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Power Maps and Completions of Free Groups and of the Modular Group
191, 252]264 Ž1997. JA966925
JOURNAL OF ALGEBRA ARTICLE NO.
Power Maps and Completions of Free Groups and of the Modular Group John G. Thompson Department of Mathematics, Uni¨ ersity of Florida, 201 Walker Hall, Gaines¨ ille, Florida 32611-8000 Communicated by Walter Feit Received September 24, 1996
1. INTRODUCTION AND STATEMENT OF RESULTS If G is a group and n is an integer, the nth power map is Pn : G ª G, PnŽ x . s x n. Thus PnŽ G . s PynŽ G ., the set of nth powers of elements of G. If K is a normal subgroup of G and x g G, we say that x is an nth power Žmod K . if and only if there is y g G such that xy1 y n g K. Let PnŽ G mod K . be the set of all x in G such that x is an nth power Žmod K ., so that PnŽ G mod K . is the preimage in G of PnŽ GrK .. If K is a family of normal subgroups of G, we say that x is an nth power Žmod K . if and only if x is an nth power Žmod K . for all K g K, and we set Pn Ž G mod K . s
F Pn Ž G mod K . . Kg K
Thus, PnŽ G mod K . = PnŽ G .. For each group G, let Fin G be the set of all normal subgroups of G of finite index, and let G c be the completion of G in the profinite topology. THEOREM ŽLubotzky.. If F is a free group and n is an integer, then PnŽ F mod Fin F . s PnŽ F .. Set G s SLŽ2, Z., and for each N g N, set GN s
½ž
a c
b g G a y 1 ' b ' c ' d y 1 ' 0 Ž mod N . , d
5
/
the principal congruence subgroup of level N. Also, Z p and Q p denote the p-adic integers and the p-adic numbers, respectively. 252 0021-8693r97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
253
POWER MAPS AND COMPLETIONS
THEOREM 1. If n g N, 8 ¦ n, A g G, and A is an nth power Žmod GN . for all N s 1, 2, . . . , then there is B g SLŽ2, Q. such that B n s A. THEOREM 2. Ži. Žii.
If n is an integer ) 2, then there is A g G such that
A f PnŽ G . and A g PnŽ SLŽ2, Q..; A is an nth power Žmod GN . for all N s 1, 2, . . . .
Putting these results together, we get that Fin G and K, defined in the Corollary, behave differently with respect to power maps. COROLLARY. Ži. Žii.
Let K s GN < N s 1, 2, . . . 4 , n g N.
PnŽ G mod Fin G . s PnŽ G .. If n ) 2, then PnŽ G mod K . ; PnŽ G ..
2. PROOF OF LUBOTZKY’S THEOREM AND A LEMMA The proof, due to Alex Lubotzky, is based on an idea in Lyndon and Schupp w3, p. 26x. Let F be a free group, f g F, n g N, and suppose that f is an nth power Žmod K . for every normal subgroup K of F of finite index. A moment’s thought shows that we may assume that F is of finite rank r ) 1, that f is cyclically reduced with respect to a basis X of F, and that, for each x g X, f f ² X y x 4:. Write f s a1 ? ??? ? a k , a i g X j Xy1 , a i a iq1 / 1. We proceed by induction on k s length f ; in case k s 0, the Theorem visibly holds, as 1 s 1n for all n g Z. Write f s x 1m 1 ??? x rm r f 0 , m i g Z, f 0 g F9, X s x 1 , . . . , x r 4 . Let p be a prime exceeding n and all < m i <. Let C be a group of order p with generator c. As r ) 1, there are i, 1 F i F r, and f g homŽ F, C . such that f Ž x i . s c, f Ž f . s 1. Thus, f Ž u. s f Ž ¨ y1 u¨ . for all u, ¨ g F. Replacing f by a cyclically reduced conjugate of f " 1 , we assume without loss of generality that a1 s x i . Relabelling X, we also assume without loss of generality that i s 1. Let F0 s ker f , and let T s x 1j N 0 F j - p4 . Then T is a set of coset representatives for F0 in F, and T has the Schreier property that each initial segment of an element of T is in T. Let : F ª T be defined by u g T l F0 u. Then F0 is free on y1
U s u g F < u / 1 and u s txtx
for some t g T , x g X 4 .
If t g T and x g X, set y1
t Ž t , x . s txtx
,
t Ž t , xy1 . s t Ž txy1 , x .
y1
.
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JOHN G. THOMPSON
Thus k
fs
Ł t Ž a1 , . . . , aiy1 , ai . ,
is1
and as t Ž1, a1 . s 1, the U-length of f is - k. Let K 0 be a normal subgroup of F0 of finite index and let K s F x g F K 0x. Then K is a normal subgroup of F of finite index, so there is g g F such that fy1 g n g K. In particular g n g F0 . Since g p g F0 , and since Ž n, p . s 1, we get g g F0 . Since K : K 0 , we have fy1 g n g K 0 , so f, as an element of F0 , is an nth power Žmod K 0 .. This holds for all relevant K 0 , so, by induction on k, we get f g PnŽ F0 .. Since PnŽ F0 . : PnŽ F ., and since PnŽ F . s PynŽ F ., while F K g K K s 14 Žso that the Theorem holds for n s 0., the proof is complete. LEMMA 1. Suppose p is a prime, G is a group, and F is a normal free subgroup of G of index p. Suppose that CG Ž x . is cyclic for all x g F y 14 . If m g N and n s p m , then PnŽ G mod Fin G . s PnŽ G .. Proof. Since PnŽ G mod Fin G . = PnŽ G ., it suffices to show that Ž Pn G mod Fin G . : PnŽ G .. Suppose g g PnŽ G mod Fin G .. For each K g Fin G, choose g K in G such that gy1 g Kn g K. Set h K s g Kp . Then h K g F and gy1 h nKr p g K. Since Fin G l Fin F is cofinal in Fin F, it follows from Lubotzky’s theorem that g g Pn r p Ž F ., so there is h in F such that g s h n r p. In particular, g g F. If g s 1, then g g PnŽ G ., so we may assume that g / 1. Pick an element a g G, a f F. Set u s ay1 ga. Thus, g, u4 : F. Case 1.
g and u are F-conjugate.
Choose f in F with fy1 uf s g. Set f 0 s af. Thus f 0 g CG Ž g . and f 0 f F. By hypothesis, CG Ž g . is cyclic. As < CG Ž g .rCF Ž g .< s p, it follows that CF Ž g . : Pp Ž G .. In particular, h g Pp Ž G ., so g g PnŽ G .. Case 2.
g and u are not F-conjugate.
