On the rarity of quasinormal subgroups

On the Rarity of Quasinormal Subgroups

JOHNCOSSEY(*) - STEWARTSTONEHEWER(**)

ABSTRACT- For each primepand positive integern, Berger and Gross have defined a finitep-groupGˆHX, whereHis a core-free quasinormal subgroup of ex- ponentp^{n 1}andXis a cyclic subgroup of orderpⁿ. These groups are universal in the sense that any other finitep-group, with a similar factorisation into sub- groups with the same properties, embeds inG. In our search for quasinormal subgroups of finitep-groups, we have discovered that these groupsGhave re- markably few of them. Indeed whenpis odd, those lying inHcan have exponent onlyp,p^{n 2}orp^{n 1}. Those of exponentpare nested and they all lie in each of those of exponentp^{n 2}andp^{n 1}.

1. Introduction.

A subgroupQof a groupG, such thatQHˆHQfor all subgroupsHof G, is said to be quasinormal (sometimespermutable) inGand we write Q qn G. The concept was introduced by Ore in 1937 (see[6]) and in [7] he proved that, in finite groups, quasinormal subgroups are always subnormal.

He also proved that quasinormal subgroups are modular. Recall that a subgroupMof a groupGismodularif, for all subgroupsXandY ofG,

hX;Mi \Yˆ hX;M\Yi ifX4Y and

hX;Mi \Y ˆ hX\Y;Mi ifM4Y:

Indeeda subgroup Q of a finite group G is quasinormal in G if and only if Q is modular and subnormal in G. Thus the concepts of modularity and

(*) Indirizzo dell'A.: Mathematics Department, School of Mathematical Sciences, Australian National University, Canberra, ACT 0200.

E-mail: John.Cossey@anu.edu.au

(**) Indirizzo dell'A.: Mathematics Institute, University of Warwick, Coventry CV4 7AL, England.

E-mail: S.E.Stonehewer@warwick.ac.uk

quasinormality coincide in finite p-groups. In 1973 Maier and Schmid proved in [5] that if Gis a finite group andQ qn G withQcore-free, i.e.

Q_Gˆ1, thenQ lies in the hypercentre of G. It follows fairly easily from this that the complexities of the embedding ofQinGreduce to the case whereG is a p-group, i.e. where Q is a modular subgroup. Bearing in mind that modularity is a property invariant under subgroup lattice isomorphisms, the quasinormal subgroups of a finitep-groupGare surely relevant when the structure ofGis approached via its lattice of subgroups.

It was shown in [10] that givenQ qn GandQ abelian, then Qⁿqn G, providedn is odd or divisible by 4 (even whenGis infinite). Apart from this, very little appears to be known about which subgroups ofQare also quasinormal inG.

ClearlyQ qn Gif and only ifQXis a subgroup for each cyclic subgroup X of G. Thus the situationGˆQX is an obvious starting point for in- vestigations. The case when G here is a finite p-group (for p an odd prime), andQis an abelian quasinormal subgroup ofG(withXcyclic), is studied in [2]. It is shown that there are two composition series of G passing throughQ, all the members of which are quasinormal in G. The attempt to remove the hypothesis that Q is abelian has produced the present work. We have discovered a situation very far removed from the abelian case. In fact there is a family of finitep-groupsG(for each odd primep) withQ qn Gsuch that

(i) there are subgroupsL3M4Q, withL,M qn G;

(ii) Ghas no quasinormal subgroup strictly betweenLandM;

(iii) Lhas exponentpandMhas exponentp^m; and

(iv) mand the nilpotency class (even derived length) ofM=Lcan be greater than any given positive integer.

These groupsGwere constructed by Berger and Gross in [1] and they are defined as follows. Letnbe a positive integer andpbe an odd prime.

We defineGnto be the additive groupZof integers modulopⁿZ. Letxnbe the permutation ofGn given by

x_n: pⁿZ ‡ `7!p<sup>n</sup>Z ‡ ` ‡ 1:

Then x_n belongs to the symmetric group of degree pⁿ. For 04m4n, define

x_n;mˆx^p_n^{n m}; of orderp^m:

Let X_nˆ hx_ni, a cyclic group of order pⁿ; and let X_n;mˆ hx_n;mi, the subgroup of orderp^m. DefineD_n;mto be the set of elements in the additive

groupGn of orderp^m. So

Dn;mˆ flp^{n m}j04l4p^m 1; pj⁴⁴lgfor 14m4n

andDn;0ˆ f0g. ThenjDn;mj ˆp^{m 1}(p 1) for 14m4n. The permutation x_n;m shifts byp^{n m}and fixes the setD_n;m‡1, i.e. the set of lp^{n m 1} (pj⁴⁴l) between 0 andpⁿ 1, for 04m4n 1. Thusx_n;macting on the setD_n;m‡1 (of cardinality p^m(p 1)) is the product ofp 1 disjoint cycles p_n;m;i of lengthp^m, 14i4p 1, each cycle addingp^{n m}. So

p_n;m;iˆ(ip^{n m 1}; ip^{n m 1}‡p^{n m};. . .; ip^{n m 1}‡(p^m 1)p^{n m}):

Now for 04m4n 1, define A_n;mˆY^{p 1}

iˆ1

p^c_n;m;iⁱ X^{p 1}

iˆ1

c_iˆ0

:

Each permutation in this set is a product of disjoint cycles, each containing an integer of the formlp^{n m 1}(pj⁴⁴l) and shifting by a multiple ofp^{n m}. In fact An;m is an abelian group isomorphic to the direct product of p 2 cyclic groups of order p^m. (See [1], Lemma 3.1 (1).) Its elements fix every integer not inDn;m‡1.

