E. Groups and Graphs 477- i.e., w = td^1 std^2 s .. · stdn for some di, ... , dn E {1, 2}. It follows that
w(O, oo) c (-oo, 0) and hence w =/:- 1 in PSL(2, Z). (This is the so-called
"table tennis lemma" of Klein.) One can now check that s and t generate
SL(2,Z). We note that [~I] and[~~] generateacopyofthefreegrouplF2
(which has index 12 in SL(2, Z)).
Note that SL(2,JR)/P ~ JRP^1 as an SL(2,JR)-space, where Pis the sub-
group of upper triangular matrices. Indeed, SL(2, JR) acts transitively on
the projective line JRP^1 and P coincides with the stabilizer of oo E JRP^1.
Example E.11. Let p, q > 1 be a relatively prime pair of natural numbers.
The Baumslag-Solitar group
r p,q =(a ' z I z-^1 aPz. = aq)
is isomorphic to the HNN-extension associated with pZ, qZ c Z. It contains
Tp,q = (x, y I xP = yq) ~ Z *pZ=qZ Z - the torus knot group - which is
nonamenable (because p, q =/:- 1 and (p, q) =/:-(2, 2)) and has an infinite center
(generated by xP). The group r p,q has several interesting properties. Let
cp : r ---t r be the endomorphism given by
cp(a) = aP and cp(z) = z.
The endomorphism cp is surjective since p, q > 1 are relatively prime, but it
is not injective since the commutator [a, z-^1 az] =/:-e is in ker cp. (A group
which admits a surjective and noninjective endomorphism is said to be non-
Hopfian.) This implies that r p,q is not residually finite. Indeed, for any
finite-index subgroup A::::; I'p,q, the subgroup cp-^1 (A) has the same index in
r p,q as A. But since r p,q is finitely generated, the set of subgroups of index
n is finite for every n. It follows that the surjective endomorphism cp acts
bijectively on the set of subgroups of index n, and thus ker cp is contained in
the intersection of all the finite-index subgroups.
Let Nk = ker cpk be an increasing sequence of normal subgroups in r p,q,
and let N 00 = LJNk. Then, I'p,q/Nk ~ I'p,q for every k. Let us show that
r IN 00 is meta-abelian. Let (}" : r p,q ---t ( z) ~ z be a homomorphism given by
O"(a) = e and O"(z) = z. It is not hard to see that O" factors through I'p,q/N 00 •
We observe that kerO" is generated as a group by {z-nazn : n E Z}. But,
z-nazn = aqn in I'p,q/Npn; hence kerO" =(a) in I'p,q/N 00 •
We present a result from [12]. Let S = {a,a-^1 ,z,z-^1 } be the standard
set of generators of I'p,q· Consider the Cayley graphs Xk = X(I'p,q, cpk(S))
and let R > 0 be arbitrary. Then, one can find K > 0 such that
s^2 R n N oo = s^2 R n LJ Nk = s^2 R n N K.
k2:1It follows that for all k 2: K the balls BR(Xk) of radius R in Xk are identical
to BR(X 00 ), where X 00 = X(I'p,q/N 00 , S). Since I'p,q/N 00 is amenable, we