Choose L g Fin F such that gL and uL are not conjugate in FrL; L exists by Lyndon and Schupp w3, p. 26x. Set M s F x g G Lx. Then M g Fin G, and gM, uM are not conjugate in FrM but are conjugate in GrM. We may assume that g f PnŽ F .. Choose R g Fin F such that gR f PnŽ FrR .. Set S s F x g G R x. Then S g Fin G, and gS f PnŽ FrS . but gS g PnŽ GrS .. Set T s M l S. Thus, T g Fin G, T : F, and gT, uT are not conjugate in FrT but are conjugate in GrT, and gT f PnŽ FrT ., gT g PnŽ GrT .. Choose b in G such that gy1 b n g T. Thus, b g G, b f F. In addition bT g CG r T Ž gT .. Choose c g G such that cy1 ucgy1 g T. Thus, c g G,
POWER MAPS AND COMPLETIONS
255
c f F. Since < GrF < is a prime, there is an integer i, 1 F i - p, such that c s fb i for some f g F. Thus byi fy1 ufb i gy1 g T. Since T eG, we get fy1 uf ? b i gy1 byi g T. Since bT g CG r T Ž gT ., we get fy1 ufgy1 g T. This is false, as uT and gT are not conjugate in FrT. The proof is complete.
3. PRELIMINARIES TO THE PROOF OF THEOREM 1 Assume the hypotheses of Theorem 1, and let As
ž
a c
b , d
/
t s a q d.
Set K s GN < N s 1, 2, . . . 4 ,
P˜n Ž G . s Pn Ž SL Ž 2, Q . . l G.
Since PnŽ G ., P˜nŽ G ., and PnŽ G mod K . are all stable under conjugation by elements of G, it suffices to show that A 0 g P˜nŽ G . for some G-conjugate A 0 of A. For each prime p, A is an nth power Žmod Gp k . for all k s 1, 2, . . . , and since SLŽ2, Z p . is compact, Anp s A
for some A p g SL Ž 2, Z p . .
Ž 3.1.
In this section, the proof of Theorem 1 is carried out under the hypothesis that t g y2, y1, 0, 1, 24 . With the Corollary in mind, we prove more than is required in Theorem 1, namely, we prove that Ž3.2. if t g y2, y1, 0, 1, 24 , then A is an nth power in G. Case 1.
t g y1, 0, 14 .
In this case, A has order 3, 4, 6 respectively, and as A is an nth power Žmod G5 ., it follows that n is relatively prime to the order of A. Setting B s An, we get A s B n and Ž3.2. holds. Case 2.
t s 2.
In this case, we assume without loss of generality that a s d s 1, c s 0, and we must show that b ' 0 Žmod n.. This is clear if b s 0, so suppose b / 0. For each prime l dividing n, choose A l g SLŽ2, Z l . with A nl s A. This gives b ' 0 Žmod n l ., where n l is the l-part of n, so A is an nth power in G. Case 3.
t s y2 and n is odd.
Here, yA g PnŽ G mod K ., so yA is an nth power in G by the preceding case. As y1 s Žy1. n, A is also an nth power in G.
256
JOHN G. THOMPSON
Case 4.
t s y2 and n is even.
Here we assume without loss of generality that a s d s y1, c s 0. Let A 2 g SLŽ2, Z 2 . with A n2 s A. As y1 is not a square in Z 2 it follows that b s 0, n ' 2 Žmod 4., and A s Ž y10 10 . 2 . Thus, Ž3.2. holds. To complete the proof of Theorem 1, it suffices to treat the case < t < G 3.
4. CHEBYSHEF POLYNOMIALS The nth Chebyshef polynomial is TnŽ x . g Zw x x, characterized by TnŽcos u . s cos nu . Set Un Ž x . s 2Tn
x
ž / 2
.
Then UnŽ x . g Zw x x, UnŽ x . is monic, and in Zw x, xy1 x, UnŽ x q xy1 . s x n q xyn . The major ingredient in the proof of Theorem 1 involves a property of certain Chebyshef polynomials. J.-P. Serre noted that the following lemma is a consequence of w1, Theorem 1, p. 82x. LEMMA. Suppose l is a prime, m g N, n s l m , 8 ¦ n, t g Z, < t < G 3, and, for each prime p, there is c p g Z p such that UnŽ c p . s t. Then there is c g Z such that UnŽ c . s t. Proof. Suppose the statement is false. Set f Ž x . s UnŽ x . y t. Thus f Ž x . has a zero in Z p for all primes p. To prove the lemma, it suffices to find a root of f in Q, as f is a monic polynomial with integer coefficients. The first thing to pin down is Ž4.1. One of the following holds: Ži. l / 2. Žii. t G 3. Suppose Ž4.1. fails, so that l s 2 and t F y3. Since n is even, A is a square in SLŽ2, Z p . for every prime p, so U2 Ž x . y t has a zero in each Z p . Since U2 Ž x . s x 2 y 2, it follows that t q 2 is a square Žmod p . for every prime p. This is false as t q 2 - 0, and quadratic reciprocity yields a prime p such that ŽŽ t q 2.rp . s y1. From Ž4.1., we get a real number j such that
j n q jyn s t ,
Ž 4.2.
POWER MAPS AND COMPLETIONS
257
or, equivalently, f Ž j q jy1 . s 0.
Ž 4.3.
Denote by Q the set of all algebraic complex numbers, and for each k g N, let m k be the set of kth roots of 1 in Q. Set E s Q Ž mn , j . . Set Rqs jz < z g m n 4 ,
Rys jy1z < z g m n 4 ,
R s Rqj Ry, S s jz q Ž jz .
y1
< z g mn 4 .
Obviously, as j / 0, we have < Rq< s < Ry< s n.
Ž 4.4.
If Rql Ry/ B, we get j 2 g m n , which is false as < t < G 3 and Ž4.2. holds. So < R < s 2 n.
Ž 4.5.
Suppose z 1 , z 2 g m n , z 1 / z 2 , but
jz 1 q Ž jz 1 .
y1
s jz 2 q Ž jz 2 .
y1
.
Then y1 j Ž z 1 y z 2 . s jy1 Ž zy1 2 y z1 . ,
so
j2s
y1 zy1 2 y z1
z1 y z2
y1 s zy1 1 z2 ,
which is false. So if we set g Ž z . s jz q Ž jz .y1 , then S s g Ž z . < z g mn 4 ,
< S < s n.
Ž 4.6.
Since deg f s n, Ž4.6. implies that Ž4.7.
S is the set of all roots of f in Q.
For all subfields F of Q, let GF be the profinite group of all automorphisms of Q which fix F elementwise, and let OF be the ring of integers of F. If F is a Galois extension of Q, and H is a subgroup of GQ rGF , F H denotes the set of fixed points of H on F.
