Unitary subgroup of integral group rings

Publicacions Ma emá iques, Vol 36 (1992), 197-204 . A bs ac UNITARY SUBGROUP OF INTEGRAL GROUP RINGS A .A . BOVDI AND S.K . SEFIGAL Le A be a ini e abelian g oup and G= A >G (b), b 2 = 1, ab = a -1 , da E A . We ind gene a o s up o ini e index o he uni a y subg oup o 7LG . In ac , he gene a o s a e he bicyclic uni s . Fo an a bi a y g oup G, le B2(7LG) deno e he g oup gene a ed by he bicyclic uni s . We classi y g oups G such ha B 2 (7LG) is uni a y . Le 7LG be he in eg al g oup ing o an a bi a y g oup G and le : G -> U(7L) = {± 1} be an o ien a ion homomo phism . Fo each x =  1 : a g g, we pu x =  a g (g)g -1 . In pa icula , i is i ial, gEG x coincides wi h he s anda d x* . Le U(ZG) be he g oup o uni s o 7LG . Then u E U(ZG) is called -uni a y i u -l = u o u -1 = -u . All -uni a y elemen s o U(ZG) o m a subg oup U (7LG) con aining G x U(7L) . We e e o U (ZZG) as he -uni a y subg oup o U(ZZG) . In e es in he g oup U (7LG) a ose in algeb aic opology and uni a y K- heo y [4] . We a e in e es ed in he cons uc i e desc ip ion o U (ZLG) . I G is ini e cyclic, hen Bo di [1] ga e a linea ly independen se o gene a o s o a o sion ee subg oup o ini e index in U (7ZG) . This was ex ended o ini e abelian g oups by Hoechsmann-Sehgal in [3] . We gi e gene a o s up o ini e index o U (7ZG) i G is a ini e dihed al g oup . In ac , he gene a o s consis o he bicyclic uni s . The subg oup B2(7ZG) o U(ZG) gene a ed by all he bicyclic uni s o 7LG plays an impo an ole in he s udy o U(ZG) (see [5], [6]) . In Theo em 2, we cha ac e ize g oups G o which B 2 (7LG) is uni a y . This wo k was suppo ed by NSERC G an A-5300 . 198  A .A . BOVDI AND S .K . SEHGAL We need he 2 . U (ZG) o dihed al g oups Fi s , we ecall some de ini ions . Fo an elemen aE G o ini e o de n w i e á = 1 + a + - - - + an -1 . Deno e by (G) he se o all o sion elemen s o G . I a, b E G,  o(a) < oo, hen ua,b = 1 + (1 - a)bá has an in e se u~ , b = 1 - (1 - a)bá . Mo eo e , u a ,b = 1 i and only i b no malizes (a) . The elemen s u a b, a, b E G a e called bicyclic uni s o 7LG and he g oup gene a ed by hem is deno ed by B2(7LG) . We ecall [5] ha by Bl (ZG) is unde s ood he g oup gene a ed by he Bass cyclic uni s o 7LG . I is known [5] ha i G is a ini e dihed al g oup and Z is he cen e o U(ZG) hen (Z,B2(7ZG)) (equi alen ly, (Bl(7ZG),B2(7G))) is o ini e index in U(7LG) . We p o e Theo em 1 . Le G be he dihed al g oup D2 a =(a - =1= b 2 l a b= a_ 1 ) . Suppose is an o ien a ión homomo phism o G wi h ke nel (a) . Then he index (U (7ZG) : B2 (ZG» is ini e . P oposi ion . Le G be a g oup con aining a subg oup A o index 2 and an elemen b such ha G = (A, b) and b-lab = a -1 o all a E A . Suppose ha A 2 5 E 1 . I is an o ien a ion homomo phism o G wi h ke nel A, hen 1) he cen e o U (7LG) coincides wi h 2(A) x (-1), whe e 2(A)={aE (A) : a2=1} ; 2) he cen e o U(ZG) is he di ec p oduc o 2(A) x (-1) and a o sion ee abelian g oup T such ha U(ZA) = (-1) x A x T and x=x* o allxET . P oo .. Le x = xl + x2b,  xi E 7ZA be a cen al uni in 7LG . Since G is a subg oup o U(7ZG), x = b -l xb = x* + x*b  and  x = a -1 xa = xl + a-2x2b o all x E A . Then xi = xi and (1)  x2(1 - a 2 ) = 0 UNITARY SUBGROUPS  19 9 o all a E A . We wish o p o e ha x2 = 0 . _Le us suppose ha x2 =,~ 0 . ROM (1) we ob ain ha A 2is ini e . Le A 2 deno e he sum o all elemen s o A 2 . I H is a no mal subg oup o G, hen deno e by 0(G, H) he ideal o 7LG gene a ed by elemen s o he o m h - 1 wi h h E H . Clea ly, 7ZG/A(G, H) - 7L(G/H) . I X(y) is he sum o he coe icien s o y, hen he elemen x + 0(G, A) = X(x1) + X(x2)b+ 0(G, A) is i ial, because IG/AJ = 2 [7, p . 