Arch. Math. 95 (2010), 309–315c© 2010 Springer Basel AG0003-889X/10/040309-7published online August 31, 2010DOI 10.1007/s00013-010-0174-5 Archiv der Mathematik

On a group with three supersolvable subgroups of pairwiserelatively prime indices

Neil Flowers and Thomas P. Wakefield

Abstract. In this paper, we show that if G is a finite group with threesupersolvable subgroups of pairwise relatively prime indices in G and G′

is nilpotent, then G is supersolvable. Let π(G) denote the set of primedivisors of |G| and max(π(G)) denote the largest prime divisor of |G|. Wealso establish that if G is a finite group such that G has three supersolv-able subgroups H, K, and L whose indices in G are pairwise relativelyprime, q � p − 1 where p = max(π(G)) and q = max(π(L)) with L a Hallp′-subgroup of G, then G is supersolvable.

Mathematics Subject Classification (2000). Primary 20D25; Secondary20D10.

Keywords. Supersolvable group, Pairwise relatively prime indices.

1. Introduction. Let G be a finite group, π(G) denote the set of prime divisorsof |G|, max(π(G)) denote the largest prime divisor of |G|, and let the pth partof the order of G be denoted by |G|p. It is well-known that a finite group Gpossessing three abelian subgroups whose indices are pairwise relatively primeis abelian. This result also holds when abelian is replaced by nilpotent. Theresult extends to solvable groups as well, as proved in Theorem 6.4.4 of [4]. Itis natural to ask if it is possible to conclude that if G has three supersolvablesubgroups whose indices are pairwise relatively prime, then is G supersolv-able? Surprisingly, the answer to this question is no, as seen in the followingexample.

Example 1. In Theorem 4.4 of Chapter 3 of [1] it was shown there exist mono-morphisms β : Z2 → Aut(Z3) and α : (Z3 ×β Z2) → Aut(Z7 × Z7) such thatthe group G = (Z7 × Z7) ×α (Z3 ×β Z2) has three supersolvable subgroups K,L, and M of orders 2 · 3, 2 · 72, and 3 · 72. Hence their indices in G are pairwiserelatively prime. But G is not supersolvable since it was shown that G has

310 N. Flowers and T. P. Wakefield Arch. Math.

no subgroup of order 2 · 3 · 7. Hence non-supersolvable groups G with threesupersolvable subgroups of relatively prime index in G exist.

There are several results attempting to characterize groups with supersolv-able subgroups. In [5], Kegel proved the following theorem.

Theorem A. If G is a group and G = HK = HL = KL, where H and K arenilpotent subgroups and L is supersolvable, then G is supersolvable.

D. K. Friesen [3] established Theorem B.

Theorem B. If G is a group with A and B normal supersolvable subgroupsof G with relatively prime index in G, then G is supersolvable.

In [2], Klaus Doerk looked at a group with four supersolvable subgroupswhose indices are pairwise relatively prime. He proved the following surprisingresult.

Theorem C. If a group G has four supersolvable subgroups whose indices arepairwise relatively prime, then G is supersolvable.

In [7], Wang Kun-ren established Theorem D.

Theorem D. A group G is a supersolvable group whose order has at least threedifferent prime divisors if and only if there exist three maximal supersolvablesubgroups of G whose indices are three different primes.

Again, the example illustrates that Wang Kun-ren’s result does not gen-eralize to groups whose indices are a power of a prime. In this paper, weestablish two results characterizing groups with three supersolvable subgroupsof pairwise relatively prime index in G. The first result, stated as Theorem 1.1,proves that G is supersolvable if we add the condition that G′ is nilpotent.Theorem 1.2 establishes that G is supersolvable if and q � p − 1 where p =max(π(G)) and q = max(π(L)) where L is a Hall p′-subgroup of G.

Theorem 1.1. Let G be a group, H, K, and L supersolvable subgroups of Gwith pairwise relatively prime indices in G. Suppose G′ is nilpotent. Then Gis supersolvable.

Theorem 1.2. Suppose G is a group with three supersolvable subgroups H,K, and L whose indices in G are pairwise relatively prime, q � p − 1 wherep = max(π(G)) and q = max(π(L)), and L ∈ Hallp′(G). Then G is super-solvable.

Before proceeding with the proofs of the results, we present backgroundresults necessary to proceed.

2. Background results. Our proof of Theorem 1.1 relies upon the followinglemma, established by Friesen in [3].

Lemma 2.1. Let A, B be normal supersolvable subgroups of the finite group Gwith G = AB. If the indices |G : A| and |G : B| are relatively prime, then Gis supersolvable.

