2-Groups with normal noncyclic subgroups

2-GROUPS F.

WITH N.

NORMAL

NONCYCLIC

SUBGROUPS UDC 512.4

Liman

We study n e a r l y - h a m i l t o n i a n 2 - g r o u p s (H2-groups). A n o n a b e l i a n 2 - g r o u p h a v i n g at l e a s t one p r o p e r n o n c y c l i c s u b g r o u p is c a l l e d an I~2-grou p if all n o n c y c l i c s u b g r o u p s a r e n o r m a l .

A n o n a b e l i a n g r o u p h a v i n g at l e a s t one p r o p e r n o n c y c l i c s u b g r o u p is c a l l e d an Q - g r o u p if all nonc y c l i c s u b g r o u p s a r e n o r m a l . If a p - g r o u p is an l~-group, we shall b r i e f l y call it an ~ p - g r o u p . In the p r e s e n t a r t i c l e a t h e o r e m will be p r o v e d which fully d e s c r i b e s the s t r u c t u r e of I~2-groups, thus c o m p l e t i n g the i n v e s t i g a t i o n of I~-groups p o s s e s s i n g a n o r m a l s y s t e m with l o c a l l y finite f a c t o r s which has b e e n s t a r t e d in [1]. It is well known (see [2]) t h a t e v e r y infinite 2 - g r o u p contains an infinite a b e l i a n s u b g r o u p . H e n c e , an infinite I~2-group is an e x t e n s i o n of an a b e l i a n g r o u p by an a b e l i a n o r h a m f l t o n i a n g r o u p and, t h e r e f o r e , is l o c a l l y finite. It is obvious that all h a m i l t o n i a n 2 - g r o u p s , e x c e p t the q u a t e r n i o n groups, a r e H2-groups. The following t h e o r e m d e s c r i b e s the s t r u c t u r e of the n o n h a m i l t o n i a n H 2 - g r o u p s . THEOREM. A nonhamiltonian types :

2 - g r o u p r is an H 2 - g r o u p if and only if it is o f o n e of the following

1) r = {A} {B}; A 8= B 4= 1, A 4= B 2, B - l A B = A-i; 2) r = {C} {D}; C 8= Ds= 1, C 4= D 4, D-1CD = C-l; 3) r

{B}) • {C}; F 2 n = B 2 = C 2 = 1 , [F, B] = [F, C] = I, [B, C ] = f 2 n - 1 ;

4) r = (~I • {B}) • {C}; B 2 = C 2= A12 = 1, AlE ~I, ~I is a q u a s i c y c l i c 2 - g r o u p , []3, C] = A 1, [A, B] = [A, C] = 1 f o r all A 6 )I; 5) r = ( ~ • {B})• i (i = 1, 2);

~ = {S1, S2} is a q u a t e r n i o n group, B 2 = C 2 = 1, [B, C] = S12, [Si, B] = [Si, C] =

6) gt = {X} • {Y}; X 2n= Y 2m = 1, n >- 2, m ->- 1, IX, Y] = X 2n-1. 7) • = ({D} • {F}) {C}; D 4= F 4= C 4 = 1 , C 2=D2F 2, [C, F] = C 2, [D, C] = D2; 8) gt = ({D} • {F}) {C} {Y}; D 4= F 4 = C 4 = Y 4 = 1 , C 2 = y 2 = D 2 F 2, [D, C] = D 2, [C, F] = C 2, [Y, D] = y2, [F, Y] = F 2, [Y, C] = D2; 9) r = ~ x {C}; ~ is a q u a t e r n i o n g r o u p , C 2n= 1, n>-2; 10) ~t = ~ • ~; ~ is a q u a t e r n i o n g r o u p , ~l is a q u a s i c y c l i c 2 - g r o u p . E a c h of the g r o u p s o c c u r i n g in the above t h e o r e m is, o b v i o u s l y , an H2-group, and h e n c e , only the n e c e s s i t y of the t h e o r e m will be p r o v e d . We r e d u c e the p r o o f of the n e c e s s i t y to the p r o o f s of the s u b sequent lemmas. Obviously, the following a s s e r t i o n holds: A n o n - h a m i l t o n i a n H p - g r o u p does not contain an e l e m e n t a r y a b e l i a n s u b g r o u p of type (p, p, p). It will be u s e d in the p r o o f of the t h e o r e m . L E M M A 1. A n o n - h a m f l t o n i a n H2-group r e i t h e r is a g r o u p of type 1) o r 2) of the t h e o r e m a b e l i a n e x t e n s i o n of an a r b i t r a r y one of its e l e m e n t a r y a b e l i a n s u b g r o u p s , of type (2, 2).

o r is an

P r o o f . We a s s u m e that r contains a unique involution. Using the local f i n i t e n e s s of $, T h e o r e m 12.5.2 of [3], and the definition of an I~2-grou p we conclude that ~ is a g e n e r a l i z e d q u a t e r n i o n g r o u p of o r d e r 16. M a t h e m a t i c s Institute, A c a d e m y of S c i e n c e s of the U k r a i n i a n SSR. T r a n s l a t e d f r o m M a t e m a t i c h e s k i e Z a m e t k i , Vol. 4, No. 1, pp. 75-84, July, 1968. Original a r t i c l e s u b m i t t e d D e c e m b e r 25, 1967.

