see [22]).

5. If A and B are subnormal in G, then 〈A,B〉 is subnormal in G (see [5, Chapter A, Lemma 14.4]).

6. If A is subnormal in G and |G:A| is a p-number, then Op⁢(G)≤A (see [6, VIII, Proposition 1.2 (2)]).

Lemma 2.3 (see [7, Chapter 1, Lemma 4.14]).

Let A and B be proper subgroups of G such that G=A⁢B.

1. Gp=Ap⁢Bp for some Sylow p-subgroups Gp, Ap and Bp of G, A,B, respectively.

2. G=Ax⁢B and G≠A⁢Ax for all x∈G.

Lemma 2.4 (see [5, Chapter A, Lemma 1.6]).

Let H, K and N be subgroups of G. If H⁢K=K⁢H and H⁢N=N⁢H, then H⁢〈K,N〉=〈K,N〉⁢H.

Lemma 2.5 (see [15, Lemma 5]).

If H, K and N are pairwise permutable subgroups of G and H is a Hall subgroup of G, then N∩H⁢K=(N∩H)⁢(N∩K).

Proof.

Assume that this theorem is false, and let (G,H) be a counterexample with |G|+|H| as small as possible. Then, clearly, H<G. Let R=H𝔑 be the nilpotent residual subgroup of H, that is, R is the intersection of all normal subgroups N of H such that H/N is nilpotent. We proceed the proof by the following steps.

(1) HG=1. If HG≠1, then, by Lemma 2.1 (1), H/HG is completely c-permutable in G/HG, and by Lemma 2.2 (1), H/HG is subnormal in G/HG. So we have that (G/HG,H/HG) satisfies the hypothesis. The choice of (G,H) implies that H/HG is nilpotent, a contradiction. Hence (1) holds.

(2) There exists an element x∈G such that G=〈H,x〉. Assume that this is false. Then, for every element x of G, we have that M=〈H,x〉<G. Hence, by Lemma 2.1 (3), we know that H is completely c-permutable in M, and by Lemma 2.2 (2), we have that H is also subnormal in M. This shows that (M,H) satisfies the hypothesis. The choice of (G,H) implies that H/HM is nilpotent. Hence R≤HM≤H, which shows that Rx≤HMx=HM≤H. Consequently, RG=〈Rx∣x∈G〉≤H. Hence, by (1), we have that R=1, and so H is nilpotent, a contradiction. Hence (2) holds.

Let X=〈x〉.

(3) If there exists an element g of G such that H⁢Xg=G, then H is nilpotent. Since G=H⁢Xg and clearly H and Xg are all proper subgroups of G, we have that G=H⁢X by Lemma 2.3 (2). Let π⁢(G)={p1,p2,…,pn}. Then we claim that π⁢(X)={p1,p2,…,pn}. In fact, if pi∤|X| for some i, then the Sylow pi-subgroup Hpi of H is also a Sylow pi-subgroup of G. Since H is subnormal in G, we have that, for every element g of G, Hpig∩H is also a Sylow pi-subgroup of H by Lemma 2.2 (3). This shows that Hpig≤H, and so HpiG≤H, which contradicts (1). Hence π⁢(X)={p1,p2,…,pn}. Now let Xpi be the Sylow pi-subgroup of X, where i=1,2,…,n, and Ki=H⁢Xpi.

(a) For every subgroup X1 of X, we have that H⁢X1 is a subgroup of G, and so Ki is a subgroup of G for every i=1,2,…,n. Let X1 be any subgroup of X. Since H is completely c-permutable in G, there exists an element y=a⁢h such that H⁢X1y=X1y⁢H, where a∈X and h∈H. It follows from the fact that X=〈x〉 is cyclic that H⁢X1=X1⁢H, that is, H⁢X1 is a subgroup of G for every subgroup X1 of X. Hence (a) holds.