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JOHN G. THOMPSON
Since f g Qw x x, Ž4.7. implies that S is GQ -stable. This implies that R is GQ -stable, as each element of R is a root of an equation X 2 y g Ž z . X q 1, both of whose roots lie in R for all z g m n . Since Q Ž R . s Q Ž mn , j . , it follows that E is a Galois extension of Q, with Galois group G s GQ rGE . Identifying GQ rGQŽ m n . with ŽZrnZ.=, we get an exact sequence =
1 ª H ª G ª Ž ZrnZ . ª 1,
Ž 4.8.
where H s GQŽ m n .rGE . For each r g R, let f r Ž x . be the monic irreducible polynomial in Ž Q m n .w x x which satisfies f r Ž r . s 0. We check that < H < s deg f r Ž x . ,
all r g R.
Ž 4.9.
To see this, observe that as E is Galois, H permutes transitively the roots of f r Ž x ., all of which lie in E. If h g H, and r 1 is a root of f r Ž x . fixed by h , then since r 1 s j ez 1 , e g 1, y14 , z 1 g m n , it follows that h Ž j . s j . Since E s QŽ m n , j ., we get h s 1, yielding Ž4.9.. Set h s < H <. Thus, for each r g R and each z g m n , z h f r Ž zy1 x . s f rz Ž x .. Also, f r Ž0. is in OE= , as each element of R is in OE= , so f ry1 Ž x . s f r Ž0.y1 x h f r Ž xy1 .. Define the integer e by e s < fr Ž x . < r g R4 < , so that 2 n s he.
Ž 4.10.
Write
f r Ž x . < r g R 4 s F1Ž x . , . . . , Fe Ž x . 4 . For each i s 1, . . . , e, set R i s r g R < Fi Ž r . s 0 4 , q Rq i s Ri l R ,
y Ry i s Ri l R .
Next, we use a well-known theorem to get Ž4.11. of G.
E C contains an element of S for every cyclic subgroup C
POWER MAPS AND COMPLETIONS
259
To see this, let C be a cyclic subgroup of G. Set Ds
Ł Ž s y s9. 2 .
s, s9gS s/s9
By a theorem of Frobenius w2x, there is a prime p such that p does not ramify in E, p ¦ D, and such that there is a prime ideal p of OE which contains p such that C is the stabilizer of p in G, and in addition such that < OErp < s p
for all i s 1, . . . , e.
Ž 4.12.
Rq i ,
Suppose the statement is false for i, and jz 1 , jz 2 g where z 1 , z 2 g m n , and z 1 / z 2 . Thus, there is h g H with h Ž jz 1 . s jz 2 . Let C s ²h :. Thus h Ž j . s jz 3 , where z 3 s z 2 zy1 1 / 1, and it follows that h fixes no element of S. This violates Ž4.11., and so yields Ž4.12.. From Ž4.12., we get e G n. Thus, Ž4.10. implies that Ž4.13. one of the following holds: Ži. e s 2 n and h s 1. Žii. e s n and h s 2. We next argue that Ž4.14.
G is not cyclic.
This is an immediate consequence of Ž4.11.. If e s 2 n and h s 1, then since 8 ¦ n, we get that G is cyclic, against Ž4.14.. So Ž4.13. implies that e s n and h s 2. Since ŽZrnZ.= is cyclic, while G is not cyclic, it follows that G s H = G0 ,
=
G 0 ( Ž ZrnZ . .
Let F s E G 0 , so that F s QŽ'M ., where M is an integer such that
'M f QŽ m n ..
We first study the case l / 2. Let H s ²h :, G 0 s ² s :, and let t be complex conjugation on E. Thus s Ž'M . s 'M and s Ž z . s z g , z g m n , where g is an integer which is a primitive root Žmod n.. Also, h Ž'M . s y 'M , and h Ž z . s z , t Ž z . s zy1 , t Ž'M . s sg Ž M .'M . Since e s n, we get
h Ž j . s jy1z 0 ,
260
JOHN G. THOMPSON
for some z 0 g m n , and fj Ž x . s Ž x y j .Ž x y jy1z 0 .. Since G s H = ²hs :, we assume without loss of generality that s Ž j . s jz 1 , z 1 g m n . By Ž4.11., there is g Ž z 2 . g E ² s : for some z 2 g m n . This gives yg y1 y1 jz 1 z 2g q jy1zy1 1 z 2 s jz 2 q j z 2 ,
whence
z 1 s z 21y g .
Ž 4.15.
Since G9 s 1, so also t Žg Ž z 2 .. s s Žt Žg Ž z 2 ..., and since t Žg Ž z 2 .. s ., we get g Ž zy1 2 qg z 1 s zy1 . 2
Ž 4.16.
By Ž4.15. and Ž4.16., we conclude that z 1 s 1, and so z 2 s 1. Thus, s fixes g Ž1. s j q jy1 , whence M ) 0, since otherwise j q jy1 g Q, and the lemma holds. On the other hand, h Žg Ž1.. g QŽ'M ., and as h Žg Ž1.. s y1 y1 y1 . Ž y1 jy1z 0 q jzy1 , whence z 0 s 1. 0 , we get j z 0 q jz 0 s s j z 0 q jz 0 Ž Ž .. Ž . Ž . This forces h g 1 s g 1 , so g 1 is fixed by s and by h , so g Ž1. g Q, and the lemma holds. It remains to treat the case e s n, h s 2, l s 2. Since 8 ¦ n, we have n s 2 or 4. Since G is not cyclic and < G < s 2 f Ž n., it follows that n s 4. In this case, yM is not a square and G 0 s ² s :, where s Ž i . s yi, s Ž'M . s 'M , h Ž i . s i, h Ž'M . s y 'M . The three quadratic subfields of E are E ² s : , E ²h : , E ² sh : , and by Ž4.11., each of these quadratic subfields contains an element of S. As S s yS : Q=, each of the quadratic subfields contains at least two elements of S. As < S < s 4, the lemma holds in case n s 4. The proof is complete. Remark. Setting f Ž x . s U8 Ž x . y 64,514, we find that f Ž x . s Ž x 2 y 18.Ž x 2 q 14.Ž x 4 y 4 x 2 q 256.. The roots of f Ž x . are thus "3'2 , " 'y 14 , " Ž3 " 'y 7 ., so f Ž x . has no rational root. Since y7 is a square in Q 2 , 2 is a square in Q 7 , and since one of 2, y7, y14 is a square in Q p for each prime p f 2, 74 , it follows that f Ž x . has a root in Z p for all primes p. Thus, the condition 8 ¦ n in the lemma is necessary.
5. THE PROOF OF THEOREM 1 Assume the hypotheses of Theorem 1. Let As
ž
a c
b , d
/
t s a q d.
If t g y2, y1, 0, 1, 24 , then Theorem 1 holds for A by Ž3.2.. So we assume that < t < G 3.