Berger and Gross define the group

G_nˆ hx_n;A_n;mj04m4n 1i;

andH_n to be the stabiliser of 0 inG_n. So G_n ˆH_nX_n;

a finitep-group. AlsoHn has exponentp^{n 1}. The main result of [1] is:- THEOREM. The subgroup H_n is core-free and quasinormal in G_n. Moreover, if H is a core-free quasinormal subgroup of a finite p-group GˆHhxi, wherehxiis a cyclic group of order pⁿ, then there is a unique embeddingcof G in Gn such thatc(x)ˆxn andc(H)4Hn.

Our main results concern the quasinormal subgroups ofG_n that lie in Hn. In the Berger-Gross Theorem above, the primepis arbitrary. How- ever, the casepˆ2 requires much additional analysis. Accordingly in all our work here we assume that

the prime p is odd:

We shall prove

THEOREM 1. Let n52 and let Q be a non-trivial quasinormal sub- group of Gnlying in Hnof exponent p^k.Then one of the following holds:-

(i) kˆ1;

(ii) n54and kˆn 2;

(iii) n53and kˆn 1.

Moreover for n54, the quasinormal subgroups of exponent p are nested and are contained in all those of exponentp^{n 2}and exponentp^{n 1}. Thus there is a maximal quasinormal subgroup L of exponent p and a quasinormal subgroupMof exponentp^{n 2}such thatL4M and

there are no quasinormal subgroups of Gnin the interval[M=L]:

ClearlyL3M. AlsoM=Lhas exponentp^{n 3}and there is no bound to its derived length asnincreases. (See [9].)

In Theorem 2, we find all the quasinormal subgroups ofGnlying inHn

and of exponent p. Theorem 3 gives first some necessary conditions for subgroups ofGn, lying inHnand of exponentp^{n 1}, to be quasinormal inGn; then necessary and sufficient conditions for certain subgroups of exponent p^{n 1}to be quasinormal inG_n. Theorem 4 is similar, dealing with subgroups of exponentp^{n 2}. Finally in Theorem 5 we show that there are no quasi- normal subgroups of Gn (n55), lying in Hn and of exponent p^k, for 24k4n 3. Theorem 1 follows from these results.

We use standard notation for familiar concepts, together with that above introduced by Berger and Gross. Also we switch from multiplicative to additive notation for computation with modules over groups. In this connection there is a considerable amount of calculation involved and the following list should be helpful.

(a)_b;g: the cyclic permutation containinga, shifting byb, of lengthg. Hereais a non-negative integer and b and g are positive integers such that bgˆpⁿ.

fb_i;kjr_k5i51g: a basis (moduloV_{k 1}(H_n)) of the elementary abelian group Vk(Hn)=Vk 1(Hn).

B<sub>`k;k</sub>: the subgrouphb<sub>i;k</sub>j`_k5i51i.

B_k: the subgroupB_rk;k.

C_k: the centraliser inG_n of the element x_n;k, i.e. C_Gn(x_n;k), chiefly when kˆ2.

rk: the rank of the elementary abelian group Vk(Hn)=Vk 1(Hn), for 14k4n 1.

Xn: the cyclic grouphxniof orderpⁿ.

xn;m: the elementx^pn^{n m}, 04m4n.

Xn;m: the cyclic subgrouphxn;miof orderp^m.

fw_ijr15i51g: a basis of the elementary abelian group V1(Hn), and w_iˆb_1;i.

W_j: the subgrouphw_ijj5i51i(=B_j;1).

Z_p^m: a cyclic group of orderp^m.

Vk(G): the subgroup of the finitep-groupGgenerated by the elements of order at mostp^k.

2. Description of the groupsG_n. It is shown in [1], Lemma 3.14, that

Vk(Gn)ˆ hx_nⁱAn;kxⁱ_nji50iVk 1(Gn);

…1†

for 14k4n 1. Also

V_k(G_n)=V_{k 1}(G_n)is elementary abelian of rank p^{n k 1}(p 1):

…2†

We see this as follows. By [1], Lemma 3.3, if n52, then there is an epi- morphismt_n:G_n ! G_{n 1}, viz.g7!t_n(g) where

tn(g): amodp^{n 1}Z7!bmodp^{n 1}Z …3†

if g: amodpⁿZ7!bmodpⁿZ;

for all integersa,bandg2Gn. In particulartnmapsHnontoHn 1andxnto x_{n 1}. The kernel oft_nis elementary abelian, by [1], Lemma 3.3 (9), and it is precisely V₁(G_n), by [1], Lemma 3.10. AlsojG_nj ˆp^pn1, by [1], Corollary 3.15. Thus

V₁(G_n)is elementary abelian of rank p^{n 1} p^{n 2}ˆp^{n 2}(p 1):

Applyingtnrepeatedly for decreasing values ofn, we see that G_n=V_{k 1}(G_n)G_{n k‡1};

…4†

for 24k4n; and

V₁(G_n=V_{k 1}(G_n))ˆV_k(G_n)=V_{k 1}(G_n)V₁(G_{n k‡1});

…5†

this last subgroup being elementary abelian of rankp^{n k 1}(p 1), proving (2). We shall use (4) and (5) repeatedly throughout our work.

