滋賀大学学術情報リポジトリ

On  Arithmetic 

Subgroups 

with

      Finite 

Quotients

Isomorphic

Masahiko]NIWA

  Let  F(G)denote  the  set  of  isomorphism  classes  of  finite  homomorphic  images  of agroup  G.  P. F.  Pickel  shows  in〔5〕that  if G  is a finitely  generated  nilpotent  group, the  finitely  generated  nilpotent  groups  H  for  which  F(G)=F(H)lie  in  only  finitely' many  isomorphim  classes.  It seems  to be  interesting  to generalize  this result  for groups of  another  type.  The  purpose  of  this  paper  is to  study  the  analoguotユs  problem  for S-arithmetic  subgroups  of  a  certain  semi-simple  algebraic  group  defined  over  a global  field.  Of  course  these  groups  are  more  complicated  than  nilpotent  groups,  SO the  problem  is not  done  completely  analoguous  and  the  results  are  partial.  But  in  the process  we  give  several  practically  useful  statements  on  S-arithmetic  subgroups  and their  completions.

1・ 一Notations  and  assumptions.

k:  a  global  field  i.e.  a  finite  algebraic extension  of  Q  or  a  field  of  algebraic functions  of  dimension  l over  a finite  field. k":the  completion  of  k at  a place  v(of  k).

If v  is  a  finite  place,0.:the  maximal compact  subring  of  k。.

S:afinite  non-empty  set  of  places  of  k containing  all infinite  places.

0=0(S):the  ring  of  S-integers  of  k、 G=asimply  connected  semi-simple  alge-braic  group  defined  over  k.

Gk(resp.  G"):the  group  of  k一(resp.  k"一) rational  points  of  G.      Gs=H"∈s  G" We  fix  an  embedding  of  G  into  GL.  de-fined  over  k.

Go(resp.  Gc"):the  group  of  points  of  G with  values  in O(resp.0")

GA(resp.  GA(s》):the  adele(resp.  S-adele) group  of  G  i.e.  the  restricted  product  of G.'s  for  all  v  (resp.  for  all v  outside  S) relative  to  Go,,'s.      ・

We  set  up  the  following  assumptions  for (G,S.).

Assumption(S)strong  approximation 

prop-  ertyprop-  i.e.prop- GsGkprop-  is dense  in  G小

Assumption(C)congruence  subgroup  prop-  ertyprop-  i.e.  every  S-arithmetic  subgroup   contains  some  S一一congruence  subgroup.   Ramark  that  a  semi-simple  group  for

which(S)or(C)holds  is  automatically simply  connected.(see〔'23〔63)The  fol-lowing  are  known  facts  about  the・conditions for  (G,  S)・satisfying  (S)  or  (C).・The Kneser-Platonov's  strong  approximation theorem  says  that(S)holds  if  Gs  is not compact.  Also(C)holds  if G量s  split  and Sis  not  totally  imaginary  except  for  the case  of G=SL2  and  S={one  place}.(see〔33 〔η)But  if G  is not  split  it is yet  unknown

whether(C)holds  or  not.

2,一We  define  two  topologies  on  Gk  i。e. the  S-arithmetic  topology  Td(S)and  the S-congruence  topology  Tc(S).  The  assump・ tion  (C)implies'Ta(S)=T6(S).  Also  the assumption(S)implies  that  the  completion Gkof  Gkin  T.(S)is  equal  to  GA(s)when we  identify  Gk  with  a  subgroup  of  Gs;s)

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              ム

completion  r  of  r  in  T、(S)is  nothing  but the  profinite  completion  of  I'and  the  corn-pletion  1'of  P in T.(S)is  identified  with  the closure'of  1'in  GA(S),  where  1'is  a  S-arithmetic  subgroup  of  Gk.  Then  we  have' the  following  lemma.

Lemma.1.  Let  G  be  as in  n。1  and「aS-arithmetic  subgroup  of  Gk.  Then

  (1)  1'is  an  open  cempact  subgroup  of      G4(s).

  (2)  ∬T∩Gk:=f

  (3)  if  U  is  an  open  compact  subgroup     of  GA(s),r=u∩Gk  is a S-arithmetic      subgroup  of  Gk  andハ=U.