(b) Ki⁢Kj=Kj⁢Ki for every 1≤i,j≤n. If i=j, then this is clear. If i≠j, then Ki⁢Kj=H⁢Xpi⁢H⁢Xpj=H⁢Xpi⁢Xpj=H⁢Xpj⁢Xpi=H⁢Xpj⁢H⁢Xpi=Kj⁢Ki. So (b) holds.

Let Li=K1⁢K2⁢…⁢Ki-1⁢Ki+1⁢…⁢Kn, where i=1,2,…,n. Then, by (b), Li is a subgroup of G, and clearly, |G:Li| is a pi-number.

(c) L1∩L2∩⋯∩Ln=H. First we claim that Ki∩Li=H. Note that

where X1 is the Hall pi′-subgroup of X. By the Dedekind identity,

Since

divides |H⁢X1:H|, we have that |Xpi∩H⁢X1:Xpi∩H| is a pi′-number. But, as Xpi∩H⁢X1 is a pi-group, this implies that Xpi∩H⁢X1=Xpi∩H. Then we have Ki∩Li=H⁢(Xpi∩H⁢X1)=H.

Since

and

we have that L1∩L2∩⋯∩Ln-1=Kn by induction. So

Hence (c) holds.

(d) Li is subnormal in G for every i=1,2,…,n. Since H is subnormal in G, there exists a subnormal subgroup chain

But, as G=H⁢X, we get Hi=H⁢(Hi∩X). We claim that Hi⁢Xp is a subgroup of G, where Xp is the Sylow p-subgroup of X, p∈{p1,…,pn}, i=0,1,…,m. In fact, since Xp⁢H=H⁢Xp by (a) and X is cyclic, Xp⁢(Hi∩X)=(Hi∩X)⁢Xp. So, by Lemma 2.4, we know that Xp⁢Hi=Hi⁢Xp, that is, Hi⁢Xp is a subgroup of G.

Next we claim that Hi-1⁢Xp is normal in Hi⁢Xp, where i=1,2,…,m. Since Hi=H⁢(Hi∩X), we have that g=h⁢y for any element g∈Hi, where h∈H and y∈Hi∩X. From H≤Hi-1, it follows that H≤NG⁢(Hi-1⁢Xp). Hence (Hi-1⁢Xp)g=(Hi-1⁢Xp)h⁢y=(Hi-1⁢Xp)y=Hi-1y⁢Xp. But Hi-1 is normal in Hi, and y∈Hi∩X, so (Hi-1⁢Xp)g=Hi-1y⁢Xp=Hi-1⁢Xp. This implies that Hi-1⁢Xp is normal in Hi⁢Xp. Thus G has a subnormal subgroup chain

This shows that Ki is subnormal in G. Therefore, by Lemma 2.2 (5), Li is subnormal in G for every i=1,2,…,n.

(e) H is nilpotent. Since |G:Li| is a pi-number and Li is subnormal in G by (d), we have that Opi⁢(G)≤Li by Lemma 2.2 (6). Hence

by (c), so, by (1), Op1⁢(G)∩Op2⁢(G)∩⋯∩Opn⁢(G)=1. Since Ki=H⁢Xpi, H is subnormal in Ki by Lemma 2.2 (2). Then, by Lemma 2.2 (6), we get that Opi⁢(Ki)≤H, which shows that H/Opi⁢(Ki) is a pi-subgroup, and so is nilpotent. Hence R≤Opi⁢(Ki), where i=1,2,…,n. It follows that

and so H is nilpotent. Hence (3) holds.

(4) If there exists an element g of G such that T=H⁢Xg=Xg⁢H is a proper subgroup of G, then H is nilpotent. Let Y=Xg.