By [1], Lemma 3.10,

V1(Gn)ˆV1(Hn)Xn;1: Also, by (1),

V₁(G_n)ˆ hx_nⁱA_n;1xⁱ_nji50i;

an elementary abelian group of rankp^{n 2}(p 1), by (2). By [1], Lemma 3.2 (4), xn;1lies in the centre of Gn:

Let

r₁ˆp^{n 2}(p 1) 1ˆrank ofV₁(H_n);

…6†

for n52. Recall that Hn is core-free in Gn and so V1(Gn) is an in- decomposableXn-module. We find a canonical basis forV1(Gn) as follows.

Define

w_r1ˆx^p_n^{n2 1}(p_n;1;1¹ p_n;1;2)x_n^{(pn2 1)}2V₁(G_n):

…7†

We introduce the following shorthand notation for cyclic permutations of integers modulopⁿZwhich just add a fixed integer:-

(a)_b;gˆ(a; a‡b; a‡2b;. . .; a‡(g 1)b);

…8†

wherebgˆpⁿ. Then

p_n;1;1¹ pn;1;2ˆ(p^{n 2})_pn¹1;p(2p^{n 2})_p^{n 1}_;p2An;1:

Note that the two cycles here are disjoint, coming from different orbits of x_n;1. Also

w_r1ˆ(1)_pn¹1;p(1‡p^{n 2})_pn1;p;

where again the two cycles are disjoint. Therefore wr1 fixes 0 and so wr1 2V1(Hn).

Forr1 15i50, define

w_iˆ[w_i‡1;x_n] …9†

inductively. Then dropping the suffixes in (8) for the moment and writing rˆr₁andxˆx_n, we have

w_{r 1}ˆ(1)(1‡p^{n 2}) ¹(2) ¹(2‡p^{n 2})ˆ(1)(2) ¹(1‡p^{n 2}) ¹(2‡p^{n 2}), wr 2ˆ(1) ¹(2)²(3) ¹(1‡p^{n 2})(2‡p^{n 2}) ²(3‡p^{n 2}),

w_{r j}ˆ(1)^{( 1)j‡1}(2)^?. . .(j‡1‡p^{n 2}),

w₁ˆ(1) ¹(2)^?. . .(p^{n 1} 1), w0ˆ(1)(2)^?. . .(p^{n 1}).

Any 2 cycles above are either the same or disjoint. Hence they commute.

Also w_i, for r₁(ˆr)5i51, fixes 0 and so belongs to V₁(Hn). Clearly

w_r1; w_{r1 1};. . .; w₀ are linearly independent as elements of the module

V₁(G_n), since each contains a cycle that does not occur in its predecessors.

The rank ofV1(Gn) isr1‡1, by (2). Therefore

fw_ijr15i51gis a basis for V1(Hn) …10†

and

fw_ijr₁5i50gis a basis for V₁(G_n):

…11†

Clearly V1(Hn) is an indecomposable Xn-module. For if not, then the centraliser of Xn in Gn is not cyclic and therefore intersects V1(Hn) non-trivially. Then the normal closure of this intersection in G_n would lie in H_n, giving a contradiction. Also we must have w₀2 hx_n;1i and x_n;1ˆ(1)(2). . .; (p^{n 1}), with cycles of length p, shifting by p^{n 1}. Thus

w0ˆxn;1ˆx^p_nⁿ¹: …12†

NOTE. For anyi,j, using additive module notation, we have [wi;x^p_n^j]ˆwi(x^p_n^j 1)ˆwi(xn 1)^pj ˆw_{i p}^j: …13†

Herew_i, fori50, is taken to be 1.

Next, for 24k4n 1, we find a basis for V_k(Gn)=V_{k 1}(Gn) as in- decomposableX_n-module of rank

p^{n k 1}(p 1)ˆr_k‡1;

…14†

say (see (2)). Notation will necessarily becomes more complicated. By [1], Corollary 3.11,

Vk(Gn)ˆVk(Hn)Xn;k: …15†

Thus by analogy with (7), define

b_rk;kˆx^p_n^{n k 1}(p_n;k;1¹ p_n;k;2)x_n^{(pn k 1 1)}

for 14k4n 1. Then using the notation (8), we have p_n;k;1¹ p_n;k;2ˆ(p^{n k 1})_pn k¹ ;p^k(2p^{n k 1})_p^{n k}_;p^k2A_n;k and

b_rk;kˆ(1)_pn k¹ ;p^k(1‡p^{n k 1})_p^{n k}_;p^k: …16†

Sinceb_rk;k fixes 02Gn, it belongs toV_k(Hn). Whenkˆ1, we have b_r1;1ˆw_r1:

Under the isomorphism (5),b_rk;k moduloV_{k 1}(G_n) corresponds to (1)_pn k¹ ;p(1‡p^{n k 1})_pn k;p

…17†

inV1(Gn k‡1), since in the setZ=p^{n k‡1}Zon whichGn k‡1acts,p^{n k‡1}0.

Also (17) is br1;1in G_{n k‡1}; and under the homomorphism (3), xn7!xn 1. Therefore defining

b_i;kˆ[b_i‡1;k;x_n] …18†

forr_k 15i50, we see from (10) and (11) that

fbi;kjrk5i51gis a basis forVk(Hn) moduloVk 1(Gn);

even moduloV_{k 1}(Hn), since (as forkˆ1)b_i;k 2Hnifk6ˆ0. Also fb_i;kjr_k5i50gis a basis forV_k(G_n) moduloV_{k 1}(G_n):

…19†

Note thatb_i;1ˆw_i, for alli. Also

tnmapsb_i;2inGn tob_i;1(ˆw_i) inGn 1; …20†

for alli. We shall use the `w' -notation when we wish to emphasise the fact that we are consideringV₁(G_n) asX_n- (orG_n-)module.