Vol. 95 (2010) On a group with three supersolvable subgroups 311

To prove Theorem 1.2, we will rely upon the following lemmas. Lemma 2.2is stated as Theorem 6.4.4 in [4].

Lemma 2.2. If a group G possesses three solvable subgroups whose indices arepairwise relatively prime, then G is solvable.

The following result appears as 7.2.14 in [6].

Lemma 2.3. If G/H is supersolvable and H is cyclic, then G is supersolvable.

Lemma 2.4 appears as Exercise 9.3.18 in [6].

Lemma 2.4. If G/Φ(G) is supersolvable, then G is supersolvable.

We will also use the Schur–Zassenhaus Theorem.

Schur–Zassenhaus Theorem. Let N be a normal Hall subgroup of G. Then1. N is complemented in G.2. If, in addition, at least one of N or G/N is solvable, then G acts transi-

tively on the complements of N in G by conjugation.

Lemma 2.5. If G is supersolvable and 1 �= N � G, then N contains a normalcyclic subgroup of G isomorphic to Zp for some prime p.

Proof. Since G is supersolvable, there exists a normal series

G = G0 � G1 � · · · � Gn = 1

with Gi/Gi+1 cyclic for all 0 ≤ i ≤ n − 1. But then

N = N ∩ G0 � N ∩ G1 � · · · � N ∩ Gn = 1

with Gi ∩ N � N for all 0 ≤ i ≤ n andGi ∩ N

Gi+1 ∩ N=

Gi ∩ N

Gi+1 ∩ (Gi ∩ N)∼= (Gi ∩ N)Gi+1

Gi+1≤ Gi

Gi+1,

which is cyclic. Let i be the maximal index such that N ∩Gi �= 1. Then N ∩Gi

is a normal cyclic subgroup of G contained in N . As N ∩ Gi is cyclic, it has asubgroup H isomorphic to Zp for some prime p. As H char (N ∩ Gi) � G, weget H � G. �3. Proofs of main results. First we will show that if G has three supersolvablesubgroups of pairwise relatively prime indices in G and G′ is nilpotent, thenG is supersolvable.

Theorem 1.1. Let G be a group, H, K, and L supersolvable subgroups of Gwith pairwise relatively prime indices in G. Suppose G′ is nilpotent. Then Gis supersolvable.

Proof. Proceed by induction on |G|. First suppose that there are primes p �= qsuch that pq divides |G′| and let P ∈ Sylp(G′) and Q ∈ Sylq(G′). Since G′ isnilpotent, P �G′ and Q�G′. Thus P char G′ �G and so P �G and similarlyQ � G.

Suppose first that H ≤ P . Without loss of generality, suppose p � |G : K|.Let Gp ∈ Sylp(G) such that P ≤ Gp. Since |G|p = |K|p, Gp

x ≤ K for some

312 N. Flowers and T. P. Wakefield Arch. Math.

x ∈ K. Now P x ≤ Gpx ≤ K, but P = P x since P � G. Hence P ≤ Gp

x ≤ Kand so P ≤ K. But then H ≤ K so that G = HK = K is supersolvable.

Hence we may assume that H � P . Similarly, K � P and L � P . LetG = G/P and ˜G = G/Q. Now HP/P , KP/P , and LP/P are supersolvablesubgroups of G/P with pairwise relatively prime indices in G/P . By inductionG is supersolvable and similarly ˜G is supersolvable. Hence G× ˜G is supersolv-able. Let φ : G → G × ˜G be defined by (g)φ = (g, g) for all g ∈ G. Then φ isa homomorphism and Kern φ ≤ P ∩ Q = 1. Hence G ∼= (G)φ ≤ G × ˜G and soG is supersolvable.

We are left with the possibility that G′ is a p-group. Then G′ ≤ Gp, whereGp ∈ Sylp(G). Thus Gp�G. Without loss of generality, assume that p � |G : H|and p � |G : K|. Then |G|p = |H|p = |K|p and Gp � G, so Gp ≤ H ∩ K. ButG′ ≤ Gp, so G′ ≤ H and G′ ≤ K. Hence H � G and K � G. We have that Hand K are normal subgroups of G, H and K are both supersolvable and haverelatively prime indices in G. By Lemma 2.1, G is supersolvable. �

We now prove Theorem 1.2.

Theorem 1.2. Suppose G is a group with three supersolvable subgroups H,K, and L whose indices in G are pairwise relatively prime, q � p − 1 wherep = max(π(G)) and q = max(π(L)), and L ∈ Hallp′(G). Then G is super-solvable.

