On a problem of G. Sa,nsone

by MORRIS I~-EW~A~

$ammary.- It is proved that the members of a cert<ti~t family of free tsubgroups of index 6~ in the modular group P are congruence groups i f and oltly i f n ~ 1~ 2, 4~ 8.

l. In t roduc t ion . - Let l ' denote the 2),< 2 modular group, consist ing of all l inear fract ional t ransformations

(1) ~' ~ - - a z + b c~ + d

where a, b, c, d are rat ional integers and a d - - b e - - - - - 1 . Then P may be regarded as the mult ipl icative group of 2 X 2 rational integral matrices of determinant l, in which a matr ix is identified with its negative. Let n be a positive integer and let F (n) denote the p r i n c i p a l c o n g r u e n c e group of level n ;

that is, the tota[ity of sttbstitutions (1) such that

a ~ d - - = + - - - 1 (rood n), b - ~ c ~ - O (mod n).

In his opening address [2] to the Conference on Fini te Groups held at Florence in April, 1960 Professor SA•SONE ment ioned several unsolved pro- blems concerning groups of l inear fractional transformations. The first of these was the ari thmetic charac ter iza t ion of the subgroups F6,~ of the modular group, generated by the n q - 1 matrices

t; :If;" 7] i;+ r (2) , , (1 ~ r ~ n - - 1). - - 1 - - 4r

The purpose of this note is to determine the s t ructure of these groups. In par t icular we shall show that if n =~= 1, 2, 4, 8 then r6, is not a c o n g r u e n c e

28 MORRIS ~NEW~A~: On a problem of G. Sa/nsone

group, and so cannot be characterized by finitely many polynomial con. gruences in the entries of the elements of P6,~. We shalt also show that

]:6,, D F (2n}, n - - 1, 2, 4, 8

so that snob a characterizat ion is indeed possible in these cases. We set

Then the matrices (2) may be replaced by

(3) S ~ , S'2"W~'S - ~ ' (0 ~ r ~ n - - 1).

( I: 1 It is only necessary to rep lace by

2. Remarks on free groups. - We first make some elementar3~ remarks about free groups of rank 2.

Lel G - - I x , Yl be the free group generated by x and y. For every posi- tive integer n define the "exponential congruence group,, G,, as the subgroup of G consisting of all words of G for which the sum of the exponents of x is divisible by n. Then G,, is a normal subgroup of G, and G/G, is the cyclic group In G. t of order n. Since G is of rank 2 and G. of index n in G, it follows from the RE[DEM:EISTER-SCKREIER formula that the rank of G,, is

1 q- n(2 - - 1)---1 q-n .

It is easily proved that the elements

(4) ~ , x ~ 'yx -~ ( 0 ~ r ~ n - - 1 )

MORRIS NEW~IAN: On a problem of G. Scnso¢ve 29

of G,, are generators of G,,; and since there are jus t n + 1 of them, m u s t be free genera tors of G , . Thus G, may be described as the free group of rank n + 1 on the free generators (4).

We apply these remarks to the principal congruence group ]? 12). It is known that P 12) is a free group of r~nk 2 with generators S ~, WL If we make the correspondence

~c ~-~ S*, y ~--> W 2

we obtain from the preceding discussion

THEOREM 1. - The group ~ , is a normal subgroup of P (2) of index n in P (2), and is the free gronp on the n + 1 free generators (3). F v, consists precisely o f those wards iu S ~, W ~ for which the sum of the exponents o f S 2 is divisible by n.

We remark that I" (2) is of index 6 in P, so that I'6,, is of index 6n in P. W e also remark that ]~G,, is not a normal subgroup of r , for n > 3. For example

wTi= 11 0

anp W ' e 1?6, while S -~ does not.

3. P r o o f tha t F6, is not a congruence group for n:4= 1, 2, 4, 8. - In this section we apply a method originated by L REI~Et~ [1] to provo that 1'0. is not a congruence group for n=l= 1, 2, 4, 8. A subgroup ~ of F is a congruence group if and only if it contains a principal congruence group r (m) of level m; and the least such integer m is jus t the least common mult iple of the

ampli tudes )> of the parabolic elements of ~ (any parabolic element P of r is conjugate to S a for some k; Ik[ is called the ampli tude of P). An exposi- tion of this result with applicat ions to certain subgroups of [' is given by K. WOI~Ia~AI~RT in a for thcoming article in the Illinois J. MA~I~. For the group ]36,, it is easily seen from Theorem 1 that m must be 2n. Thus we have

LEM~A 1 - The group 1'~, is a congruence group if and only i f

(5) re,, D P (2n).

We also note

30 ~:[ORRIS NEWSMAN: On #~ pt'oblem of G. S(tn~so~vc

LElVIi~A 2. - I f m/n~ th n V_ 6rod P6,. W e are now prepared to prove

TIIEOREM 2. - Suppose that n :4= 1, 2, 4, 8. The~ F~,, is not a co~tgrue~we group.