By [1], Lemma 3.1 (4),

bi;kcommutes withxn;k; …21†

for alln 15k51 andrk5i50. Let

B_kˆ hb_i;kjr_k5i51i;

14k4n 1. ThenB_khas exponentp^kand Y^{n 1}

kˆ1

B_k

X_n ˆG_n: …22†

LEMMA1. Let14k4n 1.Then

(i) hBk;Xn;ki Z_p^k Z_p^k (with rk‡1factors); and (ii) [w_i;B_kX_n;k]ˆ1, for p^{n k} 15i51,24k4n 1.

PROOF. (i) By (21), [B_k;X_n;k]ˆ1. Let C_kˆC_Gn(x_n;k). So, writing XˆX_n,hB_k;Xi4C_kandC_kˆ(H_n\C_k)X. Then by (15),

Vk(Ck)ˆVk(Hn\Ck)Xn;kˆDk;

say. SinceDk is a characteristic subgroup ofCk, it follows that the derived subgroupD⁰_k is normalised byX. ButD⁰_k4Hnand so (D⁰_k)^Gn4Hn. There- fore sinceH_nis core-free inG_n, we must haveD⁰_kˆ1, i.e.

D_k is abelian:

…23†

However,hBk;Xn;ki4Dkand sohBk;Xn;kiis abelian of exponentp^k. Thus (i) follows from (19). Observe that

B_kˆ hb_rk;ki hb_{rk 1;k}i hb_1;ki;

which is homogeneous of exponentp^k. (ii) Letk52. By (13),

[w_i;x_n;k]ˆw_{i p}n k ˆ1;

for 14i4p^{n k} 1. Thereforewi2Dk and (ii) follows from (23). p We have bases (19) for the V-layers of G_n (as X_n-modules) and the homomorphism t_n: G_n ! G_{n 1} maps these bases to those of G_{n 1} (see (20)). In fact the bases that we have chosen have another very useful property withinGn itself. We claim first that

b^p_r2;2ˆb_r2;1(ˆwr2):

…24†

For, from (16) withkˆ2, where the two cycles are disjoint, b^p_r2;2ˆY^{p 1}

iˆ0

(1‡ip^{n 2})_pn¹1;p(1‡p^{n 3}‡ip^{n 2})_p^{n 1}_;p: …25†

The cycles in (25) are disjoint and hence commute. By (11) and (19) (with kˆ2),

V2(Gn)ˆB1B2Xn;2:

Therefore by Lemma 1 (i), and using the notation of that Lemma, D2ˆ(D2\B1)B2Xn;2

andD2is abelian, by (23), and is normalised byXn. Thus again by Lemma 1 (i), D^p₂ ˆB^p₂X_n;1is elementary abelian

of rank r₂‡1 and is an X_n-submodule of V₁(G_n). Hence, setting W_jˆ hw_ijj5i51iforr₁5j50, we must have

b^p_r2;22D^p₂\HnˆWr2: …26†

Now from the equations following (9) and a simple calculation, we have b_r2;1ˆ(1)_pn¹1;p(2)^?_pn1;p. . .(1‡p^{n 3}‡(p 1)p^{n 2})_pn1;p:

…27†

Observe that the last cycle here is the same as the last cycle in (25). But from (26),

b^p_r2;2ˆ^rY^{2 1}

iˆ0

b^a_r2ⁱ _i;1;

for integersa_i. Therefore, again by the equations following (9), we must have ar2 1 ˆar2 2ˆ ˆa1ˆ0

and so

b^p_r2;22 hb_r2;1i:

Thus (24) follows by comparing the exponents of (1)_p^{n 1}_;pin (25) and (27).

From (18) we deduce that

b^p_{r2 1;2}ˆ[b_r2;2;x_n]^pˆ[b^p_r2;2;x_n];

sinceB^X₂ⁿ4D2, which is abelian. Therefore, by (24) and (9), b^p_{r2 1;2}ˆ[br2;1;xn]ˆbr2 1;1:

Continuing in this way, we see that

b^p_i;2ˆbi;1(ˆwi);

…28†

r₂5i51. Also, by (18),

b_0;2ˆ[b_1;2;x_n]x_n;2 mod V₁(G_n);

…29†

using (12) and (20). Moreoverhb_0;2;x_n;2i4C₂ and V₁(G_n)\C₂ˆW_pn2 1X_n;1 (by (13)). Therefore

b^p_0;2ˆxn;1(ˆb0;1):

…30†

Combining the above results, we can now obtain useful information about adjacentV-layers inGn. For 24k4n 1, we have from (4)

V2(G_{n k‡1})V2(Gn=V_{k 1}(Gn))ˆV_k‡1(Gn)=V_{k 1}(Gn):

…31†

Thus we deduce from (28) and (30) that

b^p_i;k‡1bi;k mod Vk 1(Hn);

…32†

forr_k‡15i51; and

b^p_0;k‡1xn;kb0;k mod Vk 1(Gn):

More information about adjacent V-layers of G_n will follow from our next key result.

LEMMA2. Let24k4n 1and for04i4p^{k 2}(p 1), let V_iˆW_ip^{n k} ₁:

Then B_kcentralises the series

V₁(H_n)ˆW_rˆV_pk2(p 1)> >V_i>V_{i 1}> >V₀ˆ1 …33†

of length p^{k 2}(p 1).

