Abstract A NOTE ON SUPERSOLUBLE MAXIMAL SUBGROUPS AND THETA-PAIRS

A NOTE ON SUPERSOLUBLE MAXIMAL SUBGROUPS AND THETA-PAIRS JAMES C. BEIDLEMAN AND HOWARD SMITH A B-pair for a maximal subgroup Mof agroup G is a pair (A, B) of subgroups such that B is a maximal G-invariant subgroup ofA with B but not A contained in M. B-pairs are considered here in some groups having supersoluble maximal subgroups.


. Introduction
Let M be a maximal subgroup of the group G.An ordered pair (A, B) of subgroups of G is called a B-pair for M if B is a G-invariant subgroup of A such that (i) B <_ M but A ~M and (ii) A/B contains properly no nontrivial normal subgroup of G/B .
The set of all B-pairs for M is denoted by B(M) (see [3]).A partial order is defined on B(M) by means of (A, B) <_ (C, D) if and only if A <_ C. In this case B <_ D also.It is then clear what is meant by saying that (A, B) is a maximal B-pair for M .If (A, B) is in 6(M) and A < G then A/B is a chief factor of G.
This brief note is concerned with 8-pairs in relation to the property of supersolubility.Our principal result is Theorem 1, which bears some relation to Theorem 1 of [1] .It will be seen that Theorem 1 is an easy consequence of Theorem 2. The concepts and results found here can be found in [4] .
Let Fit(G) denote the Fitting subgroup of the group G.The máin results presented here are as follows.
Theorem 1 .Let G be a group with a supersoluble maximal subgroup M and suppose that Fit(G) n M is a maximal subgroup of Fit(G) .Then G is supersoluble.Theorem 2 .Let G be a group and M a supersoluble maximal subgroup of G not containing Fit(G) .If 6(M) has a maximal pair (A, B) such that A/B is cyclic and A is subnormal in G then G is supersoluble.
An argument similar to that employed in proving Theorem 2 allows us to establish the following result (the proof of which is omitted) .Theorem 3 .Let G be a group and M a supersoluble maximal subgroup not containing Fit(G) .If A/B is cyclic for each maximal pair (A, B) in B(M) then G is supersoluble.

. Proofs
We require two preliminary lemmas .Lemma 1.Let G be a group and M a maximal subgroup of finite índex in G. Let (A, B) be a maximal 0-pair for M .Then, given any G-invariant subgroup N of finite índex in M, there exists a maximal 0-pair (C/N, DIN) for M/N such that CID is isomorphic to a normal section of A/B .Further, if A is subnormal in G then C may be chosen subnormal in G .
Proof. .If N <_ B then (A/N, B/N) is a maximal member of B(M/N) and there is nothing to prove .Suppose that N is not contained in B. Then N is not contained in A, otherwise A = BN <_ M, a contradiction .Let K be the normal core of AN n M in G. Then BN <_ K. Since A < AN and (A, B) is maximal, (AN, K) is not in B(M) .Let H/K be a minimal G-invariant subgroup of (the finite group) ANIK.Then (H, K) belongs to B(M) and is contained in some maximal member (C, D) of 0(M) .Now C = HD is normal in G and CID = HD/D -HIHnD, an image of H/K, which is, in turn, normal in the image AN/K of A/B .Finally, (C/N, DIN) is a maximal member of B(M/N) .Note that C = A in the case where N < B, while if N ~B then C < G.
Note that some of the ideas in the proof of Lemma 2.1 of [3] are used to establish Lemma 1.
Lemma 2. Let G be a group and M a polycyclic maximal subgroup of G not containing Fit(G) .Then G is polycyclic.
Proof. .Let N be a nilpotent normal subgroup of G not contained in M .Then G = MN and so G is soluble .We may assume M is corefree in G. Then G is a soluble primitive group and it is known that G has a unique non-trivial abelian normal subgroup A which satisfies G = MA, A n M = 1 and A = CG (A) .Thus A is a simple ZM-module and, by a result of Roseblande [5, p. 308], A is finite .Therefore, G is polycyclic .
Proof of Theorem 2: By Lemma 2, G is polycyclic and so, by a theorem of Baer [6, 11 .11], it suffices te prove that every finite image GIN of G is supersoluble.Clearly we may assume that N <_ M and hence, by Lemma 1, that G is finite .Suppose that G is not supersoluble and let T be a nontrivial normal subgroup of G.By Lemma Proof ofTheorem 1 : By Lemma 2, G is polycyclic and so, by a result of Hirsch [4, 5.4 .19],O(G) <_ Fit(G) and Fit(G)/0(G)=Fit(G/0(G)) .Hence, by a result of Lennox [2], we may assume that O(G) = 1.Since G is polycyclic, F = Fit(G) is nilpotent ([4, p .129]) and consequently every maximal subgroups of F is normal and of prime index in F. Therefore, F is abelian and hence F n M is normal in G and of prime index in F. It follows that (F, F n M) E B(M) .Let (A, B) be a maximal member of 9(M) containing (F, F n M) .Then either FB = A or B <_ F = A.
In either case, A/B is cyclic and A is normal in G.By Theorem 2, G is supersoluble .
1 and an obvious induction, G/T is supersoluble .Thus G has a unique minimal normal subgroup W and G/W is supersoluble .If O(G) :~1 then W <_ O(G) and G is supersoluble, by a result of Huppert [4, 9.4.5] .Thus O(G) = 1 and Fit(G) = W, by a result of Gaschütz [4, 5.2 .15] .Since W ~M we see that MG = 1 and hence B = 1 and A is cyclic and subnormal in G. Thus A <_ W. Certainly (W 1) belongs to B(M) and so, by maximality, A = W. Thus G is supersoluble and we have the required contradiction .