First we assume that Y is a p-group. By Lemma 2.2 (2), H is subnormal in T, also by Lemma 2.1 (3), we have that H is completely c-permutable in T. This implies that (T,H) satisfies the hypothesis. Then, as T=H⁢Y is a proper subgroup of G, H/HT is nilpotent by the choice of (G,H), so R≤HT. Since H is subnormal in G, there exists a proper normal subgroup M of G such that H≤M. Then, by (2), G=〈H,x〉=M⁢X, and so G=M⁢Y by Lemma 2.3. Also, by Lemma 2.1 (3), we know that H is completely c-permutable in M, and by Lemma 2.2 (2), H is subnormal in M. Hence (M,H) satisfies the hypothesis, which shows that H/HM is nilpotent by the choice of (G,H). It follows that R≤HM. If HM=1 or HT=1, then R=1, which implies that H is nilpotent, a contradiction. Hence HM≠1 and HT≠1. Since

we have HM≤TG. By Lemma 2.2 (6), Op⁢(TG)≤Op⁢(T)≤H. Then, by (1), Op⁢(TG)=1, so TG is a p-group. It follows from HM≤TG that HM is a p-group, and so Hp is normal in H for H/HM is nilpotent, where Hp is a Sylow p-subgroup of H. Let R1 be the nilpotent residual subgroup of HT. Then R1 is a characteristic subgroup of HT, and so Y≤NG⁢(HT)≤NG⁢(R1). Since R≤HT, HT/R is nilpotent, and so R1≤R. Note that

Hence R1=1 by (1). This implies that HT is nilpotent. Let q be any prime divisor of |H|, which is different from p, and let Q be a Sylow q-subgroup of H. Since T=H⁢Y and Y is a p-group, Q is also a Sylow q-subgroup of T. Now, since H is subnormal in T, for any element t of T, Qt∩H is also a Sylow q-subgroup of H by Lemma 2.2 (3). Hence Qt≤H for any t∈T. It follows that QT≤H, and so QT≤HT. Consequently, Q is also a Sylow q-subgroup of HT. But, since HT is nilpotent, Q is normal in HT, and so Q is normal in H. This shows that every Sylow subgroup of H is normal in H. Therefore, H is nilpotent.

Now we assume that Y is not a p-group. Since H is subnormal in G, there exists a proper normal subgroup K of G such that H≤K, and so G=K⁢X=K⁢Y by (2). Then G/K≅Y/(Y∩K) is abelian, so G′≤K. Therefore, there exists a proper normal subgroup M of G such that H≤K≤M and G=M⁢Gp, where Gp is a Sylow p-subgroup of G. Since H is completely c-permutable in G, there exists an element g∈〈H,Gp〉 such that H⁢Gpg=Gpg⁢H. Let P=Gpg. Then, by Lemma 2.3 (2), G=M⁢Gp=M⁢P. Let T=H⁢P. If T=G, then, because H is subnormal in G and |G:H|=|T:H|=|P:P∩H| is a p-number, we get Op⁢(G)≤H by Lemma 2.2 (6). By (1), Op⁢(G)=1, that is, G is a p-group, and so H is nilpotent, a contradiction. Hence we can assume that T is a proper subgroup of G. Then, by the same discussion as above, we can obtain that H is nilpotent.

(5) The final contradiction. Since H is completely c-permutable in G, there exists an element g∈〈H,X〉=G such that H⁢Xg=Xg⁢H. If G=H⁢Xg, then, by (3), H is nilpotent. If H⁢Xg is a proper subgroup of G, then, by (4), H is also nilpotent. The contradiction completes the proof. ∎

Proof.

Assume that the assertion is false, and let G be a counterexample of minimal order. We proceed the proof by the following steps.

(1) If N is a minimal normal subgroup of G, then G/N is p-supersoluble, and so N is noncyclic. This is clear by Lemma 2.1 (1) and the choice of G.

(2) Op′⁢(G)=1. If Op′⁢(G)≠1, then, by (1), we know that G/Op′⁢(G) is p-supersoluble, and so G is p-supersoluble, a contradiction. Hence we have (2).