POWER MAPS AND COMPLETIONS
261
Set C s CG LŽ2, Q .Ž A.. Then C j 04 is a commutative field which is a quadratic extension field of Q ? I, and I, yI 4 is the set of elements of finite order in C. For each prime l, write n s n l ? n l9 , where n l is the l-part of n. Since A is an nth power Žmod GN . for all N s 1, 2, . . . , it follows that A is an n l th power Žmod GN . for all N s 1, 2, . . . , so for each prime p, there is AŽ p, l . g SLŽ2, Z p . such that AŽ p, l . n l s A. Set t Ž p, l . s tr AŽ p, l ., and set f l Ž x . s Un lŽ x . y t. Then t Ž p, l . is a root of f l Ž x . in Z p , and so by the lemma on Chebyshef polynomials, there is c l g Q such that f l Ž c l . s 0. Set 0 B˜l s y1
1 c l g SL Ž 2, Q . .
ž
/
Thus tr B˜ln l s t, so there is X l g GLŽ2, Q. such that
˜n l Xy1 l Bl X l s
ž
0 y1
1 . t
/
Since tr A s t, there is Yl g GLŽ2, Q. such that Yly1
ž
0 y1
1 Y s A. t l
/
nl ˜ Ž . Set Bl s Yly1 Xy1 l Bl X l Yl . Then Bl g SL 2, Q and Bl s A. Thus the group generated by Bl < l is a prime4 is a subgroup of CS LŽ2, Q .Ž A., so is abelian. Choose integers r l such that Ý l < n r l n l9 s 1. Set B s Ł l Blr l . Then
Bn s
Ł Blr n l
l9 n l
s
l
Ł Ar n l
l9
s A,
l
and the proof is complete.
6. THE PROOF OF THEOREM 2 First, assume that n s l m , m g N, l a prime. Case 1.
l / 2.
Let p be a prime ' 1 Žmod n.. Let ¨ be a positive integer which has order n Žmod p ., and let w be a positive integer such that ¨ w ' 1 Žmod p .. Write ¨ w y 1 s xy, where x, y are integers and p ¦ x. Write ¨
B0 s y
ž
x w ,
/
A 0 s B0n .
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JOHN G. THOMPSON
Then A 0 g Gp , B0 g G. Set p 0
0 1
p As 0
0 1
Bs
ž ž
y1
B0
p 0
0 ¨ s yp 1
A0
p 0
0 . 1
xrp , w
/ ž / ž / ž / y1
/
Then B n s A g G, and B g SLŽ2, Z r ., for every prime r / p. Also A s Ž ac db ., where a ' d ' 1 Žmod p ., c ' 0 Žmod p .. Since l / p, it follows that A is an nth power in SLŽ2, Z p ., as the nth power map induces a permutation of Vp s u g SL Ž 2, Z p . < u s
½
ž
u0 u2
u1 , u 0 ' u 3 ' 1 Ž mod p . , u3
/
5
u 2 ' 0 Ž mod p . . So A is an nth power in SLŽ2, Z r . for all primes r. By the Chinese remainder theorem, A is an nth power Žmod GN . for all N s 1, 2, . . . . Obviously, tr A f y2, y1, 0, 1, 24 , so I, yI 4 is the set of all elements of finite order in CS LŽ2, Q .Ž A.. As B n s A and B f G, it follows that A f PnŽ G .. Case 2.
l s 2.
Since n ) 2, we have m G 2. Let B0 s Bs
ž ž
1 1
1 , 2
3 0
0 1
A 0 s B04 s
/ / ž y1
B0
3 0
0 , 1
ž
13 21
21 , 34
/
A s B4.
/
Then Bs
1 3
1 3
ž / 2
,
As
ž
13 63
7 . 34
/
Then A is a fourth power in SLŽ2, Z 3 ., since the fourth power map in SLŽ2, Z 3 . induces a permutation of V3 . Since B g SLŽ2, Z r . for all primes r / 3, it follows that A is a fourth power in SLŽ2, Z r . for all primes r. my 2 Thus A2 is an nth power in SLŽ2, Z r . for all primes r, so by the my 2 Chinese remainder theorem, A2 is an nth power Žmod GN . for all 2 my 2 N s 1, 2, . . . . Since tr A ) 2, I, yI 4 is the set of all elements of finite my 2 order in CS LŽ2, Q .Ž A., and since B n s A2 , where B g SLŽ2, Q., B f G,
263
POWER MAPS AND COMPLETIONS my 2
we have A2 f PnŽ G .. This completes the proof of Theorem 2 in case n is a prime power. The general case is now immediate, since there is a prime l such that n l ) 2. 7. PROOF OF THE COROLLARY From Theorem 2, we know that PnŽ G mod K . > PnŽ G . if n ) 2, so it only remains to prove that PnŽ G . s PnŽ G mod Fin G ., or, equivalently, that PnŽ G mod Fin G . : PnŽ G .. Set Xs
ž
y1 0
y2 , y1
Ys
/
y1 2
ž
0 . y1
/
We compute that
ž ž ž
0 y1
0 y1
1 y1
0 y1
1 y1
1 0
y1
/ ž / ž / ž X
y1
0 y1
1 s Y, 0
/ / /
X
0 y1
1 s Xy1 Yy1 , y1
Y
0 y1
1 s X, y1
y1
and so F s ² X, Y :e G. We see that G2 s F = ² y I : ,
< G : F < s 12.
As is well known and easy to check, F is free of rank 2. Suppose A g PnŽ G mod Fin G .. If
l ) 3.
Set B s A12 . Then B g PnŽ F mod Fin F ., so there is C in F with C s B. Since ² A, C : : C G Ž B ., and since C G Ž B . is abelian, we get A g PnŽ G .. n
Case 2.
l s 3.
Set B s A4 ,
¦ ž
G s F,
0 y1
1 y1
/;
.
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JOHN G. THOMPSON
Thus B g PnŽ G mod Fin G .. By Lemma 1, we get B g PnŽ G ., so there is C in G with C n s B. Since ² A, C : is abelian, we have A g PnŽ G .. Case 3.
l s 2.
Since n is a positive power of 2, we conclude from Theorem 1 that A s A20 for some A 0 g SLŽ2, Q.. Since
ž
8 3
21 8
4
/
.
Then tr A s 64,514, so by the remark in Sect. 4, A is not an eighth power in SLŽ2, Q.. By the same remark, if p is a prime, then there are a p , bp g Q p such that
ž
ap
7bp
bp
ap
8
/
s A,
U8 Ž2 a p . s 64,514, and a2p y 7bp2 s 1. If p f 2, 74 , then since 2 a p g Z p , we get a p , bp g Z p . Also, 2 a2 s "Ž3 " 'y 7 ., so a2 g Z 2 whence b 2 g Z 2 . Finally, a7 s "3r '2 and 7b 72 s 92 y 1 s 72 , so b 7 g Z 7 . By the Chinese remainder theorem, A is an eighth power Žmod GN ., N s 1, 2, . . . . REFERENCES 1. E. Artin and J. Tate, ‘‘Class Field Theory.’’ 2. G. Frobenius, ‘‘Gesammelte Abhandlung,’’ Band II, pp. 719]733. 3. R. Lyndon and P. Schupp, ‘‘Combinatorial Group Theory,’’ Springer-Verlag, BerlinrNew York, 1977.