Proof.  The  assertions(1)(2)are  obvious for  any  principal  congruence  subgroups. The  assumption(C)means  that  a 

S-ad-thmetic  subgroup  l'of  Gk  contains  a princi-pal  congruence  subgroup  1', as  a subgroup of finite  index.  As!',  is open,1'is  so.  Since

1'1is  compact  and(r:r,)is  finite,  r  is compact.(1)is  also  true  for  I'.

  Put  T・=U葱9富fレ(9歪 ∈1「⊂Gk;adecompo-sition  of  finite  cosets)We  have  1'=U`g/1, and  so

    「∩Gk=(Uσ9∫1)∩Gk二U￠(9≠1∩Gk)'     =U￠ 〔9∫(「1∩Gk).〕==U乞9zr1=君

(2)  has  been  obtained.

  Let  U  be  as  in(3).  Since  any  two  open compact  subgroups  are  mutually  commensu-rable,  U∩Gk  is  commensurable  with  some S .一arithmetic  subgroup  of  Gk.  This  shows that  U∩Gk  is a  S-arithmetic  subgroup  of Gk.  Finally  the  closure  of  U∩Gk  in  G,(8) coincides  exactly  with  U  since  Gk  is  dense andUisopenin  G,,(3).      q.e.  d.

Corollary.  There  is a  bijective  correspon-dence  between  the  set  of  S-arithmetic subgroups  of  Gk  and  the  set  of  open  com-pact  subgroups  of  GA(8).

Proof,  In  view  4f  lemma.ユwe  can  obtain

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the  required  bijective  correspondence  by associating  a  S-arithmetic  subgroup  1'.of Gk  with  the  closure∬'of  1'in  G,(8).         q.e.  d。 3.一Let  F(G)(resp.  Fe(G))denote  the  set of isomorphism  classes  of finite  homomorphic images(resp.  of discrete  finite  continuously homomorphic  images)of  a group  G(resp. of _{.a  topological  group  G),  We  say  the} groups  G  and  H  have  isomorphic  finite quotients  if  F(G)=F(H).  Now  we  will establish  the  connection  between  two  e-quivalent  relations  on  the  set  of  S-arith-metic  subgroups  of.Gk  i.e.  F(r)=F(君) and  1'一 丁'. We  begin  with  the  following lemma。

1、emma.2.  Let∬ ▼be  a  S-arithmetic  sub-group  of Gk  and  1'the  closure  of I'in  G,(8). Then  F(り=F`(「).

Proof.  Let  U. be  an  open  no.mal  subgroup of  finite  index  in・ 「(Since  1'is  compact,  It may  be  called  simply  an  open  normal subgroup  of  1'). Put  U∩Gκ=r1.  It・is easily seen  that  U∩r一 」1'1 and  UI'=1',  hence 君/.r1…≡∬%U・

  Conversely  let」r'1 be  a  normal  subgroup of  finite  index  in  1'. Put  N・=君/1'1  and  let pbe  the  canonical  homomozphism  of  1' onto  N.  When  we  consider  T,,(S)as  a topology  on  1'and  the  discrete  topology  as that  on  N, the  assumption(C)implies  that Ker(P)is  open  in  I', hence  p is continuous. In  the  same  manner  of〔4〕   §3pcan  be extended  to continuous  homorphism  p  of  I' onto  the  discrete  group  N。 It's  clear  that Ker(p)is  I'1 and  so∬%r,=… ∬%r,.  Thus  we have  constructed  a bijective  correspondence between  the  set  of  normal  subgroups  of finite  index  of  」1'and  the  set  of  open normal  subgroups  of  1', which  preserves their  quotients,  This  means  F(r)=Fc(r).         q.e,  d,

  The  following  proposition  gives  a  neces-sary  and  sufficient  condition  for  two  S-arithmetic  subgroups  having  isomorphic finite  quotients.  In order  to prove  the  propo-sition  we  need  the  following  lemma _. Lemma.3  Let  T  be  a  S-arithmetic  sub-group  of  Gk。 For.each  positive  integer〃z we  let∫   the  closure  in  Td(S)of  the smallest  normal  subgroup  of  I'containing the  elements  x瓢`for  all x  in、 にThen  I'm is of  finite  index  in」 「for  all m.

Proof。  By  the  definition  I'm is a non-central normal  subgroup  of  I'which  is closed  in Ta(S).  Therefore  it follows  from〔4〕Prop.6 and  the  assumption(C)that  I'm is of  finite index  _{in  1'.        q . e. d.} Proposition.1  Let  1'and∫ 亨be  two  S-arithmetic  subgroups  of  Gk.  Then  F(∬ り= 」F(P)  if and  only  if-'一 ≦1マ.