Proof. We will proceed by induction on |G|. Suppose 1 �= Φ(G) � G. If H ≤Φ(G), then G = Φ(G)K, so G = K is supersolvable. Thus we may assumethat H � Φ(G). Similarly, K � Φ(G) and L � Φ(G). Let G = G/Φ(G). ThenHΦ(G)/Φ(G), KΦ(G)/Φ(G), and LΦ(G)/Φ(G) are supersolvable groups withpairwise relatively prime indices in G and L ∈ Hallp′(G). We need to show thatp | |G/Φ(G)|. Suppose not. Then Φ(G) contains Gp, a Sylow p-subgroup ofG. Since Φ(G) is nilpotent, Gp � Φ(G). Therefore Gp char Φ(G) � G and soGp � G. By the Schur–Zassenhaus Theorem, Gp has a complement M in G.Hence G = GpM = Φ(G)M = M . But then we get Gp = Gp∩G = Gp∩M = 1,a contradiction. Thus p | |G| and so G/Φ(G) is supersolvable by induction.Now by Lemma 2.4, G is supersolvable. Hence Φ(G) = 1.

Next we claim that if Gp ∈ Sylp(G), then Gp � G. Since L ∈ Hallp′(G),p | |G : L|. As |G : L|, |G : K|, and |G : H| are pairwise relatively prime,p � |G : H| and p � |G : K|. Hence Gp ≤ H and there exists g ∈ G suchthat Gp ≤ Kg. Then H and K supersolvable and p = max(π(G)) implies thatGp � H and Gp � Kg. As

(|G : H|, |G : Kg|) = (|G : H|, |G : K|) = 1,

Gp � HKg = G and, therefore, Gp ≤ H ∩ K.Let 1 �= N ≤ Gp be a minimal normal subgroup of G. Ultimately, we will

show Gp = N . By Lemma 2.2, G is solvable and, since N ≤ Gp, N is an ele-mentary abelian p-group. Suppose Gp �= N . We then claim that N is unique.Let N1 be a minimal normal subgroup of G such that N1 �= N . Then N1 is an

Vol. 95 (2010) On a group with three supersolvable subgroups 313

elementary group. As N1 ∩ N � G and N1 ∩ N < N , N1 ∩ N = 1 by the mini-mality of N . If N1 ∈ Sylp(G), as Gp � G, N1 = Gp. But then N ≤ Gp = N1

and so N = N1 by the minimality of N1. Hence N1 /∈ Sylp(G).As N �= Gp and N1 /∈ Sylp(G), consider G = G/N and ˜G = G/N1.

Then p | |G| and H, K, L are supersolvable subgroups of G with pairwisecoprime indices in G. If H = 1, then H ≤ N ≤ Gp ≤ K which impliesG = HK = K is supersolvable. Thus G and, by the same argument, ˜G, aresupersolvable by induction. Hence G × ˜G is supersolvable. Let φ : G → G × ˜Gbe defined by (g)φ = (g, g) for all g ∈ G. Now φ is a homomorphism andKern φ ≤ N ∩ N1 = 1. Hence G ∼= (G)φ ≤ G × ˜G and so G is supersolvable.

Therefore N = N1 and G has a unique minimal normal subgroup. Next,we show that CG(N) = N . Since Φ(G) = 1, N � Φ(G). Thus there existsa maximal subgroup M of G such that N � M . Then M < MN ≤ G soG = MN by the maximality of M . As N is abelian, N ≤ CG(N) impliesCG(N) � M . Also N � G implies CG(N) � G. Thus M � MCG(N) ≤ Gso G = CG(N)M by the maximality of M . Suppose CG(N) ∩ M �= 1. Then1 �= CG(N) ∩ M � NM = G. Therefore N ≤ CG(N) ∩ M ≤ M by the unique-ness of N . Hence we get N ≤ M , a contradiction. Thus CG(N) ∩ M = 1and

CG(N) = CG(N) ∩ G

= CG(N) ∩ MN

= N(CG(N) ∩ M)= N.

Now N �Gp and Gp is a p-group, so 1 �= N ∩Z(Gp). But Z(Gp) char Gp�G, soZ(Gp)�G. This implies N ∩Z(Gp)�G, so N ∩Z(Gp) = N by the minimalityof N . Hence N ≤ Z(Gp). But then Gp ≤ CG(N) = N . Hence Gp ≤ N , soGp = N , a contradiction to our original supposition that N < Gp. ThereforeGp = N and Gp is a minimal normal subgroup of G. Recall N = Gp ≤ H ∩K.

By Lemma 2.5 there exists L1 ≤ Gp such that L1 � H and L1∼= Zp.