JOURNAL OF ALGEBRA ARTICLE NO.
Power Maps and Completions of Free Groups and of the Modular Group John G. Thompson Department of Mathematics, Uni¨ ersity of Florida, 201 Walker Hall, Gaines¨ ille, Florida 32611-8000 Communicated by Walter Feit Received September 24, 1996
1. INTRODUCTION AND STATEMENT OF RESULTS If G is a group and n is an integer, the nth power map is Pn : G ª G, PnŽ x . s x n. Thus PnŽ G . s PynŽ G ., the set of nth powers of elements of G. If K is a normal subgroup of G and x g G, we say that x is an nth power Žmod K . if and only if there is y g G such that xy1 y n g K. Let PnŽ G mod K . be the set of all x in G such that x is an nth power Žmod K ., so that PnŽ G mod K . is the preimage in G of PnŽ GrK .. If K is a family of normal subgroups of G, we say that x is an nth power Žmod K . if and only if x is an nth power Žmod K . for all K g K, and we set Pn Ž G mod K . s
F Pn Ž G mod K . . Kg K
Thus, PnŽ G mod K . = PnŽ G .. For each group G, let Fin G be the set of all normal subgroups of G of finite index, and let G c be the completion of G in the profinite topology. THEOREM ŽLubotzky.. If F is a free group and n is an integer, then PnŽ F mod Fin F . s PnŽ F .. Set G s SLŽ2, Z., and for each N g N, set GN s
½ž
a c
b g G a y 1 ' b ' c ' d y 1 ' 0 Ž mod N . , d
5
/
the principal congruence subgroup of level N. Also, Z p and Q p denote the p-adic integers and the p-adic numbers, respectively. 252 0021-8693r97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
253
POWER MAPS AND COMPLETIONS
THEOREM 1. If n g N, 8 ¦ n, A g G, and A is an nth power Žmod GN . for all N s 1, 2, . . . , then there is B g SLŽ2, Q. such that B n s A. THEOREM 2. Ži. Žii.
If n is an integer ) 2, then there is A g G such that
A f PnŽ G . and A g PnŽ SLŽ2, Q..; A is an nth power Žmod GN . for all N s 1, 2, . . . .
Putting these results together, we get that Fin G and K, defined in the Corollary, behave differently with respect to power maps. COROLLARY. Ži. Žii.
Let K s GN < N s 1, 2, . . . 4 , n g N.
PnŽ G mod Fin G . s PnŽ G .. If n ) 2, then PnŽ G mod K . ; PnŽ G ..
2. PROOF OF LUBOTZKY’S THEOREM AND A LEMMA The proof, due to Alex Lubotzky, is based on an idea in Lyndon and Schupp w3, p. 26x. Let F be a free group, f g F, n g N, and suppose that f is an nth power Žmod K . for every normal subgroup K of F of finite index. A moment’s thought shows that we may assume that F is of finite rank r ) 1, that f is cyclically reduced with respect to a basis X of F, and that, for each x g X, f f ² X y x 4:. Write f s a1 ? ??? ? a k , a i g X j Xy1 , a i a iq1 / 1. We proceed by induction on k s length f ; in case k s 0, the Theorem visibly holds, as 1 s 1n for all n g Z. Write f s x 1m 1 ??? x rm r f 0 , m i g Z, f 0 g F9, X s x 1 , . . . , x r 4 . Let p be a prime exceeding n and all < m i <. Let C be a group of order p with generator c. As r ) 1, there are i, 1 F i F r, and f g homŽ F, C . such that f Ž x i . s c, f Ž f . s 1. Thus, f Ž u. s f Ž ¨ y1 u¨ . for all u, ¨ g F. Replacing f by a cyclically reduced conjugate of f " 1 , we assume without loss of generality that a1 s x i . Relabelling X, we also assume without loss of generality that i s 1. Let F0 s ker f , and let T s x 1j N 0 F j - p4 . Then T is a set of coset representatives for F0 in F, and T has the Schreier property that each initial segment of an element of T is in T. Let : F ª T be defined by u g T l F0 u. Then F0 is free on y1
U s u g F < u / 1 and u s txtx
for some t g T , x g X 4 .
If t g T and x g X, set y1
t Ž t , x . s txtx
,
t Ž t , xy1 . s t Ž txy1 , x .
y1
.
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JOHN G. THOMPSON
Thus k
fs
Ł t Ž a1 , . . . , aiy1 , ai . ,
is1
and as t Ž1, a1 . s 1, the U-length of f is - k. Let K 0 be a normal subgroup of F0 of finite index and let K s F x g F K 0x. Then K is a normal subgroup of F of finite index, so there is g g F such that fy1 g n g K. In particular g n g F0 . Since g p g F0 , and since Ž n, p . s 1, we get g g F0 . Since K : K 0 , we have fy1 g n g K 0 , so f, as an element of F0 , is an nth power Žmod K 0 .. This holds for all relevant K 0 , so, by induction on k, we get f g PnŽ F0 .. Since PnŽ F0 . : PnŽ F ., and since PnŽ F . s PynŽ F ., while F K g K K s 14 Žso that the Theorem holds for n s 0., the proof is complete. LEMMA 1. Suppose p is a prime, G is a group, and F is a normal free subgroup of G of index p. Suppose that CG Ž x . is cyclic for all x g F y 14 . If m g N and n s p m , then PnŽ G mod Fin G . s PnŽ G .. Proof. Since PnŽ G mod Fin G . = PnŽ G ., it suffices to show that Ž Pn G mod Fin G . : PnŽ G .. Suppose g g PnŽ G mod Fin G .. For each K g Fin G, choose g K in G such that gy1 g Kn g K. Set h K s g Kp . Then h K g F and gy1 h nKr p g K. Since Fin G l Fin F is cofinal in Fin F, it follows from Lubotzky’s theorem that g g Pn r p Ž F ., so there is h in F such that g s h n r p. In particular, g g F. If g s 1, then g g PnŽ G ., so we may assume that g / 1. Pick an element a g G, a f F. Set u s ay1 ga. Thus, g, u4 : F. Case 1.
g and u are F-conjugate.
Choose f in F with fy1 uf s g. Set f 0 s af. Thus f 0 g CG Ž g . and f 0 f F. By hypothesis, CG Ž g . is cyclic. As < CG Ž g .rCF Ž g .< s p, it follows that CF Ž g . : Pp Ž G .. In particular, h g Pp Ž G ., so g g PnŽ G .. Case 2.
g and u are not F-conjugate.