(3) G has a unique minimal normal subgroup, say N,

Since G is p-soluble, from the choice of G, we know that G is not a simple group. Let N be a minimal normal subgroup of G. Then, by (2) and the fact that G is p-soluble, we have that N is an elementary abelian p-group, so N≤Op⁢(G). Clearly, N is the unique minimal normal subgroup of G and N≰Φ⁢(G) by (1). Then there exists a maximal subgroup M of G such that G=N⁢M. It is easy to see that N∩M⊴G and CG⁢(N)∩M⊴G. Hence N∩M=CG⁢(N)∩M=1, and so CG⁢(N)=CG⁢(N)∩N⁢M=N⁢(CG⁢(N)∩M)=N. Hence we have (3).

(4) Final contradiction. From (1) and (3), we see that there exists a maximal subgroup M of G such that G=N⋊M and M≅G/N is p-supersoluble. Now let T be a maximal subgroup of M such that p∤|M:T|. Then

and so T is a 2-maximal subgroup of G and has a composite index in G which is divisible by p. Hence, by hypothesis, T is completely c-permutable in G. Let N1 be a maximal subgroup of N. Then, by (1) and (3), we know that N1≠1 and |N:N1|=p.

If T=1, then M is a cyclic subgroup of order q for some prime q≠p. It follows that N is a maximal subgroup of G. Hence N1 is a 2-maximal subgroup of G and |G:N1|=|G:N||N:N1|=pq. Then, by hypothesis, N1 is completely c-permutable in G. Thus there exists an element x∈G such that N1x⁢M=M⁢N1x. Since M is a maximal subgroup of G, we have that N1x⁢M=M or G=N1x⁢M. If N1x⁢M=M, then N1x≤M∩N=1, a contradiction. If G=N1x⁢M, then we have N=N1x⁢(N∩M)=N1x, also a contradiction. This contradiction shows that T≠1.

Since T is completely c-permutable in G, there is an element x∈G such that T⁢N1x=N1x⁢T. Let D=T⁢N1x. Since

we have that |G:D|=|N:N1x||M:T|=p|M:T|. This shows that D has a composite index in G that is divisible by p. We claim that D is a 2-maximal subgroup of G. In fact, clearly, D=T⁢N1X<T⁢N and |TN:D|=p, so D is a maximal subgroup of TN. On the other hand, since G=N⋊M and T is a maximal subgroup of M, T⁢N<G. If there exists a subgroup K of G such that T⁢N<K<G, then K=N⁢(K∩M) and T≤K∩M≤M. Hence T=K∩M or K∩M=M. In the former case, T⁢N=(K∩M)⁢N=K, a contradiction. In the latter case, M=K, and it follows that N≤K=M, also a contradiction. The contradiction shows that TN is a maximal subgroup of G. Therefore, D=T⁢N1x is a 2-maximal subgroup of G and has a composite index in G which is divisible by p. Thus, by hypothesis, D is completely c-permutable in G.

Let q be any prime divisor of |M| with q≠p, and let Mq be a Sylow q-subgroup of M. Then there exists an element y∈〈D,Mq〉 such that D⁢Mqy=Mqy⁢D. Since Mq is also a Sylow q-subgroup of G,

by Lemma 2.5. From this, we can see that Mqy≤NG⁢(N1x), i.e., q∤|G:NG(N1x)|. But q was arbitrarily chosen, so we have that |G:NG⁢(N1x)| is a p-number. Since p∤|M:T|, there exists a Sylow p-subgroup Mp of M such that Mp≤T. Clearly, N⁢Mp is a Sylow p-subgroup of G. Since

we have that T≤D≤NG⁢(N1x). Hence N⁢Mp≤NG⁢(N1x) for N abelian. Therefore, NG⁢(N1x)=G, that is, N1x is normal in G. The minimal normality of N implies that N1x=1. This final contradiction completes the proof. ∎