PROOF. By Lemma 1 (ii),BkcentralisesV1. We argue by induction oni increasing and suppose that [V_i‡1;B_k]4V_ifor somei50. Then by (13)

[V_i‡2;X_n;k]ˆV_i‡1X_n;1; and so

[V_i‡2;X_n;k;B_k]4V_i:

Since [Xn;k;Bk]ˆ1, by Lemma 1 (i), the Three Subgroup Lemma (see for example [8], Lemma 2.13) implies

[B_k;V_i‡2;X_n;k]4V_iX_n;1(3G_n):

Therefore [B_k;V_i‡2]4V_i‡1X_n;1, again by (13). Thus [Bk;Vi‡2]4Vi‡1Xn;1\HnˆVi‡1

and the Lemma follows. p

COROLLARY1. For 04i4n 2, the factor groupVi‡2(Gn)=Vi(Gn) is nilpotent of classp 1and hence regular.

PROOF. By (31) we may assume thatiˆ0. Take kˆ2 in Lemma 2.

ThenB2centralises the series (33) of lengthp 1. Also, moduloXn;1,Xn;2

centralises this series (by (13)). Therefore, sinceB2Xn;2 is abelian (by Lemma 1 (i)), the subgroup

V2(Gn)ˆB1B2Xn;2

is nilpotent of class at mostp 1. In fact the class is exactlyp 1, because of the action ofXn;2 onV1(Gn)ˆWrXn;1. The regularity follows, for ex-

ample, from [3], Corollary 12.3.1. p

3. Subgroups ofGn.

We can now describe all the quasinormal subgroups ofGn lying inHn

and of exponentp.

THEOREM 2. Let n52.Then a quasinormal subgroup of G_n of ex- ponent p, lying in Hn, has the form Wi, for some i.Moreover Wiqn Gnif and only if14i4p² 1.

PROOF. Observe from (6) that whennˆ2,i4p 2; and whennˆ3, i4p(p 1) 1. Therefore in both these cases we are saying that Wiqn GˆGnfor allifor whichWiis defined.

Let Q qn G with Q of exponent p and lying in HˆHn. Then Q4Wr1 ˆV1(H). Also QXn is a subgroup and V1(QXn)ˆQXn;1. Thus QX_n;1is anX_n-submodule ofV₁(G) and so, since indecomposable modules are uniserial (see [4], Theorem VII, 5.3),QX_n;1ˆW_iX_n;1, for somei. Hence QˆWi.

We establish first the sufficiency of the condition oni. Thereforesup- pose that i4p² 1 and letgˆhy, whereh2Handy2XˆXn, and put Kˆ hgi. We show thatW_iKis a subgroup by considering 3 possibilities.

(i)Suppose that y2X^p2ˆX_{n;n 2}. ThenycentralisesW_i, by (13). Since Wi3H, it follows thatgnormalisesQ.

(ii)Suppose thathyi ˆX. Then by [1], Lemma 3.12 (ii),V1(K)ˆXn;1. ThereforeW_iK ˆW_iXn;1K is a subgroup, sinceW_iXn;1/G.

(iii) Finallysuppose thathyi ˆX^p. Clearly we may assume thatn53.

By Corollary 1, V_{n 1}(G)=V_{n 3}(G) is regular with elementary abelian de- rived subgroup. Therefore

g^ph^py^p mod V_{n 3}(G):

Also h^p2 hb_{j;n 2}j14j4p 2i mod Vn 3(H);

by (32) and (14). Continuing takingp-th powers in this way, we obtain g^pn2 2W_{p 2}X_n;1:

…34†

MoreoverjHK:Hj ˆ jHhyi:Hj ˆ jhyij ˆp^{n 1}. Thereforeg^pn22=H.

Now considerWiK. Ifi4p 1, thenycentralisesWi(by (13)) and so again g normalises W_i, i.e. W_iK is a subgroup. On the other hand if p4i(4p² 1), then by (34)

W_iK ˆW_iX_n;1K and this is a subgroup as in (ii).

Conversely, suppose, for a contradiction, that W_iqn G where i5p². Then by (6), n54. Let gˆb_{1;n 1}x_{n;n 2} and Kˆ hgi. We have [b1;n 1;xn;n 1]ˆ1, by (21). So

g^pn2ˆb^p_1;nⁿ²₁ˆb_1;1 ˆw₁; …35†

using the argument of (iii) above. Consider the subgroupW_iK. SinceV₁(G) is elementary abelian, an element of orderpinW_iK must have the form wg1, where w2Wi and g12V1(K)ˆW1 (by (35). Thus V1(WiK)ˆ W_i3W_iK. SinceW_i3H, it follows that xn;n 2 also normalises W_i. But [w_p²;x_{n;n 2}]ˆx_n;1 (by (13)), a contradiction. This completes the proof of

Theorem 2. p

For the next theorems we need to be able to identify certain subgroups ofGn. Letn52,n 15k51 andr_k5`_k51. Define

B<sub>`k;k</sub> ˆ hb<sub>j;k</sub>j14j4`_ki;

which is isomorphic toZ_pk Z_pk (`_k copies), by Lemma 1 (i).

Note that B_{`1;1</sub>ˆW<sub>`1}.

