476 E. Groups and Graphs
(1) r* contains r as a subgroup and r* is generated by rand a distin-
guished element z of infinite order;
(2) for every a EA, we have z-^1 az = O(a) in r*;
(3) if sozd^1 s1 · · · zdnsn = e for some n 2: 1, Sk E r and dk E Z \ {O},
then there exists k 2: 1 such that either Sk EA and dk < 0 < dk+l
or Sk E e(A) and dk > 0 > dk+l·The HNN-extension r* = (r, z I z-^1 az = O(a) for a E A) can also be
described in the following manner. Let r(n), n E Z, be copies of rand write
s(n) for the element in r(n) corresponding to sin r. Let A(n) s r(n) be the
subgroup corresponding to A and consider the embedding
e(n): A (n) 3 a(n) 1--t O(a)(n+l) E r(n+l).We definer 00 as the iterated amalgamated product
. I' oo = · · · r(n-l) A(n-l)=l:J(A)Cn) r(n) A(n)=l:J(A)Cn+l) r(n+l) · · ·
= \lJ r(n) I a(n) = O(a)(n+l) for a EA and n E Z).
By universality of the amalgamated free product, the "shift"
r(n) 3 s(n) I--+ s(n-1) E r(n-1) C r oo
extends to an automorphism of r 00 • Then, the HNN-extension I'* is known
to be isomorphic to the subgroup of the semidirect product r 00 ><1Z, generated
by r(O) and z, where Z is the element in I' 00 ><I Z that implements the shift
automorphism.
Example E.10. Recall that
SL(2, Z) = { [ ~ ~ ] : a, b, c, d E Z and ad - be = 1}.
and that PSL(2, Z) = SL(2, Z) / {±I}, where I is the identity matrix. We
will show that
s = [ ~ 1 ~ ] and t = [ ~ ~1
]freely generate PSL(2, Z) and hence PSL(2, Z) ~ (Z/2Z) (Z/3Z). (Note
that (s) ~ Z/2Z and (t) ~ Z/3Z in PSL(2, Z).) This also implies that
SL(2, Z) ~ (Z/4Z) :z.; 2 z (Z/6Z). Consider the action of PSL(2, Z) on the
projective line IRP^1 = IR U { oo }:
· ,- a db ] : IR U { oo} 3 x 1--+ ax+ b E IR U { oo }.
_ c cx+dWe observe that s(-oo, 0) = (0, oo) and td(o, oo) c (-oo, 0) for d = 1, 2.
Let w be a reduced word in (s) and (t). By conjugating tor t^2 if necessary,
we may assume that both the first and the last letters of w come from (t)