Choose L g Fin F such that gL and uL are not conjugate in FrL; L exists by Lyndon and Schupp w3, p. 26x. Set M s F x g G Lx. Then M g Fin G, and gM, uM are not conjugate in FrM but are conjugate in GrM. We may assume that g f PnŽ F .. Choose R g Fin F such that gR f PnŽ FrR .. Set S s F x g G R x. Then S g Fin G, and gS f PnŽ FrS . but gS g PnŽ GrS .. Set T s M l S. Thus, T g Fin G, T : F, and gT, uT are not conjugate in FrT but are conjugate in GrT, and gT f PnŽ FrT ., gT g PnŽ GrT .. Choose b in G such that gy1 b n g T. Thus, b g G, b f F. In addition bT g CG r T Ž gT .. Choose c g G such that cy1 ucgy1 g T. Thus, c g G,
POWER MAPS AND COMPLETIONS
255
c f F. Since < GrF < is a prime, there is an integer i, 1 F i - p, such that c s fb i for some f g F. Thus byi fy1 ufb i gy1 g T. Since T eG, we get fy1 uf ? b i gy1 byi g T. Since bT g CG r T Ž gT ., we get fy1 ufgy1 g T. This is false, as uT and gT are not conjugate in FrT. The proof is complete.
3. PRELIMINARIES TO THE PROOF OF THEOREM 1 Assume the hypotheses of Theorem 1, and let As
ž
a c
b , d
/
t s a q d.
Set K s GN < N s 1, 2, . . . 4 ,
P˜n Ž G . s Pn Ž SL Ž 2, Q . . l G.
Since PnŽ G ., P˜nŽ G ., and PnŽ G mod K . are all stable under conjugation by elements of G, it suffices to show that A 0 g P˜nŽ G . for some G-conjugate A 0 of A. For each prime p, A is an nth power Žmod Gp k . for all k s 1, 2, . . . , and since SLŽ2, Z p . is compact, Anp s A
for some A p g SL Ž 2, Z p . .
Ž 3.1.
In this section, the proof of Theorem 1 is carried out under the hypothesis that t g y2, y1, 0, 1, 24 . With the Corollary in mind, we prove more than is required in Theorem 1, namely, we prove that Ž3.2. if t g y2, y1, 0, 1, 24 , then A is an nth power in G. Case 1.
t g y1, 0, 14 .
In this case, A has order 3, 4, 6 respectively, and as A is an nth power Žmod G5 ., it follows that n is relatively prime to the order of A. Setting B s An, we get A s B n and Ž3.2. holds. Case 2.
t s 2.
In this case, we assume without loss of generality that a s d s 1, c s 0, and we must show that b ' 0 Žmod n.. This is clear if b s 0, so suppose b / 0. For each prime l dividing n, choose A l g SLŽ2, Z l . with A nl s A. This gives b ' 0 Žmod n l ., where n l is the l-part of n, so A is an nth power in G. Case 3.
t s y2 and n is odd.
Here, yA g PnŽ G mod K ., so yA is an nth power in G by the preceding case. As y1 s Žy1. n, A is also an nth power in G.
256
JOHN G. THOMPSON
Case 4.
t s y2 and n is even.
Here we assume without loss of generality that a s d s y1, c s 0. Let A 2 g SLŽ2, Z 2 . with A n2 s A. As y1 is not a square in Z 2 it follows that b s 0, n ' 2 Žmod 4., and A s Ž y10 10 . 2 . Thus, Ž3.2. holds. To complete the proof of Theorem 1, it suffices to treat the case < t < G 3.
4. CHEBYSHEF POLYNOMIALS The nth Chebyshef polynomial is TnŽ x . g Zw x x, characterized by TnŽcos u . s cos nu . Set Un Ž x . s 2Tn
x
ž / 2
.
Then UnŽ x . g Zw x x, UnŽ x . is monic, and in Zw x, xy1 x, UnŽ x q xy1 . s x n q xyn . The major ingredient in the proof of Theorem 1 involves a property of certain Chebyshef polynomials. J.-P. Serre noted that the following lemma is a consequence of w1, Theorem 1, p. 82x. LEMMA. Suppose l is a prime, m g N, n s l m , 8 ¦ n, t g Z, < t < G 3, and, for each prime p, there is c p g Z p such that UnŽ c p . s t. Then there is c g Z such that UnŽ c . s t. Proof. Suppose the statement is false. Set f Ž x . s UnŽ x . y t. Thus f Ž x . has a zero in Z p for all primes p. To prove the lemma, it suffices to find a root of f in Q, as f is a monic polynomial with integer coefficients. The first thing to pin down is Ž4.1. One of the following holds: Ži. l / 2. Žii. t G 3. Suppose Ž4.1. fails, so that l s 2 and t F y3. Since n is even, A is a square in SLŽ2, Z p . for every prime p, so U2 Ž x . y t has a zero in each Z p . Since U2 Ž x . s x 2 y 2, it follows that t q 2 is a square Žmod p . for every prime p. This is false as t q 2 - 0, and quadratic reciprocity yields a prime p such that ŽŽ t q 2.rp . s y1. From Ž4.1., we get a real number j such that
j n q jyn s t ,
Ž 4.2.
POWER MAPS AND COMPLETIONS
257
or, equivalently, f Ž j q jy1 . s 0.
Ž 4.3.
Denote by Q the set of all algebraic complex numbers, and for each k g N, let m k be the set of kth roots of 1 in Q. Set E s Q Ž mn , j . . Set Rqs jz < z g m n 4 ,
Rys jy1z < z g m n 4 ,
R s Rqj Ry, S s jz q Ž jz .
y1
< z g mn 4 .
Obviously, as j / 0, we have < Rq< s < Ry< s n.
Ž 4.4.
If Rql Ry/ B, we get j 2 g m n , which is false as < t < G 3 and Ž4.2. holds. So < R < s 2 n.
Ž 4.5.
Suppose z 1 , z 2 g m n , z 1 / z 2 , but
jz 1 q Ž jz 1 .
y1
s jz 2 q Ž jz 2 .
y1
.
Then y1 j Ž z 1 y z 2 . s jy1 Ž zy1 2 y z1 . ,
so
j2s
y1 zy1 2 y z1
z1 y z2
y1 s zy1 1 z2 ,
which is false. So if we set g Ž z . s jz q Ž jz .y1 , then S s g Ž z . < z g mn 4 ,
< S < s n.
Ž 4.6.
Since deg f s n, Ž4.6. implies that Ž4.7.
S is the set of all roots of f in Q.
For all subfields F of Q, let GF be the profinite group of all automorphisms of Q which fix F elementwise, and let OF be the ring of integers of F. If F is a Galois extension of Q, and H is a subgroup of GQ rGF , F H denotes the set of fixed points of H on F.
258
JOHN G. THOMPSON
Since f g Qw x x, Ž4.7. implies that S is GQ -stable. This implies that R is GQ -stable, as each element of R is a root of an equation X 2 y g Ž z . X q 1, both of whose roots lie in R for all z g m n . Since Q Ž R . s Q Ž mn , j . , it follows that E is a Galois extension of Q, with Galois group G s GQ rGE . Identifying GQ rGQŽ m n . with ŽZrnZ.=, we get an exact sequence =
1 ª H ª G ª Ž ZrnZ . ª 1,
Ž 4.8.
where H s GQŽ m n .rGE . For each r g R, let f r Ž x . be the monic irreducible polynomial in Ž Q m n .w x x which satisfies f r Ž r . s 0. We check that < H < s deg f r Ž x . ,
all r g R.