535

Now we a s s u m e $ :~ 9 = {A} • {B}, w h e r e A 2 =B 2= 1 and, m o r e o v e r , the f a c t o r g r o u p fit/91 is h a m iltontan. Then $/9l contains a q u a t e r n i o n group $i/91 = {X91, Y~}. F i r s t of all we show that 91 is not c o n tained in the c e n t e r ~(gtl) of $1- In fact, if 91~ ~{$I), then, in view of the above r e m a r k , all involutions of $1 a r e contained in 91. In addition, r = {X, Y, ~} = {X, Y}. Indeed, if {X, Y} r ~ , then {X, Y} has a unique involution and, b y T h e o r e m 12.5.2 of [3], is a g e n e r a l i z e d q u a t e r n i o n g r o u p of o r d e r 16. But then the following d e c o m p o s i t i o n holds @,---~ {X, Y} • {N}, N ~ _ ~ , which y i e l d s that

@~/~R~- {X, Y}/{X, Y} ~ ~R. F r o m this follows that $i/91 is a dihedral group of o r d e r 8, c o n t r a r y to the definition of $1/91. Since X291 -- y291~ we have X 2 = y2N 1, N 1 E 91, and the a b e l i a n s u b g r o u p s ~Il -- {X 2, Y} and ~I2 = {X, y2} a r e m a x i m a l s u b g r o u p s of ~ . But then 8($i) = ~Ii n 9~2, and, t h e r e f o r e , the F r a t t i n i s u b g r o u p of $1 c o i n c i d e s with 8($1)- F r o m this follows that $I is a g r o u p all of w h o s e p r o p e r s u b g r o u p s a r e a b e l i a n (in the sequel g r o u p s with this p r o p e r t y will be c a l l e d M - g r o u p s ) . By the d e s c r i p t i o n of the M - g r o u p s in [4], the c o m m u t a t o r s u b g r o u p $~ has the o r d e r 2, and, h e n c e , $1/91 is an a b e l i a n g r o u p , c o n t r a r y to the definition of $1/91. C o n sequently, 91~.~($1)- Since 91 N .~(~) ~ T , we m a y a s s u m e , without l o s s of g e n e r a l i t y , that AE ~($1) and We show now that $i is a m e t a c y c l i c group. L e t ~I be a m a x i m a l s u b g r o u p of $i- We a s s u m e that ~I does not c o n t a i n e l e m e n t s of o r d e r 8. Then it cannot be a g e n e r a l i z e d q u a t e r n i o n g r o u p of o r d e r 16 and, t h e r e f o r e , b y T h e o r e m 12.5.2 of [3], c o n t a i n s a n o r m a l s u b g r o u p 911={A} x {Bl} , A 2 = Bi2= 1, w h e r e in gene r a l 911, does not coincide with 91. It is not difficult to c o n v i n c e o n e s e l f that ~I ~91. Indeed, o t h e r w i s e BE ~I and, t h e r e f o r e , $ i ~--- ~ •

{B},

g3i/~ ~_ g / { A } .

9J/{A}, h o w e v e r , is g e n e r a t e d b y e l e m e n t s of o r d e r 2, which is i m p o s s i b l e f o r q u a t e r n i o n g r o u p s . C o n sequently, ~ID91. But then ~I/91 is an a b e l i a n g r o u p of type (2, 2) w h i c h is i m p o s s i b l e in the q u a t e r n i o n g r o u p $1/91- T h u s , ~I c o n t a i n s e l e m e n t s of o r d e r 8 and, t h e r e f o r e , is a m e t a c y c l i c group. C o n s e q u e n t l y , all p r o p e r s u b g r o u p s of $1 a r e m e t a c y c l i c . Using the d e s c r i p t i o n of the n o n - m e t a c y c l i c p - g r o u p s all whose p r o p e r s u b g r o u p s a r e m e t a c y c l i c as given in [5] we c o m e to the c o n c l u s i o n that $i is m e t a c y c l i c . A s s u m e that {C} = ~ - 2 and m - - 2 o r a g r o u p of t y p e 7) o r 8) of the t h e o r e m . P r o o f . By L e m m a 1 $/91 i s a b e l i a n . We d e n o t e b y ~d/91 the l o w e r l a y e r of $/91. We s e p a r a t e l y c o n sider three cases: 1) ~I/91 i s a g r o u p of t y p e (2,2). We c h o o s e a n a r b i t r a r y n o n a b e l i a n s u b g r o u p $I of $.