Proof・   Suppose  r-r',  then  Fe,(ア)=・Fe(7') _. While  lemma。2  shows  that  F,の=Fの

and  F,(P)=F(君),  hence  F(r)==F(P) _. Conversely  suppose  F(r)=F(P) _{. We  will}

      ム   ム

show  r窪r'.  This  implies  1'一1'by  the  as・ sumption(C).

  Let  I'm be  as in  lemma.3.  If翫is  defined to be∫ γr肌,1'z2  is a finite  group  of  exponent 〃z and  every  finite  quotient  of  1'of  ex-ponent  dividing〃 ￠.is a quotient  of  1『払. In particular,  if〃i  divides  n, we  have  Tm⊃r'8, so  there  is a  canonical  epimorphismγ

7、,鵬: 」㌦ 「ザ 切.Similarly  we  have  I「'鵬  1'm  and 〆 η,鵬.Since  each  finite  quotient  of  ∫'of exponent吻is  a quotient  of翫and  similarly for∬-'and∬ 「'観, r  and  r'have  isomorphic finite  quotients  if  and  only  if  煽  is  iso-morphic  to∫'m  for  each  positive  integer  m. We  claim  that  there  must  exist  iso-morphisms  f鋭:∫ ㌔、→r'.  such  that  for each 刎nthe  following  diagram  is commutative:        γπ,伽       翫 ←

  (・)↓

・,。

  ↓・

・

  Let  f恥be  an  isomorphism  f鵬:翫 → 君 伽 We  will  say  that  f鵬extends  to搾for刎 π if there  is an  isomorphism  fη:疏 →r'n  such that  the  diagram(砦)commutes,  and  that f伽is  indefinitely  extendable  if f鴉 extends tonforeachmultiplenof  m.  Formin, any  isomorphism  f":疏 →1"n  is an  extens1on of some  f飢since  1',。 and  P.  are  the  largest quotients  ofノ'r, and∫ ㌦, respectively,  of ex囎 ponent〃2.  Thus  for  a  given  m  andany multiple  n  of m,  some  isomorphisms  of  f鱒: ノ振 → ∬■'伽 extends  to '".  Since  the  set  of isomorphisms  of几 、 with  r'伽is  finite,  some such  f拠must  ex七end  to  infinitely  many and  thus  to all multiple  n  of  m.

  While{μ}(resp.{r'm})forms  a  funda・ mental  system  of  neiborhoods  of  the  neu-tral element  of  I'(resp.  r').  Therefore  F(」D        ム

=F(刀)implies  that  1'=proj  li!n(翫,γ",鵬)

                      ム

is  isomorphic  to  r'=proj  lim(君 田,γ ∼、,鵬).

This  completes  the  proof.

4.一In  this  paragraph  we  consider _,the special  case  that  ∬'is  mapped  to  ∫"by  an inner  automorphism  of  G』(8).

Proposition.2  Let  1'(resp,  P)  be  a  S-arithmetic  subgroup  of  Gk  and  1'(resp.1「') the  closure  of 1'(resp.∫v)in  G.,(s).Suppose that!'is  mapped  to I"by  an  inner  automor-phism  of  G,,(8).  Then  T  is isomorphic  to 1マ.       ' Proof.  Let  Int(x)(x∈G4(s))be  an  inner automorphism  of G,,(8》which  maps  1'to  1'. If  U  is any  open  subgroup  of  GA(s),  the assumption(S)implies  GA(8)=UG化.  In  par-ticular  GA(s)旨rGk,  hence  we  may  put  x= ua  for  u∈1'and  a∈Gk.  Then  Int(a)maps isomorphically  r=u-1」Fu  to r'and  Gk  to Gk. As  r.一一r∩Gk  and  1"昌r'∩Gk,  Int(a)deter。 mines  an  isomorphism  of  1'onto  I".         q.e.  d. 5.一Let  1'be  a  S-arithmetic  subgroup  of Gk.  We  denote  by  S(r)the  set  of  iso・

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morphism  classes  of S-arithmetic  subgroups of  Gk  which  have  same  isomorphic  finite

quotients  as  1'. The  problem  which  we  will consider  now  is whether  S(のis  finite.  We will  give  in  this  paragraph  the  proposition which'reduces  the  above  problem  to that of  a  smaller  subgroup.  For  this  we  need. the  following  lemma.