PROOF - By Lamina 1, we need only exhibi t an e lement of r ( 2n )wh ich is not an e lement of ]~G-"

Consider the e lement

where a = 5, b ---- 10y -{- 2, c ---- 10~c -{- 2 and d = 1 ~ 4x -{- 4y ~ 20~cy. Suppose tha t x, y have been chosen so that d ~ i (mod 2nt; i. e. so tha t

(6)

Then M ~ [ l + b c

2 ( 5 x y T x + y ) - - ~ O (mod n).

b] ~--S°W c (mod 2u), and it follows that the e l emen t 11

(7)

belongs to l?(2n), and tha t the sum of the exponen t s of S ~ is - - 4 y . Thus if it is possible to choose x, y so that (6) is sat isf ied, and also so that

(s) -- 4 y _=]_: 0 (mod n),

t hen N will belong to ]? (2n) but not to Pc-. Suppose f i rs t tha t n is odd and > i. Choose y - - - - - 1. Then t61 becomes

- - 2 ( 4 ~ - b t ) ~ - 0 (rood n} which has a solut ion since n is odd, and (8) beco- mes 4_--I~0 (rood n) which is sat isf ied since n is odd and ~ 1. Thus the theorem is proved in this case. I t fol lows tha t the theorem is also t rue whenever n is divisible by an odd n u mb e r m ~ 1; for if such a group P~. were a congruence group~ L e m m a 2 would imply that 1?~,, was one also, which is impossible. Thus we have left only the cases n - - 2 a, k :> 4 to consider . W h e n k - - 4 , the choice x - - y = 2 sat isf ies both (6) and (8). W h e n k ~ 4, the choice x - - - - y - - : 2[ ~/~j sat isf ies both (6) a n d (8). This completes the proof o f the theorem.

MoRRis N~WMA~: On a problem o] G. 2a~sone 31

4. P r o o f t h a t I'~. is a congruence group for n - - 1~ 2, 4, 8. In this section we prove

THEOREM 3. - The groups ~ , , sa t is fy

(9) P~,, D ~ (2n), n = 1, 2, 4, 8.

PROOF. - We give only the proof for n - - 8 , this being the most diffi- cult case.

Consider an arbi t rary e lement of r(16). Since r (16) D F(2), such an e lement can be wri t ten as

Mt -" S ~a' W 2~ ...S "~at W "zo',

where a~, b i , . . . , at~ bt are integers. Pu t

~ = a , + . . . + a t .

Thon we mus t show that z t ~ 0 (rood 8). Since

S , . W~6 =

we find by induct ion that

Mt = [1

Fu r the rmore the re la t ionship

[1+ o02b

q- 4At 2Bt] ! I

2Ct 1 "4- 4DtJ

implies that the integers At , Bt satisfy the recurrences

(10) At - - At_~ + atbt + 4atbtAt- t + btBt_~ ,

(11) Bt = B t - , -{- at "4- 4atAt-~

32 MORRIS .NEWMa_~: On a problem of G. Sa~nsone

with the init ial condit ions

A ~ - - a i b ~, B ~ = a i .

Thus {10) implies that

(12~ At = At-~ q- atb, -b btBt-~ (rood 4)

and summing both sides of (11) we find that

t

(13) Bt = ~t + 4 Z a~A~_~. k z 2

Subst i tu t ing (13) into (12) we find that

A t ~ A t _ t

which implies by summat ion that

(14)

atbt (mod 4 b

t

At ~ ~' ~hb~ (mod 4). k~--1

Now subst i tut ing (14) into (13) we find after some calculat ion that

B~ ~ zt + 4 (at - - 1) At (rood. 8).

Since Mt s P (16!, we have that At ~ 0 (mod 4), Bt -~ 0 (mod 8). Hence (15}

(15)

~t -~ 0 (rood 8),

implies that

complet ing the proof of the theorem.

5. Concluding r e m a r k s . - The groups I~6,, form a valuable example of a family of subgroups of l: of strictly increas ing index but of genus 0. That I~6, is of genus 0 may be seen in ei ther of two ways:

The first is from the fundamenta l region of F~., since the isometric circles of the genera t ing t ransformat ions (2) and their inverses make up the boundary of the fundamen ta l reg ion The second is from the fact that the genera t ing t ransformat ions (3) are all parabolic, and it is known from

~ORR.IS ~EWMAN: 0~. q~ problem of G. Sa~sone 33

presentation theory that a subgroup of F of finite index can be generated by parabolic transformations alone only if it is of genus 0. The question arises: Are there infinitely many congruence subgroups of I ~ of genus 07 I conjecture thet the answer is in the negative.

A discussion of the groups r~+~ using the Hauptmodul k of the group r(2) my be found in the papers of R. FRICKE and G. PIoK in the Math. Annalen, 28, 1887; pp. 99-118 and 119-124, resp.

REFERENCES

[ 1 ] I ~EI~ER, Normal subgroups of the unimodular group. Illinois J. Math. 2~ 142-144 (1958). [2 ] G. SANSO~E, Problsmi insoluti neUa teoria dells sostituzioni tineari. Cony. Internaz.

di Teoria dei Gruppi :~initi (Firenze 1960) 5.19. Edizione Cremonese Rome 1960.

Annali dt Matematica 5