LEMMA 3. Let n52, r_k5`_k5p^{n k 1} 1 for 14k4n 3, rn 25

`<sub>n 2</sub>5`_{n 1}and r_{n 1}5`_{n 1}51.Then

LˆB_{`1;1</sub>B<sub>`2;2}. . .B_{`n1;n 1}X_{n;n 1} …36†

is a subgroup of Gn; and

B_{`1;1</sub>B<sub>`2;2}. . .B_{`n1;n 1}ˆL\H_n …37†

is a subgroup of H_n.

PROOF. Ifnˆ2, thenLˆW`1Xn;13G2. Ifnˆ3, thenLˆW`1B`2;2Xn;2

with r1ˆp(p 1) 15`15`2 and r2ˆp 25`251. By Lemma 1 (i), B`2;2Xn;2is a subgroup. AlsoW`1Xn;13G3. SinceLis the product of these two subgroups, againLis a subgroup.

Now suppose that the Lemma is true forn 1, somen54, and argue by induction onn. Then from (4) we obtain

LV1(Gn)ˆWr1L

is a subgroup. Recall that C₂ˆC_Gn(x_n;2). It follows from (13), (22) and Lemma 1 (i), that

C2ˆW_pⁿ² ₁Br2;2. . .Brn1;n 1Xn:

Then Wr1L\C2ˆWr1B`2;2. . .B`n 1;n 1Xn;n 1\C2

ˆW_pⁿ² ₁B`2;2. . .B`n1;n 1Xn;n 1

is a subgroup. Since `15p<sup>n 2</sup> 1, forming the product of this subgroup with the normal subgroupW<sub>`1</sub>X_n;1, we obtainL. SoLis a subgroup.

Equation (37) follows from Dedekind's Intersection Lemma. p The quasinormal subgroups of exponent p^{n 1} that we shall exhibit in Theorem 3 all have the formL\Hngiven by (37); and those of exponent p^{n 2} in Theorem 4 have the formVn 2(L\Hn). As a consequence of the next result,we see that these subgroups are actually normal inH_n.

LEMMA4. Let n53, r_k5`<sub>k</sub>5r<sub>k</sub> p<sup>2</sup>for14k4n 3, r<sub>n 2</sub>5`_{n 2}5`<sub>n 1</sub> and r<sub>n 1</sub>5`_{n 1}51.Define L as in(36).Then

(i) L is a subgroup of G_n; and (ii) for14k4n 1,

V_k(L)ˆB_{`1;1</sub>B<sub>`2;2}. . .B_`k;kX_n;k ˆL_k; …38†

say; and Lk3Gnfor14k4n 2.

Moreover, if rk5`k5rk p for14k4n 2and rn 15`n 151, then (iii) again L is a subgroup of Gnand L3Gn.

PROOF. (i) The hypotheses here imply those of Lemma 3. For, with n53 and 14k4n 3,

r_k p² (p^{n k 1} 1)ˆp^{n k 1}(p 1) p² p^{n k 1}ˆp^{n k 1}(p 2) p² 5p²(p 2) p²50:

ThereforeLis a subgroup ofGn.

(ii) It is easy to check that, for 14k4n 2,`k5`k‡1. We claim that, for 24k4n 1,

(B`k;kXn;k)<sup>p</sup>B`1;1B`2;2. . .B`k 1;k 1Xn;k 1: …39†

We do not know yet that the right side of (39) is a subgroup. In order to prove (39), suppose that kˆ2. ThenB<sub>`k;k</sub>X<sub>n;k</sub> Z<sub>p</sub><sup>2</sup> Z<sub>p</sub><sup>2</sup> (`_k‡1 copies), by Lemma 1 (i). Thus (39) follows from (28), since`<sub>1</sub>5`₂. Therefore (39) is true for nˆ3. We suppose that (39) is true forG_{n 1} (n54) and 24k4n 2 and argue by induction onn. Then for 34k4n 1, we have from (20)

(B`k;kXn;k)<sup>p</sup>Wr1B`2;2. . .B`k 1;k 1Xn;k 1: …40†

However, the left side of (40) and all but the first factor on the right belong toC2(ˆC_Gn(xn;2)). Thus intersecting (40) withC2gives

(B`k;kXn;k)<sup>p</sup>W<sub>p</sub><sup>n2</sup> <sub>1;1</sub>B`2;2. . .B`k 1;k 1Xn;k 1

B`1;1B`2;2. . .B`k 1;k 1Xn;k 1; since one checks easily that

`₁5p^{n 2} 1:

…41†

Therefore (39) is true.

Now we can prove (38). Clearly Vn 1(L)ˆLˆLn 1. We proceed by induction onkdecreasing and suppose that (38) is true for somek4n 1.

Certainly

V_{k 1}(L)B_{`1;1</sub>. . .B<sub>`k 1;k 1}X_{n;k 1}: …42†

If the inclusion (42) is strict, then there is an element g2B_{`k;k</sub>X<sub>n;k</sub> with jgj ˆp<sup>k 1</sup>andg2=L<sub>k 1</sub>. But by Lemma 1 (i),g2(B<sub>`k;k}X_n;k)^p, contradicting (39). Therefore (42) is an equality. Thus our induction argument goes through and we have proved (38).

The second part of (ii) will follow from L_{n 2}3G_n: …43†

When nˆ3,LˆW`1B`2;2X3;2 andL1ˆW`1X3;13G3. Therefore we sup- pose that n54 and that (43) is true for Gn 1 and proceed by induction.