Lemma.4  Let  1'be  a S-arithmetic  subgroup . of  Gk  and  U  an  open  compact  subgroup  of

GA㈹.  Then

(1)  there  are  only  finitely  many  subgroups   of  I'of  given  finite  index〃z.

(2)  there  are  only  finitely  many  S-arith-  metic  subgroups  of  Gk  which  contain  I'   as  a subgroup  of  given  finite  index〃z.

(3)  there  are  only  finitely. many  subgroups   of  U  of  given  finite  index〃2.

(4)  there  are  only.finitely  many  open   compact  sllbgroups  of  G41β)which  con-  tain  U  as a subgroup  of given  finite  index   〃2。

Proof.  In  view  of  lemma.3the  closure  r伽 in  T"(S)of  the  smallest  normal  subgroup of  1'containing  the  elements  xm  for  all x in  1'is  of  finite  index  in  I'. While  any subgroup  of  ∫'of  index  〃z contains  I'm_, hence  there  are  only  finitely  many  such subgroups.  This  completes  the  first  as-sertion.

  The  assertion(3)is  the  obvious  conse-quence  of(1)by  lemma.1.

  We  will  show(4).  By  taking,  if necessary, Ufor  some  open  subgroup  of  U _{, we  may} assume  that  U  is the  direct  product  11.  U. of  Iocal  facters  U♂s.  U.  is a maximal  com. pact  subgroup  of  G"for  almost  all v,  be-cause  U.  is equal  to  G。"for  almost  all  v and  G(w  is a maximal  compact  subgroup  of G,,for  almost  all v.  This  shows  that  it is enough  to  examine  local  parts  for  the purpose  of  proving  the.finiteness  of  the

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number  of  maximal  compact  subgroups  of GA(ε)containing  U.

Le㎜a.5  Let  G  be  a  simply  c。nnected semi-simple  algebraic  group  defined  over  a non-archimedian  local  field  k  and  U  an open  compact  subgroup  of  Gk.  Then  the number  of  maximal  compact  subgroups  of

Gk  which  contain  U  is finite.

Proof.  They  are  well-known  that  any  com-pact  subgroup  of  Gk  is contained  in  some ・ maximal  compact  subgroup  of  Gk  and  that

the  number  of conjugacy  classes  of maximal compact  subgroups  of Gk  is finite.  Therefore it isenough  to  prove  that  if welet  V  an open  compact  subgroup  of  Gk  containing  U then  there  are  only  finitely  many  conju-gates  of  V  containing  U.  Since  the  number of  the  conjugates  of  U  which  is contained in  V  is finite.  we  must  show  that  there  are

      ,

only  finitely  many  conjugates  of  V  which induce  same  cojugate  of  U.  This  is clearly equivalent  to  say  (N(U):N(U)∩N(V))

<QO,  which  is easily  shown  using  the  well-known  fact;(N(U):U)<QO.  This  completes the  proof.

(the  continuation  of the  proof  of  Lemma.4)    In  view  of  lemma.5and  the  previous

argument,  the  number  of  maximal  compact subgroups  of  GA(ε)containing  U  is finite. It  follows  that  for  proving(4)we  may assume'that  the  open  compact  subgroups which  appear  in(4)are  contained  in  one of finitely  many  maximal  compact  subgroups of  GA(8)containing  U.  Since  U  is of  finite index  in  its maximal  compact  subgroup,  It is  easily  seen  that  the  number  of  such subgroups  is  finite.  This  completes  the proof  of  (4).  Finally  the  assertion  (2)  is the  consequence  of(4)by  lemma.1.

        .      q.e.  d. Proposition.3  Let  1'and  d  be  two  S-arith-metic  subgroups  of  Gk'・such  that  I'⊃ 」,

If S(」)is  finite,  S(∬ つ1S  SO.

Proof.  We  denote  by  U(resp.  V)the closure  of  I'(resp.」)in  G,,(8).'Let  S'(U) (resp.  S'(V))be  the  set  of  open  compact subgroups  Ua(i∈1)(resp.  V`(i∈J))which is isomorphic  to  U(resp.  V).  Then  Prop.1 implies  that  S(f)(resp.  S(4))may  be  con-sidered  as the  quotient  of S'(U)(resp.  S'(V)) by  the  equivalent  relation  which  is defined' such  that  for  i, j∈1(resp.  i, j∈J)Us(resp. Vのis  equivalent  to UJ(resp.  v,)if  Us∩Gk (resp.  V￠ ∩Gk)is  isomorphic  to Uj∩Gk(resp. Vゴ ∩Gk).