46] . This implies ha one o he numbe s X(xl_) o X(X2) equals ±1 and he o he is ze o . F om (1) we ob ain x2 = ZA 2 , z E 7LA, X(x2) = X(z)1A 2 1, and his is possible only in he case when X(x2) = 0 . Suppose A=A 2 . Then x2 = 7 Z : a o some -y E 7 . F om he equali y aEA X(x2) = - yjAj = 0 we ob ain ,y = 0 and x2 = 0, which leads o a con adic ion .  Thus A =,A A 2 . W i e x2 = ( aici) A 2 wi h al E 7 i whe e ci's a e a ans e sal o A 2 in A . Then x1 +x2b+ A(G,A 2 ) = XI+ (~a i c i )A 2 b+0(G,A 2 ) = x1 + (¡A 2 1  aici)b+ 0(G, A 2 ) i is a uni in 7L(G/A 2 ) . Since G/A 2 is an abelian g oup o exponen wo, by Higman's heo em [7, p . 57], all uni s o 7L(G/A 2 ) a e i ial . Ob iously, E al = 0 and i al 7~ 0 o some i, hen n i ¡ A2 1 =~ ±1 . Thus, al = 0 o all i and he equali y x2 = 0 is con adic o y . Hence, .x = xI E U(ZLA) and x* = x = xi = xi . Clea ly, i x E U(7LA) and x* = x, hen x is a cen al uni o 7G . I is well known (see [2]) ha U(7 (A)) = ± (A) x T and U(7A) _ A x T, whe e e e y elemen u E T sa is ies he condi ion u = u* . The e o e he cen e o U (7G) is he di ec p oduc o subg oups ± 2 (A) and T . This is 2) o he P oposi ion . Suppose ha x = x1 + x2b is a cen al uni in U (7G) . Since G is a subg oup o U (7G), x is cen al in U(7LG) . I ollows ha x = xi and xx = x1xi = xi = ±1 . The e o e, by Higman's heo em xI = a whe e a E 2(A) . This comple es he p oo o he P oposi ion . 20 0  A .A . BOVDI AND S .K . SCI - IGAL P oo o Theo em 1 : Le G be he dihed al o o de 2n gi en by G= (a n = 1 = b 2 , J = a -1 ) . I n = 2, hen he heo em is i ial . So we may apply he las P oposi ion . Le Z be he cen e o U(ZZG) . Thenwe know ha (U(ZZG) : (B2 (7G), Z)) < oo . We ha e seen in he P oposi ion abo e ha Z1, he cen e o U (7ZG), is ini e and Z 1 < Z . I su ñces o p o e, he e o e, ha B2(7LG) is uni a y . I u,, :, 7, ~ A 0, hen o(x) = 2 and u ' , y = 1 + (1 - x)y(1 + x) . Now, y = a i x',  E = 0 o 1 . Since x(1 + x) = 1 + x, we ha e, in any case, Then u ,u = 1+(1 +x) (ax) (1-x) = 1+(1-x) a - Z (1+x) . The e o e, u=,ay U , , j = 1 + (1 - x) (a i + a`) (1 + x) = 1 as (a l + a- Z) is cen al . This comple es he p oo o he heo em . Rema k . The las heo em holds o nonabelian g oups G = (A, b) whe e A is ini e abelian and b 2 = 1, a b = a -, o all a E A . I A is an elemen a y 2-g oup, hen so is G and whe e is no hing o p o e . Suppose A2 :~ 1 . The nonlinca i educible ep esen a ions p o G a e induced om hose o A and p(ZLG) = p(D) o some dihed al subg oup D o A . The esul ollows . We need hc : ollowing . ux,y=1+(1-x) a' (1+ .x) . 3 . Uni a i y o he subg oup B 2 (7G) Theo em 2 . Le , G = (A, b) whe e A is he ke nel o he non i ial o ien a ion homomo phism : G - U(7L) . The subg oup B2(7LG) ás non i ial and -uni a y i and only i G is non-Hamil onian 'in which an elemen b =,A 1 o ini e o de can be chosen such ha one o he ollowin,g condi ions is ul illed : 1) A is an abelian g oup, he o de o he elemen b di ides 4 and bab-1 = a-1 o all a E A ; 2) A is a, Hamil onian 2-g oup, G is he semidi ec p oduc o A and (b 1 b 2 = 1), an,d e e y subg oup o A is no mal in G ; 3) A is a, Hamil