Ž 4.9.
To see this, observe that as E is Galois, H permutes transitively the roots of f r Ž x ., all of which lie in E. If h g H, and r 1 is a root of f r Ž x . fixed by h , then since r 1 s j ez 1 , e g 1, y14 , z 1 g m n , it follows that h Ž j . s j . Since E s QŽ m n , j ., we get h s 1, yielding Ž4.9.. Set h s < H <. Thus, for each r g R and each z g m n , z h f r Ž zy1 x . s f rz Ž x .. Also, f r Ž0. is in OE= , as each element of R is in OE= , so f ry1 Ž x . s f r Ž0.y1 x h f r Ž xy1 .. Define the integer e by e s < fr Ž x . < r g R4 < , so that 2 n s he.
Ž 4.10.
Write
f r Ž x . < r g R 4 s F1Ž x . , . . . , Fe Ž x . 4 . For each i s 1, . . . , e, set R i s r g R < Fi Ž r . s 0 4 , q Rq i s Ri l R ,
y Ry i s Ri l R .
Next, we use a well-known theorem to get Ž4.11. of G.
E C contains an element of S for every cyclic subgroup C
POWER MAPS AND COMPLETIONS
259
To see this, let C be a cyclic subgroup of G. Set Ds
Ł Ž s y s9. 2 .
s, s9gS s/s9
By a theorem of Frobenius w2x, there is a prime p such that p does not ramify in E, p ¦ D, and such that there is a prime ideal p of OE which contains p such that C is the stabilizer of p in G, and in addition such that < OErp < s p
for all i s 1, . . . , e.
Ž 4.12.
Rq i ,
Suppose the statement is false for i, and jz 1 , jz 2 g where z 1 , z 2 g m n , and z 1 / z 2 . Thus, there is h g H with h Ž jz 1 . s jz 2 . Let C s ²h :. Thus h Ž j . s jz 3 , where z 3 s z 2 zy1 1 / 1, and it follows that h fixes no element of S. This violates Ž4.11., and so yields Ž4.12.. From Ž4.12., we get e G n. Thus, Ž4.10. implies that Ž4.13. one of the following holds: Ži. e s 2 n and h s 1. Žii. e s n and h s 2. We next argue that Ž4.14.
G is not cyclic.
This is an immediate consequence of Ž4.11.. If e s 2 n and h s 1, then since 8 ¦ n, we get that G is cyclic, against Ž4.14.. So Ž4.13. implies that e s n and h s 2. Since ŽZrnZ.= is cyclic, while G is not cyclic, it follows that G s H = G0 ,
=
G 0 ( Ž ZrnZ . .
Let F s E G 0 , so that F s QŽ'M ., where M is an integer such that
'M f QŽ m n ..
We first study the case l / 2. Let H s ²h :, G 0 s ² s :, and let t be complex conjugation on E. Thus s Ž'M . s 'M and s Ž z . s z g , z g m n , where g is an integer which is a primitive root Žmod n.. Also, h Ž'M . s y 'M , and h Ž z . s z , t Ž z . s zy1 , t Ž'M . s sg Ž M .'M . Since e s n, we get
h Ž j . s jy1z 0 ,
260
JOHN G. THOMPSON
for some z 0 g m n , and fj Ž x . s Ž x y j .Ž x y jy1z 0 .. Since G s H = ²hs :, we assume without loss of generality that s Ž j . s jz 1 , z 1 g m n . By Ž4.11., there is g Ž z 2 . g E ² s : for some z 2 g m n . This gives yg y1 y1 jz 1 z 2g q jy1zy1 1 z 2 s jz 2 q j z 2 ,
whence
z 1 s z 21y g .
Ž 4.15.
Since G9 s 1, so also t Žg Ž z 2 .. s s Žt Žg Ž z 2 ..., and since t Žg Ž z 2 .. s ., we get g Ž zy1 2 qg z 1 s zy1 . 2
Ž 4.16.
By Ž4.15. and Ž4.16., we conclude that z 1 s 1, and so z 2 s 1. Thus, s fixes g Ž1. s j q jy1 , whence M ) 0, since otherwise j q jy1 g Q, and the lemma holds. On the other hand, h Žg Ž1.. g QŽ'M ., and as h Žg Ž1.. s y1 y1 y1 . Ž y1 jy1z 0 q jzy1 , whence z 0 s 1. 0 , we get j z 0 q jz 0 s s j z 0 q jz 0 Ž Ž .. Ž . Ž . This forces h g 1 s g 1 , so g 1 is fixed by s and by h , so g Ž1. g Q, and the lemma holds. It remains to treat the case e s n, h s 2, l s 2. Since 8 ¦ n, we have n s 2 or 4. Since G is not cyclic and < G < s 2 f Ž n., it follows that n s 4. In this case, yM is not a square and G 0 s ² s :, where s Ž i . s yi, s Ž'M . s 'M , h Ž i . s i, h Ž'M . s y 'M . The three quadratic subfields of E are E ² s : , E ²h : , E ² sh : , and by Ž4.11., each of these quadratic subfields contains an element of S. As S s yS : Q=, each of the quadratic subfields contains at least two elements of S. As < S < s 4, the lemma holds in case n s 4. The proof is complete. Remark. Setting f Ž x . s U8 Ž x . y 64,514, we find that f Ž x . s Ž x 2 y 18.Ž x 2 q 14.Ž x 4 y 4 x 2 q 256.. The roots of f Ž x . are thus "3'2 , " 'y 14 , " Ž3 " 'y 7 ., so f Ž x . has no rational root. Since y7 is a square in Q 2 , 2 is a square in Q 7 , and since one of 2, y7, y14 is a square in Q p for each prime p f 2, 74 , it follows that f Ž x . has a root in Z p for all primes p. Thus, the condition 8 ¦ n in the lemma is necessary.
5. THE PROOF OF THEOREM 1 Assume the hypotheses of Theorem 1. Let As
ž
a c
b , d
/
t s a q d.
If t g y2, y1, 0, 1, 24 , then Theorem 1 holds for A by Ž3.2.. So we assume that < t < G 3.
POWER MAPS AND COMPLETIONS
261
Set C s CG LŽ2, Q .Ž A.. Then C j 04 is a commutative field which is a quadratic extension field of Q ? I, and I, yI 4 is the set of elements of finite order in C. For each prime l, write n s n l ? n l9 , where n l is the l-part of n. Since A is an nth power Žmod GN . for all N s 1, 2, . . . , it follows that A is an n l th power Žmod GN . for all N s 1, 2, . . . , so for each prime p, there is AŽ p, l . g SLŽ2, Z p . such that AŽ p, l . n l s A. Set t Ž p, l . s tr AŽ p, l ., and set f l Ž x . s Un lŽ x . y t. Then t Ž p, l . is a root of f l Ž x . in Z p , and so by the lemma on Chebyshef polynomials, there is c l g Q such that f l Ž c l . s 0. Set 0 B˜l s y1
1 c l g SL Ž 2, Q . .