~/m={Xm}x

{Ym},

r

Y,

m}={x,

T h e n we h a v e

Y},

s i n c e , b y a s s u m p t i o n , $ d o e s not c o n t a i n a q u a t e r n i o n g r o u p . C o n s e q u e n t l y , the a b e l i a n s u b g r o u p s 9d1 = {X 2, Y} and ~I2 = {X, y2} a r e m a x i m a l s u b g r o u p s o f St- But t h e n 8($1) = ~/1 [1 ~I2, a n d , t h e r e f o r e , t h e F r a t t i n i s u b g r o u p of $1 c o i n c i d e s with t h e c e n t e r ~ ($1) of $1- T h i s y i e l d s t h a t $~ i s a n M - g r o u p of t y p e 6) of the

537

t h e o r e m with n ~ 2 and m->2. (The case n>-2 and m = 1 has been treated in L e m m a s 2 and 3.) Since @1 has been choosen a r b i t r a r i l y and since @ is locally finite, we get @ = @1. 2) ~I/3 is a group of type (2,2,2). Since @ does not contain an abelian subgroup of type (2,2,2) and since ~I is a subgroup of exponent 4, 9/is a nonabelian group of o r d e r 32. All subgroups of o r d e r 8 of ~I a r e abelian, and among the subgroups of o r d e r 16 there must be at least one M-group (otherwise, ~I is an Mgroup generated by two elements, c o n t r a r y to our assumptions). Consequently, ~I is a n o n - m e t a c y c l i c group all of whose p r o p e r subgroups are metacyclic. Usingthe description of these groups in [5] we conclude that ~Iis a group of type 7) of the theorem. It is not difficult to show that @ = ~I. 3) Finally, we consider the case where the lower l a y e r of @/3 contains an e l e m e n t a r y abelian subgroup of o r d e r 16. In ~I/3 we choose a subgroup @1/3 of o r d e r 8. Then @2 is a n o n - m e t a c y c l i c group all of whose p r o p e r subgroups are m e t a c y c l i c . As b e f o r e , we may a s s u m e that

~,

=

({D}

C a = D2F 2,

x

{ F } ) {C};

v ~ ---- F ' =

C - i D C = D-i,

C' =

1,

F - i C F = C -i.

We c o n s i d e r the factor group of @2 over {C2}. It may be r e p r e s e n t e d as follows:

$~!{C2 } = ~ = [ ( { d

~} x {c}) x (dc}l •

{id},

where d, f, and c a r e h o m o m o r p h i c images of the elements D, F, and C, such that c 2 = (de) 2= (fd) 2= 1. If ~t contains all involutions of @~=@/{CZ}, then, by L e m m a 2, ~] also contains all elements of o r d e r 4 of ~. But then all elements of o r d e r 4 of @a r e contained in @2, c o n t r a r y to our a s s u m p t i o n s . Consequently, there exists an involution yE ~ \ ~ . Using L e m m a 2 we conclude that ~=~

x {g}; [g, cl-~ [y, dc]~-~ [y, ] d ] ~ - d 2.

F r o m this we obtain the following possible relations for the g e n e r a t o r s D, F, C, and Y (where Y is an element of o r d e r 4 of the p r e - i m a g e of y): 1) C-IYC = YD2 or C - 2 Y C - YF 2, 2) D-2YD = Y or D-2YD=YC 2, 3) F-2YF = YF 2 or F - 1 y F -- YD 2. We show that {D, Y} is an M-group of o r d e r 16. In fact, if DY=YD, then {D, F, Y} must be a group of type 7) of the theorem. Since the cyclic subgroup {D} of o r d e r 4 is contained in the center of {D, F, Y}, this is not possible. Consequently, DY ~ YD. F r o m this and f r o m the fact that D 2 ~ y2 follows that {D, Y} is an M-group of o r d e r 16. It is not difficult to see that one of the subgroups {D} and {Y} is n o r m a l in {D, Y}. If Y-iDY= D -2, we obtain, using the relations given above, that D 2= C ~, c o n t r a r y to the relations holding in @i. Hence, D-2YD = y - t Similarly we establish that Y-1FY = F -2. Now we consider the f a c t o r group

where C, D, F, and Y are homomorphic images of the elements C, D, F, and Y. For the elements CF-and YDC the relation [2~,

C~] = d ~,

holds, which yields the relation C - 1 y c =YD 2. Thus, in case 3) we obtain a group of type 8). The lemma is proved. LEMMA 5. If a nonhamiltonian IT2-group @ contains a quaternion group ~ ={S 1, $2} and if its cent e r contains a subgroup 3 = {A} x {B} of type (2,2), then @ is a group of type 9) or 10) of the theorem. P r o o f . At f i r s t we note that the d i r e c t product of two quaternion groups: ~ = {BI, ]32} and ~ = {C1, C2} is not an H2-group, since it contains the n o n - n o r m a l quaternion subgroup {B2C1, B2C~}. F r o m the

538

condition 9 ~ ~(r follows that all involutions of r a r e contained in 9t. By L e m m a 1, the c o m m u t a t o r subgroup of $ i s contained in ~l, and, t h e r e f o r e , the s q u a r e of e v e r y element of $ is contained in the center of $. A s s u m e that ~ (] ~ =