Again from (20)

W_r1B_{`2;2</sub>. . .B<sub>`n2;n 2}X_{n;n 2}3G_n; and intersecting withC₂gives

W_pⁿ² ₁B_{`2;2</sub>. . .B<sub>`n 2;n 2}X_{n;n 2}3C₂:

Thus from (41) andW`1Xn;13Gn, we see thatLn 2 is normalised by C2. However, [Wr1;Ln 2]4W_{r1 p}²Xn;1, by Lemma 2, and so Wr1 normalises Ln 2. SinceWr1C2ˆGn, (43) follows.

(iii) It is easy to check that the hypotheses here imply those of parts (i) and (ii). SoLis a subgroup. Now however

[W_r1;L]4W_{r1 p}X_n;14L;

by Lemma 2, and soWr1normalisesL. Also whennˆ3, by consideringL modulo V1(G3) and intersecting with C2, we see that L3Gn. Then the induction argument used to prove (43) can be applied here to giveL3Gn

for alln. p

4. Quasinormal subgroups of large exponent.

We shall see that the quasinormal subgroupsQofGnlying inHn, other than those of Theorem 2, all have exponentp^mformˆn 2 orn 1. Also V_k(Q) modulo V_{k 1}(Hn) has `large' rank, for all 14k4m 1. We begin with those of exponentp^{n 1}.

THEOREM3. (i)Let n53 and Q be a quasinormal subgroup of G_n of exponent p^{n 1}and lying in Hn.Then for14k4n 1,

Vk(Q)B`k;k modVk 1(Hn);

…44†

and`<sub>k</sub>5r<sub>k</sub> p, for14k4n 2, and`n 151.

(ii)Again let n53and

QˆY^{n 1}

kˆ1

B_`k;k

with`<sub>k</sub>5r<sub>k</sub> p for14k4n 2and`_{n 1}51.Then Q is a subgroup of H_nof exponent p^{n 1}.Moreover Q qn G_nif and only if

`k5rk p‡1;

…45†

for14k4n 2.

PROOF. (i) Since QX_n is a subgroup, we see that V₁(QX_n) is an X_n- module contained inV₁(G_n) and therefore has the formW_iX_n;1. Also there is an elementhb_{1;n 1}2Qwithh2V_{n 2}(H_n), by Theorem 2. Thus sinceW_r1 must normaliseQ,

[W_r1;hb_{1;n 1}]4Q\W_{r1 p};

by Lemma 2. However

this commutator subgroup does not belong to Wr1 p 1:

For, otherwise we would have [Wr1;b1;n 1]4Wr1 p 1. Then conjugating by xn, we would obtain (using (18))

[W_r1;b_{1;n 1}b_{0;n 1}]4W_{r1 p 1}X_n;1: …46†

But b_{0;n 1}x_{n;n 1} mod V_{n 2}(G_n), using (29) and the homomorphismt_n. Thusx_{n;n 1}would centraliseW_r1X_n;1=W_{r1 p 1}X_n;1, a contradiction.

Thus (44) is true forkˆ1. The rest of (i) follows by induction onnfrom Gn=V1(Gn)Gn 1.

(ii) The hypotheses here imply those of Lemma 4. ThusQis a subgroup ofH_nof exponentp^{n 1}(of the form (37)).

Now suppose that Q qn G_n. Assume for the moment that

`15r1 p‡1:

…47†

Then we obtain (45), again by induction onn. So it suffices to prove (47).

Letgˆw_r1x_nandKˆ hgi. Then

V1(K)ˆ hxn;1i;

…48†

by [1], Lemma 3.12 (2). In fact this follows easily from a simple module calculation. Thus using additive notation,

g^pˆ(wr1xn)^pˆwr1(1‡x_n¹‡x_n²‡ ‡x_n^{(p 1)})xn;n 1

ˆwr1(x_n¹ 1)^{p 1}xn;n 1 wr1(xn 1)^{p 1}xn;n 1 mod Wr1 pXn;1

w_{r1 p‡1}x_{n;n 1} mod W_{r1 p}X_n;1: …49†

(The appearance ofw_{r1 p‡1}here is the clue to establishing (47).) Similarly g^p2w_{r1 p}²_‡1xn;n 2 mod W_{r1 p}²Xn;1:

(We can obtain (48) by computingg^pn1in this way.) Since`1>r1 p²‡1, it follows from (48) that

x_{n;n 2}2QK:

…50†

Alsob1;n 12Q, by hypothesis, and soQKcontains [b1;n 1;g]ˆ[b1;n 1;xn][b1;n 1;wr1]^xn: However, by Lemma 2,

[b_{1;n 1};wr1]^xn2W_{`</sub><sup>x</sup><sub>1</sub><sup>n</sup>ˆW<sub>`}^g₁4QK:

Hence

QKcontains [b1;n 1;xn]ˆg1; …51†

say. Nowg12Q^Gn4QXn;n 1, sinceQXn;n 13Gn, by Lemma 4 (iii). From (4) and (12) we have

g₁ˆx_{n;n 1}g₂;

where g₂2V_{n 2}(G_n)ˆV_{n 2}(H_n)X_{n;n 2}. Therefore g₂ˆhy, with h2 V_{n 2}(H_n) andy2X_{n;n 2}. Thusy2QK, by (50). Then

g12QXn;n 1)g22QXn;n 1)h2QXn;n 1\Hn ˆQ)g22QK) )x_{n;n 1}2QK;

by (51).

Finally from (49) we now obtain

wr1 p‡12QK\HnˆQ:

Thus`₁5r₁ p‡1, as required.

