It is s hown that if t ~ 3, the n no subgroup of SLit, Z) of finite inde x is free (in fact is not even the fre e produ c t of cyc li c groups). H e r e SL(t, Z ) is the multipli ca tive group of t X t matri ces o ver the integers o f d ete rmin a nt 1.Key word s: Co ngru e nce s ubgro up property; fr ee g roup s; fr ee products; hi ghe r modular gTOUpS.It is kn ow n that th e class ica l modul ar group f = PSL (2, Z ) possesses man y s ub groups whi c h are free: In fact (see [4])1 e ve ry normal s ubgroup of f is free, apart from the three exce ptions f, P , f 3, (fp is th e fully invariant s ubgro up of f ge nerated b y th e pth powers of the ele me nts of f). Th e r eason for thi s is that f is th e free product of tw o cyclic groups, one of ord e r 2 and the other of order 3, and the Kuros h s ub group th eore m the n implies that a s ub gro up of f is free if and only if it co ntain s 'no elements of finite orde r. Thi s res ult at once implies that e ve ry norm al s ub group of SL (2, Z) whi c h does not co ntain -/ mu st be free.It is natural to ask if a simular situati on exis ts for the hi ghe r modular groups. The purpose of thi s no te is to point out that just th e opposite is true. In fact we will show that no normal subgroup nor any s ubgroup of f, = SL(l , Z ) of finite ind e x can be free for t 3 3.If n is a ny positive integer , th en f,(n ) is the prin cipal co ngrue nce subgroup of f t of level n; that is, th e totality of matri ces A E f t s uc h that A == I mod ii. Epq stands for the tXt matrix with a 1 in po sition (p, q) and ° else wh ere.The proof mak es use of a deep res .. .!lt prov ed rece ntly by Me nnicke [2], and by Bass, Lazard , and Serre [1] , toge th e r with so me e le me ntary re mark s on free groups.We assume throu ghout the following that t 3 3. We first prove LEMMA 1. Let F be a free group. Then any two commuting elements ofF are each powers of the same element of F.PROOF. Suppose that x , yare co mmutin g elements of F. Put G= {x, y}. Th e n G is an abelian subgroup of F of rank 1 or 2. Furthermore G is free, by th e Kuros h s ubgro up th eore m. Thus G can· not be of rank 2, since a free group of rank 2 is not abelian. He nce G mu st be of rank 1 and there· fore cyclic: G= {z}. Thus x and yare each powers of z, and the le mm a is proved. LEMMA 2. Let n be any positive integer. Put A= I + nE1 2' B = 1+ nE 13. Then A, B are commuting elements of ft(n ) which are not powers of the same element C, for any CEft(n).PROOF. Th e fa c t that A and B are co mmuting elements of ft(n) is clear, since E I2E I3 = EI3EI2 =0. Suppose that there is an ele me nt CEft(n) such that A=C", B=Cs, for some integers r, s. The n rs oft 0, and A s=BI"=Crs. Since E'f2=EY3=O, we have 1 Fig ures in brac kets indi ca te th e lit e rature refere nces at the e nd of thi s pa per.143