onian 2-g oup and G is he di ec p oduc o a Hamil- onian 2-subg oup o A and a cyclic g oup (b) o o de 4 ; 4) (A) is an abelian g oup, e e y subg oup o (A) is no mal in G and bab -1 = a -1 b 47 o all aE A, whe e he in ege i depends on a, . UNITARY SUBGROUPS  20 1 Lemma . Suppose ha G has a subg oup A o index 2 uwi h G = (A, b) and o(b) < oo . Suppose u he ha A 7¿ NA((b)) and 1) (A) is abelian and all subg oups o (A) a e no mal in A ; 2) bgb -1 = g -1 o all gE A NA((b)) . Then bab -1 = a-1 o all aE A and b 4 = 1 . P oo : Le c E N A ((b)) . Choose aE A N A ((b)) . A i s , suppose c has ini e o de . Then by (2) we ha e a - 'bcb -1 = b(ac)b - = c-1a-1 I aE (A), hen by (1) we ha e bcb -1 = c -1 . I a, has in ini e o de , he e exis s an in ege n such ha anc -- ca", sin(,(, , (c) is no mal in A . B,y hypo hesis, a 'c ~ N((b)) and ln s a - 'Lbcb -1 = b(a n c)b -1 = ( .--l a-'L . I ollows ha bcb -1 = c- as desi ed . Now i is enough o p o e ha c canno ha e in ini e o de . Suppose ha o(c) = oo a ld o(a) < oo . Then he e is an n such ha cna = ac" . Clea ly, acn ~ N((b)) . We ha e a -1 bc"b -1 = b(ac n )b -1 = e-na-1 . I ollows ha bc' b -1 = -n . Tllis is impossible because c" E N((b)) . Now le o(c) = oc, o(a) = oo . The e exis s an n such ha bc" = c"b and a -1 cn = ba .c"b -1 = -n a - ' . I ollows ha [c" ; a21 = 1 . Clea ly, a 2 cn 1 N((b)) and wc ; ge which implies c 2 n = 1, a con adic ion . Since b 2 E A,  bb 2 b -1 = b -2 and we ha e b 4 = 1, comple ing he P oo o he , lenuna . P oo o Tlheo em 2 : "Necessi Y ." a -2 c' = ba 2 c"b - = a-2c-n Suppose ha B2(ZLG) is non i ial and -uni a y . Le us i s p o e ha e e y ini e subg oup (a) o A is no mal in G . Le n be he o de o (a) . I gNG((a)), hen u,,, g = 1 + (1 - a)gá q¿ 1 . Tilen o l he equali y u- = u ,9 we ha e - dg -1 ( .g)( 1 - a-1) =- ( 1 - a)gá . Mul iplying by á we ob ain n( 1- a)g - a = 0, which is impossible . The e- o e, e e y subg oup o , (A) is no mal in G . Because B 2 (7ZG) ~ 1 ;  G A 20 2  A .A . BOVDI AND S.K . SGI1GAL con ains an elemen e o ini e o de wi h (e) no no malized by A . Then e 2 E (A) and c2 is cen al in 7LG . Clea ly, U C , 9 = 1 + ,(1 - c)g(1 + e)c2 and (e) = -1 . Since u,,s is -úni a y,  u e , 9 u, ,9 = 1 and i ollows ha ( 1 ) (g+g-1 (g))(1+c)c2=c(g+g-1 (g))(1+c)c2 . Choose b E G A such ha b is a 2-elemen o leas o de and le g E A . In (1) aking c = b, g = bg -1 b 1 + 2= whene e g1 NA((b)) . We ob ain bgb-1 = g-1b 2i' o all g E A NA((b)) and (bg) 2 = (g-1b2g)`'+1 . Clea ly, bg is a 2-elemen in G A and i' is e en, o he wise he o de o bg is less han he o de o b, which is impossible . The e o e, (2)  b gb-1 = g-1b 4i o all g , e A NA((b)) . a) Suppose ha he o de o b di ides 4 . Then om (2) bgb -1 = g-1 o all gE A NA((b)) . I (A) is abelian, hen, by he Lemma, A is abelian and bab-1 = a- o all a E A . This is case 1) o he heo em . I (A) is nonabelian, hen (A) is a Hamil onian g oup and (A) =QxExT whe e Q is he qua e nion g oup o o de 8, E 2 = 1 and all elemen s o T a e o odd o de . We wisll o p o e ha A = (A) . Suppose ha g is an elemen o in ini e o de o A N((b)) . Then g 2 E CA (Q) and he e exis s an elemen w o o de 4 o Q such ha [b, u)] = 1, because e e y subg oup o Q is no mal in G . Clea ly, g 2 w 1 N((b)) and by (2) wg -2 = bwg 2 b -1 = bg 2 wb -1 = w-1g-2, which is impossible . The e o e, all elemen s o A N((b)) ha e ini e o de s . Le gbe an elemen