Conversely, suppose that `<sub>k</sub>5r<sub>k</sub> p‡1 for 14k4n 2 and`_{n 1}51.

We show thatQ qn G_n. Letgˆhy, whereh2H_n,y2X_n, and letKˆ hgi.

It suffices to show that QK is a subgroup. Let y2Xn;knXn;k 1, where 04k4n. (Assume thatXn; 1ˆ1.) We claim that

g^p2QX_{n;k 1}: …52†

Note thatQX_{n;n 1}3G_n, by Lemma 4 (iii). ThusQX_n;iis a subgroup for alli;

and

Q3Hn: …53†

Let hˆwb2b3. . .bn 1, where w2Wr1 and b_i2B_i, 24i4n 1. Con-

sideringV₁(G_n) asG_n-module, we have

g^pˆ(wb₂. . .b_{n 1}y)^pˆw((b₂. . .b_{n 1}y) ¹ 1)^{p 1}(b₂. . .b_{n 1}y)^p: Thus

g^p2W_{r1 p‡1}C₂; …54†

by Lemma 1. We prove (52) by induction onn.

Supposenˆ3. If k42, then g^p2W_{p 2}X_{3;k 1}, by Corollary 1, and so g^p2QX3;k 1. Ifkˆ3, then by (54)

g^p2W_r1X_3;2\W_{r1 p‡1}C₂ˆ(W_r1\W_{r1 p‡1}C₂)X_3;24QX_3;2:

So (52) holds fornˆ3. We proceed by induction onn, assuming thatn54 and that (52) holds inGn 1. Thus by induction (usingGn=V1(Gn)Gn 1) and (54), we have

g^p2W_r1X_n;1QX_{n;k 1}\W_{r1 p‡1}C₂4(W_r1Q\W_{r1 p‡1}C₂)X_n;1X_{n;k 1}

ˆW_{r1 p‡1}(W_r1Q\C₂)X_n;1X_{n;k 1}4W_{r1 p‡1}(W_pn2 1Q)X_n;1X_{n;k 1}; by the argument used in Lemma 3. Therefore g^p2QX_n;1X_{n;k 1}. So (52) follows ifk52. But ifk41, theng^p2H_nand sog^p2Q(ˆQX_{n;k 1}). Thus we have established (52).

Now from (52) we obtainQhg^pi4QX_{n;k 1}. But

jQhg^pij ˆ jQkhg^pi:Q\ hg^pij ˆ jQjp^{k 1}ˆ jQXn;k 1j:

Therefore

Qhg^pi ˆQX_{n;k 1}; …55†

a subgroup. By Lemma 4

V_{k 1}(QX_{n;n 1})ˆV_{k 1}(Q)X_{n;k 1}ˆN;

say, and N3G_n. If k4n 1, then by Lemma 1 (i), X_n;k centralises Q modulo N and therefore g normalises Q(by (53)). Otherwise kˆn and thenNˆQXn;n 1ˆQhg^pi, by (55); i.e.QKis a subgroup.

This completes the proof of Theorem 3. p

Next we deal with the quasinormal subgroups of exponent p^{n 2}. The argument follows that of Theorem 3, though there added complications.

Establishing the necessity of our conditions for quasinormality involves considering the product QK forKˆ hwr1b1;n 1xn;2irather than hwr1xni;

and the prime 3 causes problems.

THEOREM4. (i)Let n54and Q qn G_n, Q4H_n and Q have exponent p^{n 2}.Then for14k4n 2,

Vk(Q)B`k;k mod Vk 1(Hn);

…56†

and`k5rk p<sup>2</sup>, for14k4n 3, and`n 251.

(ii)Again let n54and

QˆY^{n 2}

kˆ1

B_`k;k;

with`<sub>k</sub>5r<sub>k</sub> p<sup>2</sup>for14k4n 3and`n 251.Then Q is a subgroup of Hn

of exponent p^{n 2}.Moreover Q qn Gnif and only if

`_k5r_k p(p 1)

for14k4n 3and`_{n 2}5p 2.

PROOF. (i) We argue exactly as in Theorem 3 (i) to obtain the form (56).

Then for the restrictions on `k we use the fact that there is an element hb1;n 22Q, with h2Vn 3(Hn). Thus [Wr1;hb1;n 2] lies inQ\W_{r1 p}² (by Lemma 2), but not inW_{r1 p}² ₁.

(ii) WithLdefined as in Lemma 4, we have QˆV_{n 2}(L)\H_n: HenceQis a subgroup.

Now suppose that Q qn G_n.We shall show that

`₁5r₁ p(p 1) …57†

and

`25p 2 whennˆ4:

…58†

Then the necessity of our conditions will follow by induction on n, as in Theorem 3. We shall prove that

w_pn2 2Q;

…59†

i.e.`₁5p^{n 2}. Assume this for the moment and let gˆwr1b1;n 1xn;2; Kˆ hgi:

Then sinceQKis a subgroup, (59) implies [w_pⁿ²;g]2QK:

Therefore substituting forgand expanding (using (13)), we get [w_p^{n 2};b_{1;n 1}]x_n;12QK:

Thusx_n;1 2QK. Considering actions on V₁(G_n) and using Lemma 2, we obtain

g^pˆwb^p_{1;n 1}x^e_n;1; where

w2W_{r1 p(p 1)}nW_{r1 p(p 1) 1}: …60†

The appearance of the suffixr1 p(p 1) in (60) is the clue to establishing