  For  each  U,  we  choose  an  isomorphism (pi of  U  onto  _{Ua . Then{仰(V)【i∈1}is  a} subset  of  S'(V),  because  the  restriction  of (pt to V  induces  an  isomorphism  of  V  onto ρε(V).We  will  show  that  for  given  i∈I there  are  only  finitely  many  j∈I  such  that ψ」(V)=∼ ρ歪(V).In  fact  (pJ・仰 一'is  an  iso-morphism  of  Us  onto  U/which  maps(pa(V) to  pゴ(V),  hence  which  induces  an  isomor-phism  between  Ui/(Pi(V)and  Uノ/ρ 」(V).  As ρ`(V)・=ρ 」(V),the  statement  is the  conse-quence  of  Lemma.4(4).

  Now  it's  clear  that  1'g=U乞 ∩Gκ 窪 乃= U'∩Gk  implies  da=・ ρ`(V)∩G虐Jj=ψ 」(V)∩ Gk. Though  J」…≧JJ does  not  imply  in general ∫㌃窪 乃,we  can  verify  that  the  number  of the  classes  by  I'i…v rj of  each  of isomorphism classes  of  Ja=卿(V)∩GA;(i∈1)is  finite.  In fact  as  (疏:ゴ`)  <co  for  any  i∈1_{,  the} number  of  isomorphism  classes  of  1't which satisfy  J3-Ja  is'finite,  which  shows  the above  statement.  Thus  we  have  shown  that the  mapping  from  S(1りto  a subset  of  S(の which  is induced  by  Vi(i∈1)is  such  that the  inverse  image  of  each  element  of  its image  has  finitely  many  elements。  While  by the  assumption  S(」)is  finite, therefore  S(r) 1SSO.         q.e.d. 6.一If  we  let U  an  open  compact  subgroup

of GA(s⊃,  U contains  a subgroup  of the  form of  n"￠8  U".  By  Prop.3  we  suppose  from now  on  that  U  is an  open  compact  subgroup of  such  form.

  Let  I'be  a S-arithmetic  subgroup  of  Gk such  that  U=fis  of  the  form  of  nU".  We denote  by  T(1')the  isomorphism  classes  of S-arithmetic  subgroups  1"of  Gk  such  that U'ロUn  for  all  v  outside  S,  where  ULr' is  of  the  form  ofU』nU㌔.  In  this  last paragraph  we  shall  prove  the  finiteness  of T(r)under  assuming  the  following  two statements.

(G)  Letk,GandSbeas  inn。1andlet I'and  1"be  two  S-arithmetic  subgroups  of Gk  such  that  1'一 丁,. If  f  is  any  isomor-phism  of  1'onto  I", f can  uniquely  extend to a  k-automorphism  of  G.

(L)  Let  G  be  a  simply  connected  semi-simple  algebraic  group  defined  over  a non-archimedian  local  field  k  and  let  U  and  U' be  two  open  compact  subgroups  of  Gk  such that  U-U'.  If  f is any  isomorphism  of  U onto  U',  f  can  uniquely  extend  to  a  k-automorphism  of  G.

  Though  the'above  statements(G)(L)

are  themselves  worth  studing  (Cf.  for example〔 ユコ), we  make  no  allusion  to their questions.

  Now  we  can  show  by  using  (G)that T(∫)is  a  subset  of  S(1').  In  fact  iftwo S-arithmetic  subgroups  I'and  1"are  isomor・ phic,(G)implies  that  any  isomorphism between  1'and  I"induces  k-automorphism of  G,  hence  k"一automorphism  of  G  for  all v.It  follows  that  it is continuous  in  the topology  on  Gk  induced  by  that  on  Gn  for each  v.  We  conclude  that  U"…=U'"for  all v,where  r=HU"and∫"=HU㌔,  This  means T(の ⊂S(r).From  now  on  we  deal  only with  T(r),