ž
/
Thus tr B˜ln l s t, so there is X l g GLŽ2, Q. such that
˜n l Xy1 l Bl X l s
ž
0 y1
1 . t
/
Since tr A s t, there is Yl g GLŽ2, Q. such that Yly1
ž
0 y1
1 Y s A. t l
/
nl ˜ Ž . Set Bl s Yly1 Xy1 l Bl X l Yl . Then Bl g SL 2, Q and Bl s A. Thus the group generated by Bl < l is a prime4 is a subgroup of CS LŽ2, Q .Ž A., so is abelian. Choose integers r l such that Ý l < n r l n l9 s 1. Set B s Ł l Blr l . Then
Bn s
Ł Blr n l
l9 n l
s
l
Ł Ar n l
l9
s A,
l
and the proof is complete.
6. THE PROOF OF THEOREM 2 First, assume that n s l m , m g N, l a prime. Case 1.
l / 2.
Let p be a prime ' 1 Žmod n.. Let ¨ be a positive integer which has order n Žmod p ., and let w be a positive integer such that ¨ w ' 1 Žmod p .. Write ¨ w y 1 s xy, where x, y are integers and p ¦ x. Write ¨
B0 s y
ž
x w ,
/
A 0 s B0n .
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JOHN G. THOMPSON
Then A 0 g Gp , B0 g G. Set p 0
0 1
p As 0
0 1
Bs
ž ž
y1
B0
p 0
0 ¨ s yp 1
A0
p 0
0 . 1
xrp , w
/ ž / ž / ž / y1
/
Then B n s A g G, and B g SLŽ2, Z r ., for every prime r / p. Also A s Ž ac db ., where a ' d ' 1 Žmod p ., c ' 0 Žmod p .. Since l / p, it follows that A is an nth power in SLŽ2, Z p ., as the nth power map induces a permutation of Vp s u g SL Ž 2, Z p . < u s
½
ž
u0 u2
u1 , u 0 ' u 3 ' 1 Ž mod p . , u3
/
5
u 2 ' 0 Ž mod p . . So A is an nth power in SLŽ2, Z r . for all primes r. By the Chinese remainder theorem, A is an nth power Žmod GN . for all N s 1, 2, . . . . Obviously, tr A f y2, y1, 0, 1, 24 , so I, yI 4 is the set of all elements of finite order in CS LŽ2, Q .Ž A.. As B n s A and B f G, it follows that A f PnŽ G .. Case 2.
l s 2.
Since n ) 2, we have m G 2. Let B0 s Bs
ž ž
1 1
1 , 2
3 0
0 1
A 0 s B04 s
/ / ž y1
B0
3 0
0 , 1
ž
13 21
21 , 34
/
A s B4.
/
Then Bs
1 3
1 3
ž / 2
,
As
ž
13 63
7 . 34
/
Then A is a fourth power in SLŽ2, Z 3 ., since the fourth power map in SLŽ2, Z 3 . induces a permutation of V3 . Since B g SLŽ2, Z r . for all primes r / 3, it follows that A is a fourth power in SLŽ2, Z r . for all primes r. my 2 Thus A2 is an nth power in SLŽ2, Z r . for all primes r, so by the my 2 Chinese remainder theorem, A2 is an nth power Žmod GN . for all 2 my 2 N s 1, 2, . . . . Since tr A ) 2, I, yI 4 is the set of all elements of finite my 2 order in CS LŽ2, Q .Ž A., and since B n s A2 , where B g SLŽ2, Q., B f G,
263
POWER MAPS AND COMPLETIONS my 2
we have A2 f PnŽ G .. This completes the proof of Theorem 2 in case n is a prime power. The general case is now immediate, since there is a prime l such that n l ) 2. 7. PROOF OF THE COROLLARY From Theorem 2, we know that PnŽ G mod K . > PnŽ G . if n ) 2, so it only remains to prove that PnŽ G . s PnŽ G mod Fin G ., or, equivalently, that PnŽ G mod Fin G . : PnŽ G .. Set Xs
ž
y1 0
y2 , y1
Ys
/
y1 2
ž
0 . y1
/
We compute that
ž ž ž
0 y1
0 y1
1 y1
0 y1
1 y1
1 0
y1
/ ž / ž / ž X
y1
0 y1
1 s Y, 0
/ / /
X
0 y1
1 s Xy1 Yy1 , y1
Y
0 y1
1 s X, y1
y1
and so F s ² X, Y :e G. We see that G2 s F = ² y I : ,
< G : F < s 12.
As is well known and easy to check, F is free of rank 2. Suppose A g PnŽ G mod Fin G .. If
l ) 3.
Set B s A12 . Then B g PnŽ F mod Fin F ., so there is C in F with C s B. Since ² A, C : : C G Ž B ., and since C G Ž B . is abelian, we get A g PnŽ G .. n
Case 2.
l s 3.
Set B s A4 ,
¦ ž
G s F,
0 y1
1 y1
/;
.
264
JOHN G. THOMPSON
Thus B g PnŽ G mod Fin G .. By Lemma 1, we get B g PnŽ G ., so there is C in G with C n s B. Since ² A, C : is abelian, we have A g PnŽ G .. Case 3.
l s 2.
Since n is a positive power of 2, we conclude from Theorem 1 that A s A20 for some A 0 g SLŽ2, Q.. Since
ž
8 3
21 8
4
/
.
Then tr A s 64,514, so by the remark in Sect. 4, A is not an eighth power in SLŽ2, Q.. By the same remark, if p is a prime, then there are a p , bp g Q p such that
ž
ap
7bp
bp
ap
8
/
s A,
U8 Ž2 a p . s 64,514, and a2p y 7bp2 s 1. If p f 2, 74 , then since 2 a p g Z p , we get a p , bp g Z p . Also, 2 a2 s "Ž3 " 'y 7 ., so a2 g Z 2 whence b 2 g Z 2 . Finally, a7 s "3r '2 and 7b 72 s 92 y 1 s 72 , so b 7 g Z 7 . By the Chinese remainder theorem, A is an eighth power Žmod GN ., N s 1, 2, . . . . REFERENCES 1. E. Artin and J. Tate, ‘‘Class Field Theory.’’ 2. G. Frobenius, ‘‘Gesammelte Abhandlung,’’ Band II, pp. 719]733. 3. R. Lyndon and P. Schupp, ‘‘Combinatorial Group Theory,’’ Springer-Verlag, BerlinrNew York, 1977.