By Czedli G.

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**Example text**

Theorem. Let the group G act without inversion on the tree X and let Y = XjG be the quotient graph. For any vertex v and edge a of Y let fJ and jj denote a vertex and edge of X lying over v and a respectively. Then G is (isomorphic to) the fundamental group of a graph (g, Y) of groups whose vertex groups and edge groups, respectively, are stabilisers of the form Gv = Stabc(fJ) and Ga = Stabc(jj). 22. Theorem. Let G be the fundamental group of a graph (9, Y) of groups and E+ (Y) an orientation of Y.

PI(U)! Pi",PI(U)! Pj-I" ,PI(U)! Pj .. ,PI(U)! , (c) IPk-l .. ·PI(U)! Pk .. ·PI(U)! , l~i~p, p+1~j~q, q+1~k~n. 0 The inequalities (a), (b) and (c) are illustrated by the following diagram in which length is plotted upwards. 6 is derived as follows. J. Collins, H. Zieschang 50 u \ ~~ l . -J. Fig. 1 many Whitehead automorphisms will replace u and v by minimal words Uo and Vo. 9, Uo can be transformed into Vo by a sequence of applications of permutation and Whitehead automorphisms in such a way that the lengths of the intermediate words obtained are the same as the length of Uo and Vo.

We will explain it with an example. 12. Example. 6 (d). Then 7r1(N2) = (a, b I a2b2). The symmetric group Sym(3) consists of the six elements 1, r = (12), s = (13), t = (23), u = (123), v = (132), where r, s, t have order 2 and u, v have order 3. Elements f(a), f(b) with f(a)2 f(b)2 = 1 determine a homomorphism f : 7r1(N2) -; Sym(3). If f(a) = 1 or f(b) = 1 then f is not transitive. If f(a) has order 2 then, up to equivalence, f(a) = r, f(b) = s. If the order of f(a) is 3 then f(b) = f(a)-I and, up to equivalence, f(a) = u, f(b) = v.