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of(algebraic)automor-phisms  of  G,  then  since  G  is  semi-simple Aut(G)has  the  structure  of  an  algebraic group  defined  over  k  and  the  group Aut(G)北(resp.  Aut(G)")of  k一(resp.  kv) ratinal  points  of  Aut(G)coincides  with  the set  of(algebraic)k一(resp.  kザ)automor-phisms  of  G.  If  we  choose  suit'ably  an embedding  of  Aut(G)'into  GL,:correspond-ing  to  the  embeddGL,:correspond-ing  of  G  IIlto GL鵬which is chosen  before,  Aut(G)o(resp.  Aut(G)") is identified  with  the  subgroup  of Aut(G)㌃ (resp.  Aut(G)η)consisting  of elements  which leave  Gc  (resp.  G6")  stable.  In  paticular every  element  of  Aut(G)4《s)is  regarded as  an  automorphism  of  GA(8).  If we  let  c'ノ. ak-automorphism  of  G,  Yi, induces  an  auto-morphism  of  Gk(For  the  sake  of  conve-nience  we  identify  the  restiction  of  O  to Gk  with  O.),  that  of  Gりfor  all  v  which maps  isomorphically  G`"to  itself  for  almost all v, and  hence  that  of  G,(s).  lf qY  is the automorphism  of  GA`8)induced  byψ,  the restriction  of  T  to  GA  is clearly  equal  to the  automorphism  O  of  Gk一.

  Let  1', r∼u=nu"and  U'=IIU',,  be  as in the  first  part  of  this  paragraph.  For  each v∈Swe  denote  by  w  an  isomorphism  of U"onto  U'"and  by  4>v the  kザautomorphism of  G  which  is  the  extension  of  Pv.(see

(L))  (For  the  sake  of  convenience  we identify  the  restriction  ofΦ"to  G"with Φ の   Since  Uσ=U'"=Go,,  for  almost  all  v, for  such  v  ch,  is  an  automorphism  of  G。 which  maps  isomorphically  G`"to  itself, hence  which  belongs  to Aut(G)`".  It follows that  the  automorphism  c1)=・(Φ")of  G4(8) belongs  to  Aut(G)8(δ).  lf NY  and(D  are  as above,Φ 『1・Ψis  an  automorphism  of  G4(8) belonging  to  Aut(G)滝(s,.

  We  will  show  here  that  I"一 → Φ 一且・Ψ determines  a  well-defined  mapping  from

号

T(1っto  K＼Aut(G)4(s)/Aut(G)滝,  where

K={F=(・ 。)∈A・

・(・)。,,,

   謡1。

欝o蹴

話

二

、。}.

  First  suppose  that  ∼〆"  is  a  different choice  of  isomorphism  of  U"to  U'ηfor  each

v.We  denote  byΦ'v  the  automorphism  of

G"induced  byψ'".  ThenΦ'.:=q)ガ Φ 『1.・ Φ'"

andΦ 『1"・ Φ ㌔is  an  automorphism  of  G"

which  leaves  U。   stable  for  each  v.  Thus

we  have  ch閏1・ Φ'is  an  element  of  K.  where       ,

Φ'二(Vv),  so  that  ch'一'・NY lies  in the  same double  coset  asΦ 一1・Ψ 。  Next  if  Vii,'is  an。 other  k-automorphism  of  G  andΨ'is  the induced  automorphism  of  Gμ(s),Ψ'is  also contained  in  Aut(G)κ,  so  that  cl)一1・NP'lies in  the  same  double  coset  as  『i・Ψ. Finally suppose  that  I'"is  a S-arithmetic  subgroup isomorphic  to  I"  by  an  isomorphism  o. Then  it follows  fro.m(G)that  a can  extend toak-automorphism  f  of  Gand  to  akv automorphism  Y。  of  G  for  each  v.  If  we call  X  the  automorphism  of  G,(8)induced by  fandletY=(Y"),wehaveXコYasan

automorphism  of  G,,(8).  We  have  also  that Y・ Φ=(Y"・ Φ 響,)maps  U=.nU"to  f"=U'』 nU""and  that  X・-V  maps  Gk  to itself.  The element  of  Aut(G)滝(8)corresponding  to these  automorphisms  is(Y・ Φ)『1・(X・ Ψ)= Φ 一1・Y-1・X・ Ψ=Φ 一1・Ψ.We  thus  have  that the  above  defined  mapping  is well-defined. Moreover  we  have  the  following.