Similarly, there exists L2 ≤ Gp such that L2 � K and L2∼= Zp. Now Gp is a

minimal normal subgroup of G containing L1 and L2, so Gp = 〈L1G〉 = 〈L2

G〉.Let Lq ∈ Sylq(L). Since q = max(π(L)) and L is supersolvable, Lq � L. By

Lemma 2.5, there exists L3 ≤ Lq such that L3 � L and L3∼= Zq.

We now show that there exist x or y in G such that L3 ≤ CG(L1x) or

L3 ≤ CG(L2y). Let R = {L1

x | x ∈ G} and S = {L2y | y ∈ G}. Then

H ≤ NG(L1) ≤ G and K ≤ NG(L2) ≤ G implies

|G : NG(L1)| | |G : H| and |G : NG(L2)| | |G : K|.As gcd(|G : H|, |G : K|)=1, we have that gcd(|G : NG(L1)|, |G : NG(L2)|)=1.Now L3 acts on R and S by conjugation. Thus we have

|G : NG(L1)| = |R| = | ∪i Oi| =∑

i

|Oi| =∑

i

|Q : QL1xi |

314 N. Flowers and T. P. Wakefield Arch. Math.

and

|G : NG(L2)| = |S| = | ∪j O′j | =

∑

j

|O′j | =

∑

j

|Q : QL2yj |

where Oi and O′j are orbits of L3 on R and L3 on S respectively. If |Oi| �= 1

and |O′j | �= 1 for all i and j, then q | |G : NG(L1)| and q | |G : NG(L2)|,

contradicting gcd(|G : NG(L1)|, |G : NG(L2)|) = 1. So there exists an xi oryj such that L3 ≤ NG(L1

xi) or L3 ≤ NG(L2yj ). Now if L3 ≤ NG(L1

xi), thenthere is a homomorphism φ : L3 → Aut(L1

xi) with kernel CL3(L1xi). Hence

|L3 : CL3(L1xi)| divides |Aut(L1

xi)| = p−1. As q � p−1, |L3 : CL3(L1xi)| = 1.

Hence L3 ≤ CG(L1xi) or, using the same argument if L3 ≤ NG(L2

yj ), weobtain L3 ≤ CG(L2

yj ).Without loss of generality, suppose L3 ≤ CG(L1

x). Since L3 � L, L3x−1

�Lx−1

. Hence, as L1 � H, we get

L3x−1 ≤ CG

(

〈L1Lx−1

〉)

= CG

(

〈L1HLx−1

〉)

= CG

(〈L1G〉)

= CG(Gp).

But then L3x−1

�Lx−1Gp = G since Lx−1 ∈ Hallp′(G). Let G = G/L3

x−1. Then

p | |G|. Further, H �= 1 and K �= 1 since p | |H| and p | |K|. As L ∈ Hallp′(G)and |G| is divisible by at least three distinct primes, L �= 1. Hence G is super-solvable by induction. Since L3

x−1is cyclic, G is supersolvable by Lemma 2.3.

Thus we have established Theorem 1.2. �

Note that a generalization of Wang Kun-ren’s Theorem D now follows fromTheorem 1.2. We have

Theorem 3.1. Suppose G is a finite group with H,K,L ≤ G supersolvable suchthat |G : H| = pa, |G : K| = qb, and |G : L| = rc = |G|r where p �= q �= r,r = max(π(G)) and max(π(L)) � r − 1. Then G is supersolvable.

Proof. As |G : L| = rc = |G|r, L is a Hall r′-subgroup of G. All the otherconditions of Theorem 1.2 are satisfied. �

References

[1] H. Bray et al., Between Nilpotent and Solvable, Polygonal Publishing House,

Passaic, NJ, 1982.

[2] K. Doerk, Minimal nicht uberauflosbare, endliche Gruppen, Math. Z. 91 (1966),

198–205.

[3] D. K. Friesen, Products of normal supersolvable subgroups, Proc. AMS. 30

(1971), 46–48.

[4] D. Gorenstein, Finite Groups, Harper & Row, New York, 1968.

Vol. 95 (2010) On a group with three supersolvable subgroups 315

[5] O. H. Kegel, Zur Struktur mehrfach faktorisierbarer endlicher Gruppen, Math.

Z. 87 (1965), 42–48.

[6] W. R. Scott, Group Theory, Prentice Hall, Englewood Cliffs, 1964.

[7] K. Wang, Finite group with two supersolvable subgroups of coprime indices,

Northeast. Math. J. 17 (2001), 219–223.

Neil Flowers

Department of Mathematics and Statistics,Youngstown State University,Youngstown, OH 44555, USAe-mail: flowers@math.ysu.edu

Thomas P. Wakefield

Department of Mathematics and Statistics,Youngstown State University,Youngstown, OH 44555, USAe-mail: tpwakefield@ysu.edu

Received: 1 June 2010