o in ini e o de om NA((b)) and le an a E A N A ((b)) . Clea ly he e exis s n such ha [g', a] = 1, because he ini e UNITARY SUBGROUPS  20 3 cyclic subg oup (a) is no mal in G . Then gna E A NA((b))and gna is o in ini e o de , which leads o a con adic ion . The e o e, (A) = A . We claim ha T= 1 . Le be an elemen o odd o de om A N((b)) . Ob iously, he e exis s an elemen w o o de 4 in Q such ha [b, w] = 1, as e e y subg oup o Q is no mal in G . Thus w ~ N((b)) and by (2) which is impossible . Nex , le be an elemen o odd o de om NA «b) ) . Because ( ) a G, [ , b] = 1 . Clea ly he e is an elemen w o o de 4 in Q such ha b-1 wb = w-1 and w ~ NA((b)) . Then which is impossible . Hence, he s uc u e o G is desc ibed in case 2) o 3) o he heo em . b) Suppose ha he o de o b is 2"  (k >_ 3) . Then by (2) b 2 belongs o he cen e o (A), Because (A) is abelian o Hamil onian . Hence, (A) is abelian and e e y subg oup o (A) is no mal in G . Then om (2) bab-1 = a -1 W o all aE A NA((b)) . Deno e by (b4 ) he subg oup gene a ed by b 4y = abab -1 , as a uns o e A NA((b)) . Pu G = G/(b 4 ),  A = A/(b 4 ) and b = b(b 4 ) . Then _ G _ sa is ies he condi io_ns o ou Lemma and i ollows ha = 1 and bab-1 = a -1 o all a E A . This is case 4) o he heo em . "Su ciencg ." Le G sa is y one o he condi ions 1)-4) o he heo em . Cléa ly, i a ini e subg oup (c) is no no mal in G, hen c E bA, (c 2 ) -- (b 2 ) and c 2 belongs o he cen e o ZG . The e o e, and and (g + g -1 )c 2 is cen al in 7LG . -1 w= b wb -1 = w-1 -1, w -1 = bw b -1 = -1w-1 u,, 9 = 1 + (1 - c)g(1 + C)2_ uc,g u ,s = 1 + (1 - c) (9 + 9 -l (9))( 1 + C)2- . Suppose ha g E A . Then (g) = 1 and (g + g -1 )c 2 is a cen al elemen . This is ob ious in cases 1), 2) and 3) . Suppose ha G sa is ies he condi ion 4) o he heo em . Then (c 2 ) = (b 2 ), and bgb-1 = g-1b4i and G/(b 4 ) is abelian . Thus b(g + 9 -1 )c 2 b -1 = (9 + 9 -1 )P = a -1 (9 + 9-1)72a 204  A .A . BOVDI AND S .K . SEIIGAL 2 . 3 . 4 . I g E bA, hen g = ba,  ( .g) = -1 and A .A . BOVD1, Uni a y subg oup o he mul iplica i e g oup o in- eg al g oup ing o a cyclic g oup, Ma h . Za ne ki 41 (4) (1987), 467-474 . A .A . BOVDI, The mul iplica i e g oup o an in eg al g oup ing, Uzhgo od, 1987 . K . HO cIISMANN AND S.K . SEIIGAL, On a heo em o Bo di, o appea . S .P . NOVIKOV, Algeb aic cons uc ion and p ope ies o He mi ian analogues o K- heo y o e ings wi h in olu ion om he iew- poin o Ha nil onian o malis n, Applica ions o di e en ial opol- ogy a,nd i e heo y o cha ac e is ic classes, II, Iz . Akad . Nauk SSSR Se . Ma . 34 (1970), 475-500 ; English ansl . i n Ma h . USSR Iz . 4 (1970) . 5 .  J . RITTER AND S .K . SEIIGAL, Gene a o s o subg oups o U(7LG)*, Con empo a y Ma h . 93 (1989), 331-347 . 6 .  J . RITTER AND S .K . SEIIGAL, Cons uc ion o uni s in in eg al g oup ings, T aces . A .M .S . 324 (1991), 603-621 . 7 .  S.K . SE11CAL, "Topics in g oup ings," M . Dekke , New Yo k, 1978 . A .A . Bo di : Ma hema ics Ins i u e Uz1igo od S a o Uni e si y Uzhgo od USSR g- I = a - b - = b-Iab4i . Clea ly, g -l c 2 = gc 2and (g + (g)g - )c2 = 0 . The e o e, Uc,g ~c,g = 1 and he bicyclic uni s a e -uni a y . Thus B2(7LG) is an -uni a y subg oup, p o ing he heo em Re e ences S .K . seligal : Depa men o Ma hema ics Uni e si y o Albe a Ed non on, Albe a CASADA T6G 2G1 P ime a e sió ebudo, el 24 de Juliol de 1991, Ba e a e sió ebudc el 14 d'Oc ub e de 1991