工emma.6  The  above  defined  mapping  from T(のto  K＼Aut(G)3く8)/Aut(G)乱is  injective. Proof.  Suppose  T'and∫ 「"are  two  S-arith-metic  subgroups  of  Gk  belonging  _{to . T(f),} whose  respective  isomorphism  classes  are sent  by  the  above  defined  mapping  to  the same  double  coset,  We  will  show  that  I" and  1'"must  be  isomorphic.

Lions,  ewe  restrict  our  attention  to the  case of  r"=1:The  general  c'ase can  be  done  in the  same  manner.  In  the  special  case  if q) and'lf  are  as  above,  by.  the  assumption Φ 陶1・Tbelongs  to  K・Aut(G)叱.  This  means that  if  we  change(P  by  composing  an element  of  K  and  lf  by  composing  an  ele-ment  of  Aut(G)κ,  we  have  ch-1・・Ψ=id.  Le. Φ=Ψ.While  by  the  choice  of  cD  and'1,,, Φ=Tis  an  automorphism  of  G,,(s)which maps  isomorphically  U=IIU"to  U'謡nU'" and  Gk  to itself.  As  U∩Gk=1'and  U'∩G席

=r',Φ=Tinduces  an  isomorphism  of  I' onto  1".     '       q. e. d.   Put  L:=n礁s  Aut(G)oη ⊂Aut(G)4(s). Since  Aut(G)is  an  algebraic  group,  it is well-known  that  L＼Aut(G)6(8)/Aut(G)k  IS finite.  We  will  show  that  K  is  com-mensurable  with  L,  which  implies  that K＼Aut(G)』(8)/Aut(G)x  is finite.

  IfHisasubgroup  of  agroupGandZ

is a group  of automorphisms  of  G,  we  will denote  by  stab(H,  Z)the  group  of  automor-phisms  in  Z  which  leave  H  stable. Lemma・7  Let  U  and  V  be  open  compact subgroups  of  GAく8)such  that  U⊃V.

  (1)  stab(U,  stab(V,  Aut(G)4(8)))is  of     finite  index  in  stab(V,  Aut(G)4(s,).

      ノ

  (2)  stab(V,  stab(U,  Aut(G)4(8)))is  of     finite  index  in  stab(U,  Aut(G)4(8)).   (3)  stab(U,  Aut(G)4(s))and

    stab(V,  Aut(G)4(s))・are  commensura-    ble  in  Aut(G)憾 《8).

  Moreover  if U,and  U'are  any  two  open compact  subgroups  of  GA(s),then

  (4)  stab(U,  Aut(G)濯(s))and

    stab(U',  Aut(G)4(s))are  commensurable     in Aut(G)4(s).

Proof.  Let  m  betheindex  of  V  in U  and let U=U1,  U2,....,Uk  be  a  list(finite  by Lemma.4(4))of  the  subgroups  of  GA(8) containing  V  as a subgroup  of  index  m.  If g

is any  element  of stab(V,  Aut(G)4(8)),  g per-mutes  the  groups  Ua. Thus  we  may  construct a;homomorphism  of  stab(V,  Aut(G)』(s)) into  a finite  permutation  group.  The  kernel of  this  homomorphism  is_{. of  finite  index  in} stab(V,  Aut(G)」(s))and  is  contained  in stab(U,  stab(V,  Aut(G)漣(s))).  This  shows (1).By  analoguous  argument,  we  may prove  (2).

  (3)  is the  immediate  consequence  of(1) and(2)since  stab(U,  stab(V,  Aut(G)A(s)))

=stab(V _{, stab(U,  Aut(G)4(s))).  By} con-sidering  U∩U',(3)implies(4)。

        q.e.  d.   Since  Lemma.7(4)implies  that  L  and K  are  commensurable,  we  have  that K＼Aut(G) _{.4(ε)/Aut(G)k  is finite。 Therefore} we  have

Proposition.4  T(r)is  finite.

Proof.  This  is the  consequence  of  Lemma.6 and  the  finiteness  of K＼Aut(G)孟(s)/Aut(G)κ .

        q。e.  d.

Reference

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  straits"de  groupes  algebriques  simples(Ann.

  of  Math.(3)971973)

〔2〕M.Kneser,  Strong 

approximation(Alge-  braicapproximation(Alge-  groups  'and  discontinuous  subgroups,

  Proc.  Symp.'Pure  Math.  Vol.9A.  M.  S.1966) 〔3〕H.Matsumoto,  Sur  les  sous-groupes 

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〔4〕M.Niwa,  Some  results  on  arithmetic  groups

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