A NOTE ON FINITE GROUP STRUCTURE INFLUENCED BY SECOND AND THIRD MAXIMAL SUBGROUPS

The structure of a finite group having specified number of second and third maximal subgroups has been investigated in the paper.

For the sake of completeness we mention below the following theorem or p-groups [Theorem 7.6, p. 304] in IIuppert [3] which will be used.
THEOREM H. Let G be a p-group and suppose all abellan normal subgroups of G are cyclic.
Then, (a) G is cyclic if p > 2 (b) if p 2, G has a cyclic normal subgroup of index 2.
We will first characterize groups having the desired number of second maximal subgroups.The followlng lemmas will be required.
LEMMA 2.1.A p-group G which has exactly one nontrlvlal second maximal subgroup is cycllc if p > 2 and has a cyclic normal subgroup of index 2 if p 2. Evidently G[ > p Let [G p n > 2 and M be a maximal subgroup G.
Then M-4__ G, I"l pn-1 > P and therefore the given second maximal subgroup H 0 .M. M 0 being the only maximal subgroup of M, it follows that M is cyclic and therefore each abellan normal subgroup of G is cycllc.From Theorem H it now follows that G is of the desired type.LEMMA 2.2.
A group G with no second maxlmal subgroup is a group of order p 2 or p or qt, p,q,t are different primes.
PROOF.If G is a p-group and G has no second maximal subgroup then evidently G p or p Now suppose G is not a p-group, then every maximal subgroup is of prime order and therefore the Sylow subgroups of G are cyclic and are of prime orders.Consequently G is supersolvable and G has a Sylow basis.If IGI is divisible by more than two primes then clearly G will have a second maximal subgroup and therefore the order of G must be qt, for some primes q and t.
THEOREM 2.1.group G having exactly one second maximal subgroup is a p-group.For p 2, G is cyclic and for p 2, G has a cyclic subgroup of index 2. PROOF.It suffices to show that G must be a p-group if it has one second maximal subgroup.Lemma 2.1 will then guarantee the structure of G as claimed.
Suppose G is not a p-group and without loss of generality we may assume G to be a counter example of least possible order.
Then every maximal subgroup of G is either of prime order or else contains the given second maximal subgroup M 0 of G.In the latter case, M is clearly cyclic of prime power order.Thus every maximal subgroup of G is cyclic and therfore every Sylow subgroup of G is cyclic and G is supersolvable.
Hence G PK, P G, P OK where P is the Sylow subgroup corresponding to the largest prime divisor p of IGI and K is a p-complement.
Let IKI I. Now K has at most one second maximal subgroup.If K and K 2 are two second maximal subgroups of K then PK and PK 2 are two different second maximal subgroups of G, a contradiction.(If PK| PK 2 then K K 2 implies that some -I u _elKl' u K 2 and it follows that u xv, x e P, v e K 2, x # e.Then uv x.But uv e K, and is therefore a p'element. Hence PK -!-PK2).if K has exactly one second maximal subgroup then K is a q-group for some prime q since G Is the minimal 2 counter example and if K has no second maximal subgroup then we need consider IKI r or st, r,s,t, are different primes.We first consider this latter case. 2 CASE I. IKI r Consider the subgroup p < u >, l<u>l r.M-P<u> is a maximal subgroup G and P is the second maximal subgroup.Hence (M) P and PIMI, a 2 contradictfon.Thus IKI r CASE II.IKI st.Let G--PST, where S and T are Sylow subgroups of G of orders s and t respectively.M PS is a maximal subgroup and P is the second maximal subgroup of G. Hence (M I) --P and PIMII, a contradiction.Therefore IKI # st.
We are thus left to consider the case when K is a q-group.
Distinguish two n second maximal subgroup.We may therefore assume IPI p n 2. Let P1 be maximal r-I in P so that IPll-p Note PI G since P is cyclic and P_ G. Consider PI E a maximal subgroup of G.
If P2 is maximal in P1 then P2 -G and P2 K and P1 are two second maximal subgroups of G, contradicting the existence of only one.m CASE B. IKI =q m 2. First suppose PI p and let K <" K. Then m-I IKII q and consider PK I.If K 2 <.K then PK 2 and K are two second maximal subgroups of G which contradicts again the existence of only one.Now suppose P p n > 2 and let Pl <" p and K <.K.Both PI K and PK are maximal subgroups of G.If P2 <" PI and K 2 <.K then P2 K and PK 2 are second maximal subgroups of G which is a contradiction and so IKI must be I. Hence G is a p-group.Thus there is no minimal counter example and the theorem follows.
lio investigate the structure of groups ith Cvo second maximal subgroups.The relieving 1emma is necessary.LEHHA 2.3.A group G vih exactly to ximal subgroups is necessarily cycltc and the order of he group is divisible by vo distlnc primes, PROOF, If either of the given xtmal subgroups Is no normal then 1 being its on norllzer fs of index 2 hlch hoever forces t o be normal.
Thus G nIlpotent and ve claim that G cannot be a p-group.Suppose he set S X[X is a p-group and X has exactly to ximal subgroups If S [Y S and Y is no cyclic} then T 0 is an elemen of S of least possible order.and * = I follows herefore ha Y0 SSI so ha Y0 is cycllc and Y0 ,T x> since T (Y0) and ve have a conradlcion.us S and eve element In S is cyc1ic.A group O wlh exactly wo second xll subgroups Is necessarily supersolvable and G Is elher a 2-group or else is order Is dlvlslble by eve primes only.

PROF.
Every xlmal subgroup of G Is either of prime order or has one or xll subgroups.erefore eve xll subgroup of O Is cycllc and consequently vry Sylov subgroup of O Is cycllc.Henc G Is supersolvable and G PK, PK I, P G where P Is he Syl p-subgroup orrespondng o Che largest prl divisor of O[ and K is a p-complement.If K I, Chert G Is a p-group and by Lem 2. is also a 2-group.No, suppose K[ , I. To prove [G is dlvlslble by wo primes ve may wlthou loss of generally assume O o a counter example of the leas posslble order.If K has one second xll subgroup then by eorem I, K is a q-group for so prl q and If g has wo second xlmal subgroups hen G being a counter example of leas posslble order, g Is elher a 2-group or ls order is dlvlslble by wo primes.Suppose K=RT where R one T are o Sylov subgroups corresponding o he prime divisor r and r

n
Suppose H and H* are he vo xima subgroups of YO and =p Disinguish Co cases: Case I. HH* (e>.Since H and H* are norl in Y0' Y0 * and n n-I n-I n 2 and Y0 Is no cycllc.I a and b are elens of YO hen aP, b, (ab are xll subgroups of Y0 and ve have a conradiclon.CASE [.H8* eP.Le T 8" and observe T G. Now consider Y0/T. is a p-group, , are vo xil subgroups of Y0/T Y0 and Y0 does no have any other xll subgroup besldes Hence G canno exis and he assertion in Che lem fullers.LE2.5. Bu Chis Implles every elemen X In S has exactly-one subgroup of index p l.e.X has go exactly one xlmal subgroup.Hence S s Pi Is herefore cycllc.Hence It follows ha G Is cycllc also and Che assertion in the lem Is proved.LE 2.6.A p-group th xactly two second xlmal subgroups is necessarlly a 2-group.PROOF.Let O be a counter example of the smalles posslble order.Then every xll subgroup ff of G has exactly one xil subgroup.(By Lemma 2.3 her is no p- group wih exactly vo ximal subgroups.)Hence H is cycllc and herefore by eorem H n fullers ha G is cycllc.Is hoverer, Implles G can have one second xll subgroup which is a conradlction.